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  • Looping Through Database Query

    - by DrakeNET
    I am creating a very simple script. The purpose of the script is to pull a question from the database and display all answers associated with that particular question. We are dealing with two tables here and there is a foreign key from the question database to the answer database so answers are associated with questions. Hope that is enough explanation. Here is my code. I was wondering if this is the most efficient way to complete this or is there an easier way? <html> <head> <title>Advise Me</title> <head> <body> <h1>Today's Question</h1> <?php //Establish connection to database require_once('config.php'); require_once('open_connection.php'); //Pull the "active" question from the database $todays_question = mysql_query("SELECT name, question FROM approvedQuestions WHERE status = active") or die(mysql_error()); //Variable to hold $todays_question aQID $questionID = mysql_query("SELECT commentID FROM approvedQuestions WHERE status = active") or die(mysql_error()); //Print today's question echo $todays_question; //Print comments associated with today's question $sql = "SELECT commentID FROM approvedQuestions WHERE status = active"; $result_set = mysql_query($sql); $result_num = mysql_numrows($result_set); for ($a = 0; $a < $result_num; $a++) { echo $sql; } ?> </body> </html>

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  • Sorting by some field and fetching whole tree from DB

    - by Niaxon
    Hello everyone, I am trying to do file browser in a tree form and have a problem to sort it somehow. I use PHP and MySQL for that. I've created mixed (nested set + adjacency) table 'element' with the following fields: element_id, left_key, right_key, level, parent_id, element_name, element_type (enum: 'folder','file'), element_size. Let's not discuss right now that it is better to move information about element (name, type, size) into other table. Function to scan specified directory and fill table work correctly. Noteworthy, i am adding elements to tree in specific order: folders first and only files. After that i can easily fetch and display whole table on the page using simple query: SELECT * FROM element WHERE 1=1 ORDER BY left_key With the result of that query and another function i can generate correct html code (<ul><li>... and so on). Now back to the question (finally, huh?). I am struggling to add sorting functionality. For example i want to order my result by size. Here i need to keep in my mind whole hierarchy of tree and rule: folders first, files later. I believe i can do that by generating in PHP recursive query: SELECT * FROM element WHERE parent_id = {$parentId} ORDER BY element_type (so folders would be first), size (or name for example) After that for each result which is folder i will send another query to get it's content. Also it's possible to fetch whole tree by left_key and after that sort it in PHP as array but i guess that would be worse :) I wonder if there is better and more efficient way to do such thing?

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  • How to drag only one image with SDK Iphone

    - by loka
    Hi! I want to create a little app that takes two images and i want to make only the image over draggable. After research, i found this solution : -(void)touchesMoved:(NSSet *)touches withEvent:(UIEvent *)event { UITouch *touch = [[ event allTouches] anyObject]; image.alpha = 0.7; if([touch view] == image){ CGPoint location = [touch locationInView:self.view]; image.center = location; } It works but the problem is that the image is draggable from its center and i don't want that. So i found another solution : - (void) touchesBegan:(NSSet*)touches withEvent:(UIEvent*)event { // Retrieve the touch point CGPoint pt = [[touches anyObject] locationInView:self.view]; startLocation = pt; [[self view] bringSubviewToFront:self.view]; } - (void) touchesMoved:(NSSet*)touches withEvent:(UIEvent*)event { // Move relative to the original touch point CGPoint pt = [[touches anyObject] locationInView:self.view]; frame = [self.view frame]; frame.origin.x += pt.x - startLocation.x; frame.origin.y += pt.y - startLocation.y; [self.view setFrame:frame]; } It works very well but when i add another image, all the images of the view are draggable at the same time.I'm a beginner with the iphone programmation and i have no idea of how i can only make the image over draggable. Thank you in advance for your help!!

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  • problem with accessing a php page

    - by EquinoX
    So I have a info.php page which is located on the folder /var/www/nginx-default, however when I go to my ip address/info.php, it always redirects me to this site: http://www.iana.org/domains/example/ is this because I have a virtual host that I called example? Here is my config for the example website: server { listen 80; server_name www.example.com; rewrite ^/(.*) http://example.com/$1 permanent; } server { listen 80; server_name example.com; access_log /var/www/example.com/logs/access.log; error_log /var/www/example.com/logs/error.log; location / { root /var/www/example.com/public/; index index.html; } } The way I access this site is by changing my /var/hosts in my macbook so that example.com is mapped to my server IP address... however now when I do xxx.xxx.xxx.xxx/info.php.. it redirects me to that site I posted above

