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  • Different meaning in the mysql code?

    - by Emre Saracoglu
    $result=mysql_query("select * from dosyabegeni where veri_id='" . get_custom_field('dwcode') . "'"); Not Working It says the number and the screen, but the application does not work veri_id='" . get_custom_field('dwcode') . "'"); veri_id='" . echo get_custom_field('dwcode') . "'"); Working veri_id='HelloTest'"); veri_id='1234567890'"); veri_id='" . $_GET['test'] . "'"); Main Codes <?php include('/home/emre2010/public_html/EntegreOz/DosyaBegeni/config.php'); $result=mysql_query("select * from dosyabegeni where veri_id='" .get_custom_field('dwcode') . "'"); while($row = mysql_fetch_array($result)) { $sira_id=$row['sira_id']; $veri_id=$row['veri_id']; $begeni=$row['begeni']; ?> <div class="reviewbox"> <div class="summarywrap"> <div class="summarywrapinner"> <div class="summary"> <div class="reviewsection"><div class="rating points"> <a href="#" class="begeni" id="<?php echo $sira_id; ?>"> <span style="color:#fff;" align="center"> <?php echo $begeni; ?> </span> </a> <p class="ratingtext">completed!</p></div> </div><div class="clear"></div> <div class="clear"></div> </div> <div class="ratingsummary"></div> <div class="clear"></div> </div> <div class="clear"></div> </div> What's the problem?

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  • actionscript 2.0 input text box

    - by user1769760
    So, here's what I'm trying to do, and I, frankly, believe it should be obvious, but I can't figure it out. I am creating a very simple Artificial Intelligence simulation. And in this simulation there's an input box at the bottom of the screen (called "input" exactly). "input" has a variable in its properties that is called "inbox" (exactly). Using a key listener the script calls up a function when the enter button is pressed. This function has 2 if statements and an else statement which dictate the responses of the AI (named "nistra"). The problem is this, When I type in what I want to say, and hit enter, it always uses the second response ("lockpick" in the code below). I have tried variations on the code but I still don't see the solution. I believe the problem is that the "typein" variable holds all the format information from the text box as well as the variable, but I could be wrong, that information is in here as well, underneath the code itself. Any help I can get would be greatly appreciated. var typein = ""; //copies the text from inbox into here, this is what nistra responds to var inbox = ""; //this is where the text from the input text box goes var respond = ""; //nistra's responses go here my_listener = new Object(); // key listener my_listener.onKeyDown = function() { if(Key.isDown(13)) //enter button pressed { typein = inbox; // moves inbox into typein nistraresponse(); // calles nistra's responses } //code = Key.getCode(); //trace ("Key pressed = " + code); } Key.addListener(my_listener); // key listener ends here nistraresponse = function() // nistra's responses { trace(typein); // trace out what "typein" holds if(typein = "Hello") // if you type in "Hello" { respond = "Hello, How are you?"; } if(typein = "lockpick") // if you type in "lockpick" { respond = "Affirmative"; } else // anything else { respond = "I do not understand the command, please rephrase"; } cntxtID = setInterval(clearnistra, 5000); // calls the function that clears out nistra's response box so that her responses don't just sit there } clearnistra = function() // clears her respond box { respond = ""; clearInterval(cntxtID); } // "typein" traces out the following <TEXTFORMAT LEADING="2"><P ALIGN="CENTER"><FONT FACE="Times New Roman" SIZE="20" COLOR="#FF0000" LETTERSPACING="0" KERNING="0">test</FONT></P></TEXTFORMAT>

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  • Multiple choice test GUI with serialization of Q&A

    - by Bobby
    I'm working on a project for school in Java programming. I need to design a GUI that will take in questions and answers and store them in a file. It should be able to contain an unlimited number of questions. We have covered binary I/O. How do I write the input they give to a file? How would I go about having them add multiple questions from this GUI? package multiplechoice; import java.awt.*; import javax.swing.*; public class MultipleChoice extends JFrame { public MultipleChoice() { /* * Setting Layout */ setLayout(new GridLayout(10,10)); /* * First Question */ add(new JLabel("What is the category of the question?: ")); JTextField category = new JTextField(); add(category); add(new JLabel("Please enter the question you wish to ask: ")); JTextField question = new JTextField(); add(question); add(new JLabel("Please enter the correct answer: ")); JTextField correctAnswer = new JTextField(); add(correctAnswer); add(new JLabel("Please enter a reccomended answer to display: ")); JTextField reccomendedAnswer = new JTextField(); add(reccomendedAnswer); add(new JLabel("Please enter a choice for multiple choice option " + "A")); JTextField A = new JTextField(); add(A); add(new JLabel("Please enter a choice for multiple choice option " + "B")); JTextField B = new JTextField(); add(B); add(new JLabel("Please enter a choice for multiple choice option " + "C")); JTextField C = new JTextField(); add(C); add(new JLabel("Please enter a choice for multiple choice option " + "D")); JTextField D = new JTextField(); add(D); add(new JButton("Compile Questions")); add(new JButton("Next Question")); } public static void main(String[] args) { /* * Creating JFrame to contain questions */ FinalProject frame = new FinalProject(); // FileOutputStream output = new FileOutputStream("Questions.dat"); JPanel panel = new JPanel(); // button.setLayout(); // frame.add(panel); panel.setSize(100,100); // button.setPreferredSize(new Dimension(100,100)); frame.setTitle("FinalProject"); frame.setSize(600, 400); frame.setLocationRelativeTo(null); // Center the frame frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); frame.setVisible(true); } }

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  • Download a file using cocoa

