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• Using diff and patch to force one local code base to look like another

  - by Dave Aaron Smith

  I've noticed this strange behavior of diff and patch when I've used them to force one code base to be identical to another. Let's say I want to update update_me to look identical to leave_unchanged. I go to update_me. I run a diff from leave_unchanged to update_me. Then I patch the diff into update_me. If there are new files in leave_unchanged, patch asks me if my patch was reversed! If I answer yes, it deletes the new files in leave_unchanged. Then, if I simply re-run the patch, it correctly patches update_me. Why does patch try to modify both leave_unchanged and update_me? What's the proper way to do this? I found a hacky way which is to replace all +++ lines with nonsense paths so patch can't find leave_unchanged. Then it works fine. It's such an ugly solution though. $ mkdir copyfrom $ mkdir copyto $ echo "Hello world" > copyfrom/myFile.txt $ cd copyto $ diff -Naur . ../copyfrom > my.diff $ less my.diff diff -Naur ./myFile.txt ../copyfrom/myFile.txt --- ./myFile.txt 1969-12-31 19:00:00.000000000 -0500 +++ ../copyfrom/myFile.txt 2010-03-15 17:21:22.000000000 -0400 @@ -0,0 +1 @@ +Hello world $ patch -p0 < my.diff The next patch would create the file ../copyfrom/myFile.txt, which already exists! Assume -R? [n] yes patching file ../copyfrom/myFile.txt $ patch -p0 < my.diff patching file ./myFile.txt

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• Parsing xml with php and xpath

  - by Hyung Suh

  Hey guys, I'm trying to parse an xml file to return a item with a specific id only, but having trouble making it work. here's what I have in php $xml_str = file_get_contents("test.xml"); $xml = simplexml_load_string($xml_str); $albid = $_GET['id']; $nodes = $xml->xpath('//library/book[@id=1]'); foreach($nodes as $node) { echo $node['title'].'<br/>'; } First, the php is not returning anything from the xml file. What would I need to fix to return the data? Also, how would I enter $albid into the xpath so that the id will be retrieved from the link? Any pointers in the right direction would be appreciated. Thanks! --and here's the sample xml file-- <library> <book id="1"> <title>PHP and MySQL</title> <author fname="miguel" lname="alvarez">Miguel Alvarez</author> </book> <book id="2"> <title>JAVA 123</title> <author fname="william" lname="vega">WIlliam Vega</author> </book>

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• ZIP Numerous 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'); while(list($username, $description, $type, $size, $content) = mysql_fetch_array($result)) { header("Content-length: $size"); header("Content-type: $type"); header("Content-Disposition: inline; filename=$username-$description.pdf"); echo $content; } $files = array('File 1 from database', 'File 2 from database'); $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($zipname); mysql_close($link); exit; ?>

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• Pass checkbox values with Jquery to PHP and display result in div

  - by user1343955

  I want to filter realtime results with jQuery (just like on this site http://shop.www.hi.nl/hi/mcsmambo.p?M5NextUrl=RSRCH). So when someones checks a checkbox the results should update realtime (in a div). Now I'm a newbie with jQuery and I've tried lots of examples but I can't get it to work. Here's my code, could anyone tell what I'm doing wrong? Thank you very much! HTML <div id="c_b"> Kleur:<br /> <input type="checkbox" name="kleur[1]" value="Blauw"> Blauw <br /> <input type="checkbox" name="kleur[2]" value="Wit"> Wit <br /> <input type="checkbox" name="kleur[3]" value="Zwart"> Zwart <br /> <br /> Operating System:<br /> <input type="checkbox" name="os[1]" value="Android"> Android <br /> <input type="checkbox" name="os[2]" value="Apple iOS"> Apple iOS <br /> </div> <div id="myResponse">Here should be the result</div> jQuery function updateTextArea() { var allVals = []; $('#c_b :checked').each(function() { allVals.push($(this).val()); }); var dataString = $(allVals).serialize(); $.ajax({ type:'POST', url:'/wp-content/themes/u-design/filteropties.php', data: dataString, success: function(data){ $('#myResponse').html(data); } }); } $(document).ready(function() { $('#c_b input').click(updateTextArea); updateTextArea(); }); PHP //Just to see if the var passing works echo var_export($_POST);

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• Access denied when trying to access my database.