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  • inserts 'Array' into mysql table

    - by Noah Smith
    i want to insert an array into a mysql table. The array is produced by script scanning all the links, converting into absolute links and then displaying them in an array. i decided to mysql_query the array into the table but now i am stuck. it only posts 'Array', instead of every row from the array into a different row. Any ideas??! <?php require_once('simplehtmldom_1_5/simple_html_dom.php'); require_once('url_to_absolute/url_to_absolute.php'); $connect = mysql_connect("xxxx", "xxxx", "xxx") or die('Couldn\'t connect to MySQL Server: ' . mysql_error()); mysql_select_db("xxxx", $connect ) or die('Couldn\'t Select the database: ' . mysql_error( $connect )); $links = Array(); $URL = 'http://www.theqlick.com'; // change it for urls to grab // grabs the urls from URL $file = file_get_html($URL); foreach ($file->find('a') as $theelement) { $links[] = url_to_absolute($URL, $theelement->href); } print_r($links); mysql_query("INSERT INTO pages (url) VALUES ('$links[]')"); mysql_close($connect);

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  • destructor being called by subclass

    - by zero
    I'm currently learning more about php objects and constructors/destructors, but i've noticed in my code that the parent class's destructor is being called twice, I thought it was because i was extending the first class to my second class and that the second class was calling it, but this is what the php docs say about that: Like constructors, parent destructors will not be called implicitly by the engine. In order to run a parent destructor, one would have to explicitly call parent::__destruct() in the destructor body. so if it is not being called by the subclass then is it because by extended the first class that i've made a reference to the parent class making it call itself twice or I'm I way off base here? the code: <?php class test{ public $test1 = "this is a test of a pulic property"; private $test2 = "this is a test of a private property"; protected $test3 = "this is a test of a protected property"; const hello = 900000; function __construct($h){ //echo 'this is the constructor test '.$h; } function x($x2){ echo ' this is fn x'.$x2; } function y(){ print "this is fn y"; } } $obj = new test("this is an \"arg\" sent to instance of test"); class hey extends test{ function hey(){ $this->x('<br>from the host with the most'); echo ' <br>from hey class'.$this->test3; } } $obj2 = new hey(); echo $obj2::hello; ?>

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  • Parse xml file with same tag multiple times iphone sdk

    - by neha
    Hi all, In my application, I have a tag multiple times. I'm using xml parser. I'm taking a corresponding element with similar name as the one in xml file in my class. So in case of: <photo>abc</photo> <photo>def</photo> What I get in photo element of my class is the second element i.e def, as the first one gets overwritten as there's only one photo element in my class. My question is am I wrong in taking similar elements in class as in case of xml? Is there any better method or a better parser? Or I'm on right path and have to do this manually by setting some flags etc? Thanx in advance.

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  • Zip multiple database PDF blob files

    - by Michael
    I have a database table that contains numerous PDF blob files. I am attempting to combine all of the files into a single ZIP file that I can download and then print. Please help! <?php include '../config.php'; include '../connect.php'; $session = $_GET['session']; $query = "SELECT $tbl_uploads.username, $tbl_uploads.description, $tbl_uploads.type, $tbl_uploads.size, $tbl_uploads.content, $tbl_members.session FROM $tbl_uploads LEFT JOIN $tbl_members ON $tbl_uploads.username = $tbl_members.username WHERE $tbl_members.session = '$session'"; $result = mysql_query($query) or die('Error, query failed'); $files = array(); while(list($username, $description, $type, $size, $content) = mysql_fetch_array($result)) { $files[] = "$username-$description.pdf"; } $zip = new ZipArchive; $zip->open('file.zip', ZipArchive::CREATE); foreach ($files as $file) { $zip->addFile($file); } $zip->close(); header('Content-Type: application/zip'); header('Content-disposition: attachment; filename=filename.zip'); header('Content-Length: ' . filesize($zipfilename)); readfile($zipname); exit(); ?>

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  • Prevent empty form input array from being posted?