    - by dododedodonl
    Hi All, I want to download a file to the downloads folder. I searched google for this and found the NSURLDownload class. I've read the page in the dev center and created this code (with some copy and pasting) this code: @implementation Downloader @synthesize downloadResponse; - (void)startDownloadingURL:(NSString*)downloadUrl destenation:(NSString*)destenation { // create the request NSURLRequest *theRequest=[NSURLRequest requestWithURL:[NSURL URLWithString:downloadUrl] cachePolicy:NSURLRequestUseProtocolCachePolicy timeoutInterval:60.0]; // create the connection with the request // and start loading the data NSURLDownload *theDownload=[[NSURLDownload alloc] initWithRequest:theRequest delegate:self]; if (!theDownload) { NSLog(@"Download could not be made..."); } } - (void)download:(NSURLDownload *)download decideDestinationWithSuggestedFilename:(NSString *)filename { NSString *destinationFilename; NSString *homeDirectory=NSHomeDirectory(); destinationFilename=[[homeDirectory stringByAppendingPathComponent:@"Desktop"] stringByAppendingPathComponent:filename]; [download setDestination:destinationFilename allowOverwrite:NO]; } - (void)download:(NSURLDownload *)download didFailWithError:(NSError *)error { // release the connection [download release]; // inform the user NSLog(@"Download failed! Error - %@ %@", [error localizedDescription], [[error userInfo] objectForKey:NSErrorFailingURLStringKey]); } - (void)downloadDidFinish:(NSURLDownload *)download { // release the connection [download release]; // do something with the data NSLog(@"downloadDidFinish"); } - (void)setDownloadResponse:(NSURLResponse *)aDownloadResponse { [aDownloadResponse retain]; [downloadResponse release]; downloadResponse = aDownloadResponse; } - (void)download:(NSURLDownload *)download didReceiveResponse:(NSURLResponse *)response { // reset the progress, this might be called multiple times bytesReceived = 0; // retain the response to use later [self setDownloadResponse:response]; } - (void)download:(NSURLDownload *)download didReceiveDataOfLength:(unsigned)length { long long expectedLength = [[self downloadResponse] expectedContentLength]; bytesReceived = bytesReceived+length; if (expectedLength != NSURLResponseUnknownLength) { percentComplete = (bytesReceived/(float)expectedLength)*100.0; NSLog(@"Percent - %f",percentComplete); } else { NSLog(@"Bytes received - %d",bytesReceived); } } -(NSURLRequest *)download:(NSURLDownload *)download willSendRequest:(NSURLRequest *)request redirectResponse:(NSURLResponse *)redirectResponse { NSURLRequest *newRequest=request; if (redirectResponse) { newRequest=nil; } return newRequest; } @end But my problem is now, it doesn't appear on the desktop as specified. And I want to put it in downloads and not on the desktop... What do I have to do?

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  • Google maps api v3: geocoding multiple addresses and infowindow

    - by user2536786
    I am trying to get infowindow for multiple addresses. Its creating markers but when I click on markers, infowindow is not popping up. Please help and see what could be wrong in this code. Rest all info is fine only issue is with infowindow not coming up. <!DOCTYPE html> <html> <head> <meta http-equiv="content-type" content="text/html; charset=UTF-8" /> <title>Google Maps Multiple Markers</title> <script src="http://maps.google.com/maps/api/js?sensor=false" type="text/javascript"></script> </head> <body> <div id="map" style="height: 800px;"></div> <script type="text/javascript"> var locations = [ ['Bondi Beach', '850 Bay st 04 Toronto, Ont'], ['Coogee Beach', '932 Bay Street, Toronto, ON M5S 1B1'], ['Cronulla Beach', '61 Town Centre Court, Toronto, ON M1P'], ['Manly Beach', '832 Bay Street, Toronto, ON M5S 1B1'], ['Maroubra Beach', '606 New Toronto Street, Toronto, ON M8V 2E8'] ]; var map = new google.maps.Map(document.getElementById('map'), { zoom: 10, center: new google.maps.LatLng(43.253205,-80.480347), mapTypeId: google.maps.MapTypeId.ROADMAP }); var infowindow = new google.maps.InfoWindow(); var geocoder = new google.maps.Geocoder(); var marker, i; for (i = 0; i < locations.length; i++) { geocoder.geocode( { 'address': locations[i][1]}, function(results, status) { //alert(status); if (status == google.maps.GeocoderStatus.OK) { //alert(results[0].geometry.location); map.setCenter(results[0].geometry.location); marker = new google.maps.Marker({ position: results[0].geometry.location, map: map }); google.maps.event.addListener(marker, 'mouseover', function() { infowindow.open(marker, map);}); google.maps.event.addListener(marker, 'mouseout', function() { infowindow.close();}); } else { alert("some problem in geocode" + status); } }); } </script> </body> </html>

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  • Export to word from php doesn't seem to work for me...

    - by chandru_cp
    I exported data from php page to word document but the problem is the header is not available in all pages.... Header is present in the first page but not in the next pages of the word document..... Here is my code, function changeDetails() { $bType = $this->input->post('textvalue'); if($bType == "word") { $this->load->library('table'); $data['countrytoword'] = $this->AddEditmodel1->export(); $this->table->set_heading('Name','Country','State','Town'); $out = $this->table->generate($data['countrytoword']); header("Content-Type: application/vnd.ms-word"); header("Expires: 0"); header("Cache-Control: must-revalidate, post-check=0, pre-check=0"); header("Content-disposition: attachment; filename=$cur_date.doc"); echo '<br><br>'; echo '<strong>CountryList</strong><br><br>'; print_r($out); } } <? if(isset($countrytoword)) { ?> <table align="center" border="0"> <tr> <td> Name </td> <td> Country </td> <td> State </td> <td> Town </td> </tr> <? foreach($countrytoword as $dsasffd) { ?> <tr> <td><?= $dsasffd['dbName'] ?></td> <td><?= $dsasffd['dbCountry']; ?></td> <td><?= $dsasffd['dbState']; ?></td> <td><?= $dsasffd['dbTown']; ?></td> <? } } ?> </tr> </table>

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  • EJB / JSF java.lang.ClassNotFoundException: com.ericsantanna.jobFC.dao.DAOFactoryRemote from [Module "com.sun.jsf-impl:main" from local module loader