  - by Sergio Tapia

  Here's my code: <html> <head> </head> <body> <?php $user = mysql_real_escape_string($_GET["u"]); $pass = mysql_real_escape_string($_GET["p"]); $query = "SELECT * FROM usario WHERE username = '$user' AND password = '$pass'"; mysql_connect(localhost, "root", ""); @mysql_select_db("multas") or die( "Unable to select database"); $result=mysql_query($query); if(mysql_numrows($result) > 0){ echo 'si'; } ?> </body> </html> And here's the error I get when I try to run it Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: Access denied for user 'ODBC'@'localhost' (using password: NO) in C:\xampp\htdocs\useraccess.php on line 7 Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: A link to the server could not be established in C:\xampp\htdocs\useraccess.php on line 7 Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: Access denied for user 'ODBC'@'localhost' (using password: NO) in C:\xampp\htdocs\useraccess.php on line 8 Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: A link to the server could not be established in C:\xampp\htdocs\useraccess.php on line 8 Warning: mysql_numrows() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\useraccess.php on line 16

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• Uninitialized array offset

  - by kimmothy16

  Hey everyone, I am using PHP to create a form with an array of fields. Basically you can add an unlimited number of 'people' to the form and each person has a first name, last name, and phone number. The form requires that you add a phone number for the first person only. If you leave the phone number field blank on any others, the handler file is supposed to be programmed to use the phone number from the first person. So, my fields are: person[] - a hidden field with a value that is this person's primary key. fname[] - an input field lname[] - an input field phone[] - an input field My form handler looks like this: $people = $_POST['person'] $counter = 0; foreach($people as $person): if($phone[$counter] == '') { // use $phone[0]'s phone number } else { // use $phone[$counter] number } $counter = $counter + 1; endforeach; PHP doesn't like this though, it is throwing me an Notice: Uninitialized string offset error. I debugged it by running the is_array function on people, fname, lname, and phone and it returns true to being an array. I can also manually echo out $phone[2], etc. and get the correct value. I've also ran is_int on the $counter variable and it returned true, so I'm unsure why this isn't working as intended? Any help would be great!

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• Enter ID instead of name on submit (form)

  - by Derek

  In my activities table, I have a user ID and a project ID. When a user (of admin level) creates an activity they select from a drop down menu a project. Here is the select query to draw up appropriate values: $sql = "SELECT usersprojects_tb.projectid, projects.projectname FROM projects INNER JOIN usersprojects on projects.projectid = usersprojects.projectid WHERE usersprojects.userid = '".$_SESSION['SESS_USERID']."'"; And for the tag with the dropdown menu, I have this: <?php echo $row['projectname']?> I have tried submitting the form with 'projectid' here instead and the project ID is stored successfully in my activies table. However, the user needs to see the project names (IDs arent exactly user-friendly!) And with 'projectname' as displayed, they can select the names of the available projects (to associate an activity with) but the project ID is not stored, how I link this up, so that when the project name is sent, the ID for this project is stored properly in my activities table. I'm also having the exact same problem with the users drop down. As the admin user selects a user from the drop down to assign the task to. I exactly what I want, but I think I may be using the wrong syntax! Any help is much appreciated. Thanks.

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• How to parse rss from a php page, using jQuery/jFeed?

  - by ricebowl

  I'm trying to fumble my way through parsing rss sensibly, using jQuery and jFeed. Because of the same origin policy I'm pulling the BBC's health news feed into a local page (http://www.davidrhysthomas.co.uk/play/proxy.php). Originally this was just the same proxy.php script as available in the jFeed download package, but due to my host's disabling allow_url_fopen() I've amended the php to the following: $url = "http://newsrss.bbc.co.uk/rss/newsonline_uk_edition/health/rss.xml"; $ch = curl_init(); curl_setopt($ch, CURLOPT_URL, $url); curl_setopt($ch, CURLOPT_HEADER, 0); curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); $data = curl_exec($ch); echo "$data"; curl_close($ch); Which seems to generate the same/comparable contents as the original fopen on my local machine. Now that seems to be working, I'm looking at setting the jFeed script up to work with the page and, to my embarrassment, don't see how. I understand that, at the least, this should work: jQuery.getFeed({ url: 'http://www.davidrhysthomas.co.uk/play/proxy.php', success: function(feed) { alert(feed.title); } }); ...but, as I'm sure you anticipate, it doesn't. What non-output there is, is available for your perusal here: http://www.davidrhysthomas.co.uk/play/exampleTest.html. And I honestly don't have a clue what to do about it. If anyone could offer some pointers, tips, hints, or, at a pinch, a quick slap around the cheeks and a 'pull yourself together!' it'd be much appreciated... Thanks in advance =)