    - by user355295
    Sorry if this has been answered somewhere; I'm not quite sure how to phrase the problem to even look for help. Anyway, I have a form with three text input boxes, where the user will input three song titles. I have simple PHP set up to treat those input boxes as an array (because I may want, say, 100 song titles in the future) and write the song titles to another document. <form method="post"> <input type="text" name="songs[]" value="" /> <input type="text" name="songs[]" value="" /> <input type="text" name="songs[]" value="" /> <button type="submit" name="submit" value="submit">Submit</button> </form> <?php if (isset($_POST['submit'])) { $open = fopen("test.html", "w"); if(empty($_POST['songs'])) { } else { $songs = $_POST['songs']; foreach($songs as $song) { fwrite($open, $song."<br />"); }; }; }; ?> This correctly writes the song titles to an external file. However, even when the input boxes are empty, the external file will still be written to (just with the <br />'s). I'd assumed that the if statement would ensure nothing would happen if the boxes were blank, but that's obviously not the case. I guess the array's not really empty like I thought it was, but I'm not really sure what implications that comes with. Any idea what I'm doing wrong? (And again, I am clueless when it comes to PHP, so forgive me if this has been answered a million times before, if I described it horribly, etc.)

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  • Kepping object in memory (iPhone SDK)

    - by Chris
    I am trying to create a UIImageView called theImageView in the touchesBegan method that I can then then move to a new location in touchesMoved. Currently I am receiving an "undeclared" error in touchesMoved where I set the new location for theImageView. What can I do to keep theImageView in memory between these two methods? - (void)touchesBegan:(NSSet *)touches withEvent:(UIEvent *)event { ... UIImageView *theImageView = [[UIImageView alloc] initWithImage:[UIImage imageNamed:@"image.png"]]; theImageView.frame = CGRectMake(263, 228, 193, 300); [theImageView retain]; ... } - (void)touchesMoved:(NSSet *)touches withEvent:(UIEvent *)event { ... theImageView.frame = CGRectMake(300, 300, 193, 300); ... }

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  • How to evaluate json member using variable ?

    - by Miftah
    Hi i've got a problem evaluating json. My goal is to insert json member value to a function variable, take a look at this function func_load_session(svar){ var id = ''; $.getJSON('data/session.php?load='+svar, function(json){ eval('id = json.'+svar); }); return id; } this code i load session from php file that i've store beforehand. i store that session variable using dynamic var. <?php /* * format ?var=[nama_var]&val=[nilai_nama_var] */ $var = $_GET['var']; $val = $_GET['val']; $load = $_GET['load']; session_start(); if($var){ $_SESSION["$var"] = $val; echo "Store SESSION[\"$var\"] = '".$_SESSION["$var"]."'"; }else if($load){ echo $_SESSION["$load"]; } ?> using firebug, i get expected response but i also received error uncaught exception: Syntax error, unrecognized expression: ) pointing at this eval('id = json.'+svar); i wonder how to solve this

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  • javascript unable to locate a form using the ID tag

    - by ihake
    Here's my problem: I'm trying to set up a simple mobile contact form with a captcha built in. The page I'm working on can be found here: http://m.lancasterpainting.com/contact.php I'm using the following php contact form: http://www.html-form-guide.com/contact-form/php-email-contact-form.html I want to first say that I'm not the only one to run into this problem. After googling the issue, I've found multiple people struggling with this, but no-one seems to have an answer. Now for the problem... As you can see if you visit the page, each time the page is accessed, an error appears that says "Error: couldnot get Form object contact_form". I cannot--for the life of me--figure out why the javascript can't find the form I pass it. I call the function that generates this error at the top of the page: var frmvalidator = new Validator("contact_form"); The form I'm referencing is as follows in the HTML code: <div data-role="page" data-theme="e" id="contact_form" name="contact_form" data-position="inline"> ... And the function that is called that generates the error can be found in an external .js file here: http://m.lancasterpainting.com/scripts/gen_validatorv31.js Is there something that I am simply not seeing? Why can't the javascript locate the form? Thanks so much to anyone that helps with this.

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  • Only first word of two strings gets added to db

    - by dkgeld
    When trying to add words to a database via php, only the first word of both strings gets added. I send the text via this code: public void sendTextToDB() { valcom = editText1.getText().toString(); valnm = editText2.getText().toString(); t = new Thread() { public void run() { try { url = new URL("http://10.0.2.2/HB/hikebuddy.php?function=setcomm&comment="+valcom+"&name="+valnm); h = (HttpURLConnection)url.openConnection(); if( h.getResponseCode() == HttpURLConnection.HTTP_OK){ is = h.getInputStream(); }else{ is = h.getErrorStream(); } h.disconnect(); } catch (Exception e) { // TODO Auto-generated catch block e.printStackTrace(); Log.d("Test", "CONNECTION FAILED 1"); } } }; t.start(); } When tested with spaces and commas etc. in a browser, the php function adds all text. The strings also return the full value when inserted into a dialog. How do I fix this? Thank you.