    - by Eric Sant'Anna
    I'm in my first time using EJB and JSF, and I can't resolve this: 20:23:12,457 Grave [javax.enterprise.resource.webcontainer.jsf.application] (http-localhost-127.0.0.1-8081-2) com.ericsantanna.jobFC.dao.DAOFactoryRemote from [Module "com.sun.jsf-impl:main" from local module loader @439db2b2 (roots: C:\jboss-as-7.1.1.Final\modules)]: java.lang.ClassNotFoundException: com.ericsantanna.jobFC.dao.DAOFactoryRemote from [Module "com.sun.jsf-impl:main" from local module loader @439db2b2 (roots: C:\jboss-as-7.1.1.Final\modules)] I'm getting this when I do an action like a selectOneMenu or a commandButton click. DAOFactory.class @Singleton @Remote(DAOFactoryRemote.class) public class DAOFactory implements DAOFactoryRemote { private static final long serialVersionUID = 6030538139815885895L; @PersistenceContext private EntityManager entityManager; @EJB private JobDAORemote jobDAORemote; /** * Default constructor. */ public DAOFactory() { // TODO Auto-generated constructor stub } @Override public JobDAORemote getJobDAO() { JobDAO jobDAO = (JobDAO) jobDAORemote; jobDAO.setEntityManager(entityManager); return jobDAO; } JobDAO.class @Stateless @Remote(JobDAORemote.class) public class JobDAO implements JobDAORemote { private static final long serialVersionUID = -5483992924812255349L; private EntityManager entityManager; /** * Default constructor. */ public JobDAO() { // TODO Auto-generated constructor stub } @Override public void insert(Job t) { entityManager.persist(t); } @Override public Job findById(Class<Job> classe, Long id) { return entityManager.getReference(classe, id); } @Override public Job findByName(Class<Job> clazz, String name) { return entityManager .createQuery("SELECT job FROM " + clazz.getName() + " job WHERE job.nome = :nome" , Job.class) .setParameter("name", name) .getSingleResult(); } ... TriggerFormBean.class @ManagedBean @ViewScoped @Stateless public class TriggerFormBean implements Serializable { private static final long serialVersionUID = -3293560384606586480L; @EJB private DAOFactoryRemote daoFactory; @EJB private TriggerManagerRemote triggerManagerRemote; ... triggerForm.xhtml (a portion with problem) </p:layoutUnit> <p:layoutUnit id="eastConditionPanel" position="center" size="50%"> <p:panel header="Conditions to Release" style="width:97%;height:97%;"> <h:panelGrid columns="2" cellpadding="3"> <h:outputLabel value="Condition Name:" for="conditionName" /> <p:inputText id="conditionName" value="#{triggerFormBean.newCondition.name}" /> </h:panelGrid> <p:commandButton value="Add Condition" update="conditionsToReleaseList" id="addConditionToRelease" actionListener="#{triggerFormBean.addNewCondition}" /> <p:orderList id="conditionsToReleaseList" value="#{triggerFormBean.trigger.conditionsToRelease}" var="condition" controlsLocation="none" itemLabel="#{condition.name}" itemValue="#{condition}" iconOnly="true" style="width:97%;heigth:97%;"/> </p:panel> </p:layoutUnit> In TriggerFormBean.class if comments daoFactory we get the same exception with triggerManagerRemote, both annotated with @EJB. I'm don't understand the relationship between my DAOFactory and the "Module com.sun.jsf-impl:main"... Thanks.

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  • How to add Items to GridView in C# Windows Store App (Win8)

    - by flexage
    To keep things simple let's just say I have a c# Windows Store Project for Windows 8 that has the following items: GridView (PlatformsGrid) List«PlatformSearchResult» (allPlatforms) DataTemplate (PlatformDataTemplate) in standardstyles.xaml allPlatforms is a collection of "PlatformSearchResult"objects populated from an online API, and has the following 3 properties: ID Name Alias I am able to add a new item to the gridview for each object that exists in my allPlatforms collection, however the items are blank and do not show the data from my objects. A quick summary of the current code looks like this: XAML Markup: <!-- Platforms Content --> <GridView x:Name="PlatformsGrid" Grid.Row="1" CanReorderItems="True" CanDragItems="True" ItemTemplate="{StaticResource PlatformDataTemplate}" > <GridView.ItemsPanel> <ItemsPanelTemplate> <WrapGrid MaximumRowsOrColumns="2" VerticalChildrenAlignment="Top" HorizontalChildrenAlignment="Center" /> </ItemsPanelTemplate> </GridView.ItemsPanel> </GridView> Data Template <!-- Platform Item Template --> <DataTemplate x:Key="PlatformDataTemplate"> <Grid Background="#FF939598" Height="250" Width="250"> <Image Source="/SampleImage.png" Stretch="UniformToFill"/> <StackPanel Orientation="Vertical" Background="#CC000000" Height="90" VerticalAlignment="Bottom"> <TextBlock Text="{Binding Name}" Margin="10,3,0,0" Width="242" Height="62" TextTrimming="WordEllipsis" TextWrapping="Wrap" HorizontalAlignment="Left"/> <TextBlock Text="{Binding Alias}" Margin="10,2,0,0" Width="186" Height="14" TextTrimming="WordEllipsis" HorizontalAlignment="Left" FontSize="9" Opacity="0.49"/> </StackPanel> </Grid> </DataTemplate> Controlling Function private async void FetchItemInfo_Loaded(object sender, RoutedEventArgs e) { // Get List of Top Games List<PlatformSearchResult> allPlatforms = new List<PlatformSearchResult>(); allPlatforms = await GamesDB.GetPlatforms(); // Dynamically Create Platform Tiles foreach (PlatformSearchResult platform in allPlatforms) { PlatformsGrid.DataContext = platform; PlatformsGrid.Items.Add(platform); } } How do I get the added items to show the appropriate object properties (ignoring the image for now), I'm just interested in populating the content of the TextBlocks. I appreciate any help anyone can provide! Thanks, Alex.

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  • php parse error always on the last line

    - by is0lated
    I'm trying to read a comment that is stored in a mysql table. For some reason I always get a parse error on the last line of the file even if the last line is blank. I'm not sure if it's relevant but the connect.php works for putting the comment into the database. I'm using wampserver to host it and coding it by hand. I think that's it's something to do with the while loop, when I comment out the while(){ and the } near the end I just get a few missing variable errors as you would expect. I'm quite new to php coding so I'm pretty sure the problem will be something simple that I've either missed or not understood properly. Anyway, here's my code: <?php include "connect.php"; ?> <?php $sql = "SELECT * FROM main"; $result = mysql_query($sql) or die("Could not get posts from table"); while($rows=mysql_fetch_array($result)){ ?> <table bgcolor="green" align="center"> <tr> <td></td> </tr> <tr> <td><strong> <? echo $rows['name']; ?> </strong></td> </tr> <tr> <td> <? echo $rows['email']; ?> </td> </tr> <tr> <td> <? echo $rows['comment']; ?> </td> </tr> </table> <? } ?> Thanks for the help. :)

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  • I always get stuck here... Divs not behaving properly (alignment issues)

    - by user345501
    Hi, I don't know why, after encountering this problem dozens of times, the answer always seems different and I can't seem to work my way through the problem-solving process, but here I am again with misaligned divs. I've got 3rows encasing columns. each row is to have (at least) 3 columns (and probably some nested divs down the line, but I'm not even there yet). I'm trying to make a fluid chunk in the center ultimately, with pretty corners. However, my top row is already showing signs of misbehaving. .O Please help with my silly questions! Cheers and thanks in advance! <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> </head> <style type="text/css"> #wrap { margin:auto; width:80%; height:75%; border: solid #066 1px;} #row1 { width:100%; height:10%; background:#F20; } #r1c1 { float:left; width:05%;} #r1c2 { float:left; width:80%} #r1c3 { clear:both; width:05%; } #row2 { float:none; width:100%; background:#0C6; } #r2c1 {} #r2c2 {} #r2c3 {} #row3 { width:100%; height:15%; background:#00F; clear:both; } #r3c1 {} #r3c2 {} #r3c3 {} </style> <body> <div id="wrap"> <div id="row1"> <div id="r1c1">LEFT</div> <div id="r1c2">CENT</div> <div id="r1c3">RIGHT</div> </div> <div id="row2"> MIDDLE </div> <div id="row3"> BOTTOM </div> </div> </body> </html>

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  • mysql_fetch_array() expects parameter 1 to be resource problem

    - by user225269
    I don't get it, I see no mistakes in this code but there is this error, please help: mysql_fetch_array() expects parameter 1 to be resource problem <?php $con = mysql_connect("localhost","root","nitoryolai123$%^"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("school", $con); $result = mysql_query("SELECT * FROM student WHERE IDNO=".$_GET['id']); ?> <?php while ($row = mysql_fetch_array($result)) { ?> <table class="a" border="0" align="center" cellpadding="0" cellspacing="1" bgcolor="#D3D3D3"> <tr> <form name="formcheck" method="get" action="updateact.php" onsubmit="return formCheck(this);"> <td> <table border="0" cellpadding="3" cellspacing="1" bgcolor=""> <tr> <td colspan="16" height="25" style="background:#5C915C; color:white; border:white 1px solid; text-align: left"><strong><font size="2">Update Students</td> <tr> <td width="30" height="35"><font size="2">*I D Number:</td> <td width="30"><input name="idnum" onkeypress="return isNumberKey(event)" type="text" maxlength="5" id='numbers'/ value="<?php echo $_GET['id']; ?>"></td> </tr> <tr> <td width="30" height="35"><font size="2">*Year:</td> <td width="30"><input name="yr" onkeypress="return isNumberKey(event)" type="text" maxlength="5" id='numbers'/ value="<?php echo $row["YEAR"]; ?>"></td> <?php } ?> I'm just trying to load the data in the forms but I don't know why that error appears. What could possibly be the mistake in here?

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  • Windows MAchine Debugging

    - by PrettyFlower
    I've been learning how to program for Windows for some time now and am getting pretty comfy with COM. I had thought to go over to Linux and do some C++ programming there and I wished to run Rosetta Commons so I installed Fedora. I had tried installing Ubuntu a few months ago and things got messy. I had a glitch, maybe caused by one of the live cd creators, my video card or something I don't know. Who Crashed suggested it was my video card and I had regular messages about ntfs.sys and page file issues. At any rate I just installed Fedora and the same thing is happening again. I would like to think with the twenty five years of doing this that I might finally make some headway into debugging my system. I think I may have overlooked a lot of what could be done in favor of simply uninstalling, reinstalling and formatting and starting from scratch. I have opened up the folder windows debugging tools, quite accidentally and just before I was going to clean sweep again, and I found KD and WinDbg. I had never seen these before and I felt that maybe I should look into this. I am quite familiar with the modern machine that is known as the computer, I know what a Kernel is and am now pretty familiar with at the very least Windows Operating System Services. I wish to begin tracking my own machines errors. I understand that most kernel debugging is done on a second machine but I don't have one. And also I understand the goal of the debugger seems to be less about run of the mill errors and more about development time strategies but I'm sure there is more to this. This is my first go at this and I thought maybe I could get some suggestions on where to go from here. I would really like to learn ways to fix my machine and also maybe pick up some tricks on the dev side as well. I hope this isn't too broad a question or too generalized. I'm really just looking for the keywords and an overview of the more routine strategies used. thx

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  • Creating a multiplatform webapp with HTML5 and Google maps