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  • Adding an integer at the end of an input's name to get a specific url

    - by Gadgetster
    I am trying to get a url where I can retrieve the selected values from. For example, if I put a check mark on a and b then sumbit, I will get: index.php?category=1&&category=2 I want to get this instead: index.php?category0=1&&category1=2 So that I can later get this specific value with $_GET['category0'] Is there a way to add a counter for the selected checkboxes and add 0,1,2,3.. at the end of the name of its input? <form action="" method="get"> <!-- this will be a php loop instead of hardcored which will retrieve data from the db --> <label><input type="checkbox" name="category" value="1">a</label> <label><input type="checkbox" name="category" value="2">b</label> <label><input type="checkbox" name="category" value="3">c</label> <label><input type="checkbox" name="category" value="4">d</label> <label><input type="checkbox" name="category" value="5">e</label> <input type="submit"> </form>

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  • how to fetch data using jquery

    - by user3566029
    I have tried to connect my index.php file in 000webhost.com using jQuery. From the site menu I have connected with 000webhost using FTP details available in 000webhost.com. When I tried to read data from jQuery it doesn’t working. Anyone please tell me what should I need to change? Should I need to connect my Dreamweaver database to webhost? If yes please explain? This is what I have done. index.php $mysql_host = "mysql0.000webhost.com"; $mysql_database = "a000000_mydb"; $mysql_user = "a000000_root"; $mysql_password = "******"; $conn=@mysql_connect ( $mysql_host,$mysql_user,$mysql_password )or die('aa'); mysql_select_db($mysql_database,$conn) or die('eoor on db'); $quer=mysql_query("SELECT * FROM sam"); $res=array(); while($row=mysql_fetch_row($quer)) { $res[]=$row; } print(json_encode($res)); return json_encode($res); mysql_close(); index.html $.get('public_html/index.php', function( data ) { alert( 'Successful' +data); });

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  • retrieve only sub-pages (wordpress)

    - by Radek
    I want to list all sub-pages only one level though of one particular page. I was reading Function Reference/get pages and thought that $pages = get_pages( array( 'child_of' => $post->ID, 'parent' => $post->ID)) ; will do the trick but it is not working. It lists all pages on the same level like the page I call that code from. If I omit parent option I will get all pages even with sub-pages that I want. But I want only sub-pages. The whole function is like function about_menu(){ if (is_page('about')){ $pages = get_pages( array( 'child_of' => $post->ID, 'parent' => $post->ID)) ; foreach($pages as $page) { ?> <h2><a href="<?php echo get_page_link($page->ID) ?>"><?php echo $page->post_title ?></a></h2> <?php } } } and mine is second one

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  • Help me to complete my UITableView to Navigation Controller [iPhone SDK]

    - by Momeks
    I built an application. On the one my views I used TableView. So now I want to change this Table view to a navigation controller. 1- How can I change UITable view to Navigation Controller. I add navigation codes but I got some alert! [I know I must identify my navigation delegate, but HOW ?] Here is my Mapping Views! AppNameViewController FirstViewController [on this view I used table view and I want change to nav] SecondViewController ThirdViewControllerController

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  • Parameter error with Mysqli

    - by Morgan Green
    When I run this Query I recieve Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in /home/morgan58/public_html/wow/includes/index/index_admin.php on line 188 SELECT * FROM characters WHERE id=5 Warning: mysqli_free_result() expects parameter 1 to be mysqli_result, boolean given in /home/morgan58/public_html/wow/includes/index/index_admin.php on line 194 The Query is running and it is strying to select the correct information, but for on the actual output it's giving me a fetch_array error; if anyone can see where the error lies it'd be much appreciated. Thank you. <?php $adminid= $admin->get_id(); $characterdb= 'characters'; $link = mysqli_connect("$server", "$user", "$pass", "$characterdb"); if (mysqli_connect_errno()) { printf("Connect failed: %s\n", mysqli_connect_error()); exit(); } $query = "SELECT * FROM characters WHERE id=$adminid"; $result = mysqli_query($link, $query); while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) { echo $query; echo $row['name']; } mysqli_free_result($result); mysqli_close($link); ?>

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  • ajax form handling an array