    - by Bart L.
    I'm struggling how to develop a webapp for Android and iOS. My first app was a simple todo app which was easy to test in my browser and it only used html, javascript and css. However, I have to create an app which uses Google Maps Api to get the location. I created a simple html5 page to test which places a marker on a map. It works fine when testing it on my local server. But when I create an .apk file for Android, the app doesn't work. So I'm wondering, isn't it possible to use it like this? Do I have the use the phonegap libraries to use their geolocation library? And if so, how do you handle the development of a webapp in phonegap for multiple OS? Do you have to install an Android environment and an iOS environment to each include the right phonegap library and to test them properly? Update: I use the following code on my webserver and it works perfectly. When I upload it in a zip-folder to the photogap cloud and install the APK file on my phone, it doesn't work. <!DOCTYPE html> <html> <head> <meta charset="utf-8"> <title>Simple Geo test</title> <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.8/jquery.min.js"></script> </head> <body> <script type="text/javascript" src="http://maps.google.com/maps/api/js?sensor=true"></script> <script> function success(position) { var mapcanvas = document.createElement('div'); mapcanvas.id = 'mapcontainer'; mapcanvas.style.height = '200px'; mapcanvas.style.width = '200px'; document.querySelector('article').appendChild(mapcanvas); var coords = new google.maps.LatLng(position.coords.latitude, position.coords.longitude); var options = { zoom: 15, center: coords, mapTypeControl: false, navigationControlOptions: { style: google.maps.NavigationControlStyle.SMALL }, mapTypeId: google.maps.MapTypeId.ROADMAP }; var map = new google.maps.Map(document.getElementById("mapcontainer"), options); var marker = new google.maps.Marker({ position: coords, map: map, title:"You are here!" }); } if (navigator.geolocation) { navigator.geolocation.getCurrentPosition(success); } else { error('Geo Location is not supported'); } </script> <article></article> </body> </html>

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  • How do I create a class in Javascript?

    - by William
    This is what I got so far, and it's not working at all :( <!DOCTYPE html> <html lang="en"> <head> <title>Class Test</title> <meta charset="utf-8" /> <style> body { text-align: center; background-color: #ffffff;} #box { position: absolute; left: 610px; top: 80px; height: 50px; width: 50px; background-color: #ff0000; color: #000000;} </style> <script type="text/javascript"> document.onkeydown=function(event){keyDown(event)}; document.onkeyup=function(event){keyUp(event)}; var box = 0; function Player () { var speed = 5; var x = 50; var y = 50; } function update() { box.style.left = this.x + "px"; box.style.top = this.y + "px"; box.innerHTML = "<h6 style=\"margin: 0px 0px 0px 0px; padding: 0px 0px 0px 0px;\">X: "+ this.x + "<br /> Y: " + this.y + "</h6>"; } var player = new Player(); var keys = new Array(256); var i = 0; for (i = 0;i <= 256; i++){ keys[i] = false; } function keyDown(event){ keys[event.keyCode] = true; } function keyUp(event){ keys[event.keyCode] = false; } function update(){ if(keys[37]) player.x -= player.speed; if(keys[39]) player.x += player.speed; player.update(); } setInterval(update, 1000/60); </script> </head> <body> <div id="box" ></div> <script type="text/javascript"> box = document.getElementById('box'); box.innerHTML = "<h6 style=\"margin: 0px 0px 0px 0px; padding: 0px 0px 0px 0px;\">X: "+ player.x + "<br /> Y: " + player.y + "</h6>"; </script> </body> </html>

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  • How can I obtain the IPv4 address of the client?

    - by Dr Dork
    Hello! I'm prepping for a simple work project and am trying to familiarize myself with the basics of socket programming in a Unix dev environment. At this point, I have some basic server side code setup to listen for incoming TCP connection requests from clients after the parent socket has been created and is set to listen... int sockfd, newfd; unsigned int len; socklen_t sin_size; char msg[]="Test message sent"; char buf[MAXLEN]; int st, rv; struct addrinfo hints, *serverinfo, *p; struct sockaddr_storage client; char ip[INET6_ADDRSTRLEN]; . . //parent socket creation and listen code omitted for simplicity . //wait for connection requests from clients while(1) { //Returns the socketID and address of client connecting to socket if( ( newfd = accept(sockfd, (struct sockaddr *)&client, &len) ) == -1 ){ perror("Accept"); exit(-1); } if( (rv = recv(newfd, buf, MAXLEN-1, 0 )) == -1) { perror("Recv"); exit(-1); } struct sockaddr_in *clientAddr = ( struct sockaddr_in *) get_in_addr((struct sockaddr *)&client); inet_ntop(client.ss_family, clientAddr, ip, sizeof ip); printf("Receive from %s: query type is %s\n", ip, buf); if( ( st = send(newfd, msg, strlen(msg), 0)) == -1 ) { perror("Send"); exit(-1); } //ntohs is used to avoid big-endian and little endian compatibility issues printf("Send %d byte to port %d\n", ntohs(clientAddr->sin_port) ); close(newfd); } } I found the get_in_addr function online and placed it at the top of my code and use it to obtain the IP address of the client connecting... // get sockaddr, IPv4 or IPv6: void *get_in_addr(struct sockaddr *sa) { if (sa->sa_family == AF_INET) { return &(((struct sockaddr_in*)sa)->sin_addr); } return &(((struct sockaddr_in6*)sa)->sin6_addr); } but the function always returns the IPv6 IP address since thats what the sa_family property is set as. My question is, is the IPv4 IP address stored anywhere in the data I'm using and, if so, how can I access it? Thanks so much in advance for all your help!

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  • refactoring this function in Java