    - by moata_u
    am trying to handle an array comes from php file after submitting the form data , the value of data after submitting the form is = ARRAY but i cant use this array in any way , any idea how can i handle this array !!!! Javascript : $('#file').live('change',function(){ $('#preview').html(''); $('#preview').html('<img src="loader.gif" />'); $('#data').ajaxForm(function(data){ $(data['toshow']).insertBefore('.pic_content').hide().fadeIn(1000); }).submit(); }); PHP : .... ....etc echo json_encode(array('toshow'=>somedata,'data'=>somedata)); data come from php file {"toshow":"\r\n\t\t\t\t\r\n\t\t<table class=\"out\">\r\n\t\t\t<tr ><td class=\"img\"><a title=\"2012-06-02 01-22-09\" rel=\"prettyPhoto\" href=\"img\/2012-06-02 01-22-09.284.jpg\"><img src=\"img\/thumb\/2012-06-02 01-22-09.284.jpg\"\/><\/a><\/td><\/tr>\r\n\t\t\t\r\n\t\t\t<td>\r\n\t\t\t\t<table cellSpacing=\"1\" cellPadding=\"0\">\r\n\t\t\t\t\t<tr><td class=\"data\"><span class=\"click\">2012-06-02 01-22-09<\/span><\/td><\/tr>\r\n\t\t\t\t\t<tr><td class=\"data\"><span class=\"click\">Download<\/span><\/td><\/tr>\r\n\t\t\t\t\t<tr><td class=\"data\"><a href=\"img\/2012-06-02 01-22-09.284.jpg\"><span class=\"click\">View<\/span><\/a><\/td><\/tr>\r\n\t\t\t\t<\/table>\r\n\t\t\t<\/td>\r\n\t\t\t<\/tr>\r\n\t\t<\/table>","span":"<span class='text'><img src='greencheck.png'\/>2012-06-02 01-22-09 Uploaded ,File Size =152Kb <\/span>"}

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  • How can I make a new CTP (CakePHP) file in NetBeans?

    - by John Isaacks
    I have found a lot of info about adding CTP file support for NetBeans, but this is usually talking about code highlighting and treating a ctp file like a php file. This can be done at: Tools - Options - Miscellaneous - Files I have done this. However, when I try to create a NEW ctp file. I do not have the option. I tried going to Tools -> Templates to add a ctp template. There is no "new" button just an "add" button that looks for a file. I created a file on my desktop called cake_template.ctp on my desktop. I added it to the PHP templates in the template manager. I called the template "PHP Cake Template". Still when I go to create a new file, the option is not there. I restarted NetBeans, still the same. I just want to create a new .ctp file, it shouldn't be this difficult. Does anyone know how? I am using version 6.9.1

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  • How to document an existing small web site (web application), inside and out?

    - by Ricket
    We have a "web application" which has been developed over the past 7 months. The problem is, it was not really documented. The requirements consisted of a small bulleted list from the initial meeting 7 months ago (it's more of a "goals" statement than software requirements). It has collected a number of features which stemmed from small verbal or chat discussions. The developer is leaving very soon. He wrote the entire thing himself and he knows all of the quirks and underlying rules to each page, but nobody else really knows much more than the user interface side of it; which of course is the easy part, as it's made to be intuitive to the user. But if someone needs to repair or add a feature to it, the entire thing is a black box. The code has some minimal comments, and of course the good thing about web applications is that the address bar points you in the right direction towards fixing a problem or upgrading a page. But how should the developer go about documenting this web application? He is a bit lost as far as where to begin. As developers, how do you completely document your web applications for other developers, maintainers, and administrative-level users? What approach do you use, where do you start, do you have a template? An idea of magnitude: it uses PHP, MySQL and jQuery. It has about 20-30 main (frontend) files, along with about 15 included files and a couple folders of some assets. So overall it's a pretty small application. It interfaces with 7 MySQL tables, each one well-named, so I think the database end is pretty self-explanatory. There is a config.inc.php file with definitions of consts like the MySQL user details, some from/to emails, and URLs which PHP uses to insert into emails and pages (relative and absolute paths, basiecally). There is some AJAX via jQuery. Please comment if there is any other information that would help you help me and I will be glad to edit it in.

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  • problem when uploading file

    - by Syom
    i have the form, and i want to upload two files. here is the script <form action="form.php" method="post" enctype="multipart/form-data" /> <input type="file" name="video" /> <input type="file" name="picture" > <input type="submit" class="input" value="?????" /> <input type="hidden" name="MAX_FILE_SIZE" value="100000000" /> </form> form.php: <? print_r($_FILES); $video_name = $_FILES["video"]["name"]; $image_name = $_FILES["picture"]["name"]; echo "video",$video_name; echo "image",$image_name; //returns Array ( ) videoimage ?> when i try to upload the file greater than 10MB, it doesn't happen. i try in many browsers. maybe i must change some field in php.ini? but i haven't permission to change them on the server. so what can i do? thanks

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