    - by Joel
    Hi folks, I'm learning Java, and I know one of the big complaints about newbie programmers is that we make really long and involved methods that should be broken into several. Well here is one I wrote and is a perfect example. :-D. public void buildBall(){ /* sets the x and y value for the center of the canvas */ double i = ((getWidth() / 2)); double j = ((getHeight() / 2)); /* randomizes the start speed of the ball */ vy = 3.0; vx = rgen.nextDouble(1.0, 3.0); if (rgen.nextBoolean(.05)) vx = -vx; /* creates the ball */ GOval ball = new GOval(i,j,(2 *BALL_RADIUS),(2 * BALL_RADIUS)); ball.setFilled(true); ball.setFillColor(Color.RED); add(ball); /* animates the ball */ while(true){ i = (i + (vx* 2)); j = (j + (vy* 2)); if (i > APPLICATION_WIDTH-(2 * BALL_RADIUS)){ vx = -vx; } if (j > APPLICATION_HEIGHT-(2 * BALL_RADIUS)){ vy = -vy; } if (i < 0){ vx = -vx; } if (j < 0){ vy = -vy; } ball.move(vx + vx, vy + vy); pause(10); /* checks the edges of the ball to see if it hits an object */ colider = getElementAt(i, j); if (colider == null){ colider = getElementAt(i + (2*BALL_RADIUS), j); } if (colider == null){ colider = getElementAt(i + (2*BALL_RADIUS), j + (2*BALL_RADIUS)); } if (colider == null){ colider = getElementAt(i, j + (2*BALL_RADIUS)); } /* If the ball hits an object it reverses direction */ if (colider != null){ vy = -vy; /* removes bricks when hit but not the paddle */ if (j < (getHeight() -(PADDLE_Y_OFFSET + PADDLE_HEIGHT))){ remove(colider); } } } You can see from the title of the method that I started with good intentions of "building the ball". There are a few issues I ran up against: The problem is that then I needed to move the ball, so I created that while loop. I don't see any other way to do that other than just keep it "true", so that means any other code I create below this loop won't happen. I didn't make the while loop a different function because I was using those variables i and j. So I don't see how I can refactor beyond this loop. So my main question is: How would I pass the values of i and j to a new method: "animateBall" and how would I use ball.move(vx + vx, vy + vy); in that new method if ball has been declared in the buildBall method? I understand this is probably a simple thing of better understanding variable scope and passing arguments, but I'm not quite there yet...

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  • Div not filling width of floated container (css expert needed

    - by Rayden
    I know there are many variations of this question posted, but none I've found quite provide an answer that works for this case. I basically have two left floated divs. Inside those two divs are div headers and tabled content. I want the Div headers (Hour/Minute) to stretch to the width of the tabled content, but they only do this in FF and Chrome, not IE7. IE7 is my works official browser so the one I need it to work with the most. Here is the CSS: #ui-timepicker-div { padding:0.2em; } #ui-timepicker-hours { float:left; } #ui-timepicker-minutes { margin:0 0 0 0.2em; float:left; } .ui-timepicker .ui-timepicker-header { padding:0.2em 0; } .ui-timepicker .ui-timepicker-title { line-height:1.8em; text-align:center; } .ui-timepicker table { margin:0.15em 0 0 0; font-size:.9em; border-collapse:collapse; } .ui-timepicker td { padding:1px; width:2.2em; } .ui-timepicker th, .ui-timepicker td { border:0; } .ui-timepicker td a { display:block; padding:0.2em 0.3em 0.2em 0.5em; text-align:right; text-decoration:none; } Here is the HTML (did not include tabled content): <div style="position: absolute; top: 252.667px; left: 648px; z-index: 1; display: none;" class="ui-timepicker ui-widget ui-widget-content ui-helper-clearfix ui-corner-all" id="ui-timepicker-div"> <div id="ui-timepicker-hours"> <div class="ui-timepicker-header ui-widget-header ui-helper-clearfix ui-corner-all"> <div class="ui-timepicker-title">Hour</div> </div> <table class="ui-timepicker"> </table> </div> <div id="ui-timepicker-minutes"> <div class="ui-timepicker-header ui-widget-header ui-helper-clearfix ui-corner-all"> <div class="ui-timepicker-title">Minutes</div> </div> <table class="ui-timepicker"> </table> </div> </div>

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  • how to check if there is one selected radio button among the group?

    - by assehj
    Good day... I am here again to ask your idea on solving this problem. I have displayed questions using while loop. Each questions have 5 radio buttons each. The problem is, I have to check if one among the group of radio button is clicked. How to do this one using php or javascript? I have here my sample program.. Thanks in advance... $ctr = 1; $partIDTemp = ""; while($rowItems = mysql_fetch_array($resultItems)){ if( $partIDTemp != $rowItems['partID'] ){ if($ctr != 1) echo "<tr ></tr>"; echo "<tr style='font-family: Arial, Helvetica, sans-serif; font-weight:bold'><td colspan='3'>" . $rowItems['description'] . "</td></tr>"; } echo "<tr bgcolor=white style='font-family: Arial, Helvetica, sans-serif;'><td align='center'>" . $ctr . "</td>"; $rows =$rowItems['WPID']; echo "<td>" . $rowItems['descriptions']. "</td>"; $response = "response_".$rows; echo "<td><input type='radio' name='$response' value='5' /> <input type='radio' name='$response' value='4' /> <input type='radio' name='$response' value='3' /> <input type='radio' name='$response' value='2' /> <input type='radio' name='$response' value='1' /> </td>"; $ctr++; $partIDTemp = $rowItems['partID']; }

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  • Simple jquery Fade Slideshow fails on certain browsers

    - by cmay
    So I have a simple slideshow on my website which just shows one image then shows another until it reaches the end or the user hits skip in which case it shows index.html. The site is served on apache2 with Django. The slideshow works perfectly on most machines, but certain machines it shows some images twice and other images not at all and the timing is off. I am using jquery 1.4.3. Below is the section of html where I push the image urls from the database to the javascript {% for image in latest_images %} {% thumbnail image.image_file "800x600" crop="center" as im %} <script>FadeImageList.push("{{im.url}}");</script> {% endthumbnail %} {% endfor %} Below is the full javascript file var FadeImageList = []; var fadeDuration = 2000; var fadeImgID = '#slideShow'; var homePageID = '#homePage'; var menuID = '#menu'; var skipFlag = 0; $(document).ready(function(){ $(homePageID).fadeOut(50); PlaySlideshow(FadeImageList); }); var PlaySlideshow = function(FadeImageList){ var newImgSrc = FadeImageList.shift(); $('#skip').click(function(){$('#loader').show();skipFlag = 1;}); if(((typeof(newImgSrc) !== "string") || (skipFlag === 1))){ EndSlideShow(); return; } else{ $(fadeImgID).fadeOut(fadeDuration,function(){ $(fadeImgID).attr('src', newImgSrc); $(fadeImgID).fadeIn(fadeDuration,function(){ PlaySlideshow(FadeImageList); }); }); } }; var EndSlideShow = function(fadeSettings){ $(fadeImgID).fadeOut(400,function(){ $(homePageID).fadeIn(400); $("#skip").fadeOut(400); $('#loader').hide(); }); }; The strange thing is I've had it work and fail on identically version numbered browsers on the same os but on different machines. It consistently either works or fails on a machine. I've had it fail in ie 7,8 firefox 3.6.3 and chrome. I've also had it succeed in ie6,7,8 firefox 3.6.3,3.4.2,3.1 and chrome.

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  • CSS / HTML - Image will not show up

    - by weka
    Ugh, ok. I've been up all night working this thing and now an image won't show. It's so darn annoying. Trying to get this .png image to show up on a simple PHP webpage. I just wanna go to sleep X_X CSS: <style> .achievement { position:relative; width:500px; background:#B5B5B5; float:left; padding:10px; margin-bottom:10px; } .icon { float:left; width:32px; height:32px; background: url("images/trophy.php") no-repeat center; padding:05px; border:4px solid #4D4D4D; } .ptsgained { position:absolute; top:0; right:0; background:#79E310; color:#fff; font-family:Tahoma; font-weight:bold; font-size:12px; padding:5px; } .achievement h1 { color:#454545; font-size:12pt; font-family:Georgia; font-weight:none; margin:0;padding:0; } .achievement p { margin:0;padding:0; font-size:12px; font-family:Tahoma; color:#1C1C1C; } .text { margin-left:10px; float:left; } </style> HTML: <div class="achievement"> <span class="ptsgained">+10</span> <div class="icon"></div> <div class="text"><h1>All Around Submitter</h1> <p>Submit and have approved content in all 6 areas.</p> </div> </div> What am I doing wrong, guys? :\

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  • problem with logout script in php

    - by user225269
    I'm a beginner in php, and I am trying to create a login and logout. But I am having problems in logging out. My logout just calls for the login form which is this: <? session_start(); session_destroy(); ?> <table width="300" border="0" align="center" cellpadding="0" cellspacing="1" bgcolor="#CCCCCC"> <tr> <form name="form1" method="post" action="checklogin.php"> <td> <table width="100%" border="0" cellpadding="3" cellspacing="1" bgcolor="#FFFFFF"> <tr> <td colspan="3"><strong>Member Login </strong></td> </tr> <tr> <td width="78">Username</td> <td width="6">:</td> <td width="294"><input name="myusername" type="text" id="myusername"></td> </tr> <tr> <td>Password</td> <td>:</td> <td><input name="mypassword" type="text" id="mypassword"></td> </tr> <tr> <td>&nbsp;</td> <td>&nbsp;</td> <td><input type="submit" name="Submit" value="Login"></td> </tr> </table> </td> </form> </tr> </table> My problem is, when I try to press the back button in the browser. Whoever user is using it can still access what is not supposed to be accessed when a user hasn't logged in. Do I need to add a code on the user page? I have this code on the user page: <? session_start(); if(!session_is_registered(myusername)){ header("location:main_login.php"); } ?> What can you recommend that I would do so that a script will prompt to enter the username and password again when a user clicks on the back button.

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  • PHP - not sure how to ask - regarding variables and $_POST

    - by Phil
    I have a PHP form. The form works but I'm trying to test to see if a value other than the first item has been selected. I can't figure out how to write the If statement. $products = array( '' => 1, 'Item 2' => 2, 'Item 3' => 3, 'Item 4' => 4, 'Item 5' => 5, 'Item 6' => 6 ); $html = generateSelect('products', $products); function generateSelect($name = '', $options = array()) { $html = '<select name="'.$name.'">'; foreach ($options as $option => $value) { $html .= '<option value='.$value.'>'.$option.'</option>'; } $html .= '</select>'; return $html; } In my table, the drop down box is displayed: <tr> <td style="width:{$left_col_width}; text-align:left; vertical-align:center; padding:{$cell_padding}; font-weight:bold; {$product[3]}">{$product[0]}</td> <td style="text-align:left; vertical-align:top; padding:{$cell_padding};"><select name="{$product[1]}"> <option value="1"></option> <option value="2">Item 2</option> <option value="3">Item 3</option> <option value="4">Item 4</option> <option value="5">Item 5</option> <option value="6">Item 6</option> </select></td> </tr> I use the following if statement to check to see if someone has entered a phone number. if they have not entered a phone number, then the "Phone:" text turns red. How do I do an if statement similar to this to verify that someone has selected a product option from the drop down box? if(!empty($_POST['phone'])) { $phone[2] = clean_var($_POST['phone']); if (function_exists('htmlspecialchars')) $phone[2] = htmlspecialchars($phone[2], ENT_QUOTES); } else { $error = 1; $phone[3] = 'color:#d20128;'; } it seems simple but I can't figure it out.

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  • Why am I getting ClassNotFoundExpection when I have properly imported said class and am looking at it in its directory?

    - by Strider
    This is my Javac compiling statement: javac -cp "C:\java\code\j3D\j3dcore.jar;C:\java\code\j3D\j3dutils.jar;C:\java\code\j3D\vecmath.jar" Simple.java compiles with no problems. The three jar files (j3dcore, j3dutils, and vecmath) are the essential jar's for my program (or at least I am led to believe according to this official tutorial on J3D For the record I ripped this code almost line from line from this pdf file. jar files are correctly located in referenced locations When I run my Simple program, (java Simple) I am greeted with Exception in thread "main" java.lang.NoClassDefFoundError: javax/media/j3d/Cavas3d Caused by: java.lang.ClassNotFoundExpection: javax.media.j3d.Canvas3D Currently I am staring directly at this Canvas3D.class that is located within j3dcore.jar\javax\media\j3d\ wtfisthis.jpg Here is the source code: //First java3D Program import java.applet.Applet; import java.awt.BorderLayout; import java.awt.Frame; import java.awt.event.*; import com.sun.j3d.utils.applet.MainFrame; import com.sun.j3d.utils.universe.*; import com.sun.j3d.utils.geometry.ColorCube; import javax.media.j3d.*; import javax.vecmath.*; import java.awt.GraphicsConfiguration; public class Simple extends Applet { public Simple() { setLayout(new BorderLayout()); GraphicsConfiguration config = SimpleUniverse.getPreferredConfiguration(); Canvas3D canvas3D = new Canvas3D(config); add("Center", canvas3D); BranchGroup scene = createSceneGraph(); scene.compile(); // SimpleUniverse is a Convenience Utility class SimpleUniverse simpleU = new SimpleUniverse(canvas3D); // This moves the ViewPlatform back a bit so the // objects in the scene can be viewed. simpleU.getViewingPlatform().setNominalViewingTransform(); simpleU.addBranchGraph(scene); } // end of HelloJava3Da (constructor) public BranchGroup createSceneGraph() { // Create the root of the branch graph BranchGroup objRoot = new BranchGroup(); // Create a simple shape leaf node, add it to the scene graph. // ColorCube is a Convenience Utility class objRoot.addChild(new ColorCube(0.4)); return objRoot; } public static void main(String args[]){ Simple world = new Simple(); } }` Did I import correctly? Did I incorrectly reference my jar files in my Javac statement? If I clearly see Canvas3D within its correct directory why cant java find it? The first folder in both j3dcore.jar and vecmath.jar is "javax". Is the compiler getting confused? If the compiler is getting confused how do I specify where to find that exact class when referencing it within my source code?

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  • How can I obtain the IP address of my server program?

    - by Dr Dork
    Hello! This question is related to another question I just posted. I'm prepping for a simple work project and am trying to familiarize myself with the basics of socket programming in a Unix dev environment. At this point, I have some basic server side code and client side code setup to communicate. Currently, my client code successfully connects to the server code and the server code sends it a test message, then both quit out. Perfect! That's exactly what I wanted to accomplish. Now I'm playing around with the functions used to obtain info about the two environments (server and client). I'd like to obtain my server program's IP address. Here's the code I currently have to do this, but it's not working... int sockfd; unsigned int len; socklen_t sin_size; char msg[]="test message"; char buf[MAXLEN]; int st, rv; struct addrinfo hints, *serverinfo, *p; struct sockaddr_storage client; char s[INET6_ADDRSTRLEN]; char ip[INET6_ADDRSTRLEN]; //zero struct memset(&hints,0,sizeof(hints)); hints.ai_family = AF_UNSPEC; hints.ai_socktype = SOCK_STREAM; hints.ai_flags = AI_PASSIVE; //get the server info if((rv = getaddrinfo(NULL, SERVERPORT, &hints, &serverinfo ) != 0)){ perror("getaddrinfo"); exit(-1); } // loop through all the results and bind to the first we can for( p = serverinfo; p != NULL; p = p->ai_next) { //Setup the socket if( (sockfd = socket( p->ai_family, p->ai_socktype, p->ai_protocol )) == -1 ) { perror("socket"); continue; } //Associate a socket id with an address to which other processes can connect if(bind(sockfd, p->ai_addr, p->ai_addrlen) == -1){ close(sockfd); perror("bind"); continue; } break; } if( p == NULL ){ perror("Fail to bind"); } inet_ntop(p->ai_family, get_in_addr((struct sockaddr *)p->ai_addr), s, sizeof(s)); printf("Server has TCP Port %s and IP Address %s\n", SERVERPORT, s); and the output for the IP is always empty... server has TCP Port 21412 and IP Address :: any ideas for what I'm missing? thanks in advance for your help! this stuff is really complicated at first.

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  • rake db:create not working for legacy rails app (2.3.5) using MySQL (5.5.28)

    - by ridicter
    I'm a new Rails Developer, and I'm working on a legacy Rails app. Whenever I run the rake db:create command, I get an error that the database couldn't be created. I have found many StackOverflow questions related to this, but in troubleshooting nearly all permutations of solutions, I couldn't resolve the issue. I created the three Dbs (dev, prod, test), created the user with all access privileges to these dbs, and ran rake db:create. I'm running Mac OS X Lion, MySQL 5.5.28, Rails 2.3.5, Ruby 1.8.7. Here are my settings development: adapter: mysql encoding: utf8 database: adva_development username: adva password: **** host: localhost socket: /tmp/mysql.sock Here's the error: Couldn't create database for {"adapter"=>"mysql", "username"=>"adva", "host"=>"localhost", "encoding"=>"utf8", "database"=>"adva_development", "socket"=>"/tmp/mysql.sock", "password"=>"****"}, charset: utf8, collation: utf8_unicode_ci (if you set the charset manually, make sure you have a matching collation) I have done the following troubleshooting: Verified user and password are correct, and the user has access to the DB. (Double checked user access with SELECT * FROM mysql.db WHERE Db = 'adva_development' \G; User has all privileges.) Verify the socket is correct. I don't really understand sockets, but I can plainly see it at /tmp/mysql.sock. Checked collation and character set. I found out I had created the DB in latin charset and collation, so I recreated them. I ran show variables like "collation_database"; and show variables like "character_set_database"; and came back with utf8 and utf8_unicode_ci respectively. I followed the instructions in this question. After uninstalling mysql gem, I ran the following but came up with the same error: gem install --no-rdoc --no-ri mysql -- --with-mysql-dir=/usr/local/mysql-5.5.28-osx10.6-x86_64/bin --with-mysql-config=/usr/local/mysql-5.5.28-osx10.6-x86_64/bin/mysql_config Following Matt's suggestion, here's what a rake --trace db:create reveals: ** Invoke db:create (first_time) ** Invoke db:load_config (first_time) ** Invoke rails_env (first_time) ** Execute rails_env ** Execute db:load_config ** Execute db:create Couldn't create database for {"database"=>"adva_development", "adapter"=>"mysql", "host"=>"127.0.0.1", "password"=>"woof2adva", "username"=>"adva", "encoding"=>"utf8"}, charset: utf8, collation: utf8_unicode_ci (if you set the charset manually, make sure you have a matching collation) After 3 days and six or seven hours, I have pretty much run out of options. I tried various random things, like replacing localhost with 127.0.0.1 to no avail. Could there be something wrong related to my specific environment? Mac OS X Lion + MySQL 5.5.28? I plan on trying on setting up everything in a Linux environment. Thanks!

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