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• Creating self-referential tables with polymorphism in SQLALchemy

  - by Jace

  I'm trying to create a db structure in which I have many types of content entities, of which one, a Comment, can be attached to any other. Consider the following: from datetime import datetime from sqlalchemy import create_engine from sqlalchemy import Column, ForeignKey from sqlalchemy import Unicode, Integer, DateTime from sqlalchemy.orm import relation, backref from sqlalchemy.ext.declarative import declarative_base Base = declarative_base() class Entity(Base): __tablename__ = 'entities' id = Column(Integer, primary_key=True) created_at = Column(DateTime, default=datetime.utcnow, nullable=False) edited_at = Column(DateTime, default=datetime.utcnow, onupdate=datetime.utcnow, nullable=False) type = Column(Unicode(20), nullable=False) __mapper_args__ = {'polymorphic_on': type} # <...insert some models based on Entity...> class Comment(Entity): __tablename__ = 'comments' __mapper_args__ = {'polymorphic_identity': u'comment'} id = Column(None, ForeignKey('entities.id'), primary_key=True) _idref = relation(Entity, foreign_keys=id, primaryjoin=id == Entity.id) attached_to_id = Column(Integer, ForeignKey('entities.id'), nullable=False) #attached_to = relation(Entity, remote_side=[Entity.id]) attached_to = relation(Entity, foreign_keys=attached_to_id, primaryjoin=attached_to_id == Entity.id, backref=backref('comments', cascade="all, delete-orphan")) text = Column(Unicode(255), nullable=False) engine = create_engine('sqlite://', echo=True) Base.metadata.bind = engine Base.metadata.create_all(engine) This seems about right, except SQLAlchemy doesn't like having two foreign keys pointing to the same parent. It says ArgumentError: Can't determine join between 'entities' and 'comments'; tables have more than one foreign key constraint relationship between them. Please specify the 'onclause' of this join explicitly. How do I specify onclause?

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• Error handling in PHP

  - by Industrial

  Hi guys, We're building a PHP app based on Good old MVC (codeigniter framework) and have run into trouble with a massive chained action that consists of multiple model calls, that together is a part of a big transaction to the database. We want to be able to do a list of actions and get a status report of each one back from the function, whatever the outcome is. Our first initial idea was to utilize the exceptions of PHP5, but since we want to also need status messages that doesnt break the execution of the script, this was our solution that we came up with. It goes a little something like this: $sku = $this->addSku( $name ); if ($sku === false) { $status[] = 'Something gone terrible wrong'; $this->db->trans_rollback(); return $status; } $image= $this->addImage( $filename); if ($image=== false) { $error[] = 'Image could not be uploaded, check filesize'; $this->db->trans_rollback(); return $status; } Our controller looks like this: $var = $this->products->addProductGroup($array); if (is_array($var)) { foreach ($var as $error) { echo $error . '<br />'; } } It appears to be a very fragile solution to do what we need, but it's neither scalable, neither effective when compared to pure PHP exceptions for instance. Is this really the way that this kind of stuff generally is handled in MVC based apps? Thanks!

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• php POST response

  - by amit

  I am making a POST to another php file that I want to render in the browser. so clearly, using jquery POST wont work, since it works through AJAX. echo '<div class="showcase_URL"><a class="purl" name="'.$row->num.'" href="#">PROJECT URL</a></div>'; what I have till now in javascript is: $(".showcase_URL a").click(function() { //alert($(this).attr("name")); var number = $(this).attr("name"); $.post( "app-display.php", {app: number}, function(data){ top.location.href = 'app-display.php'; }); }); what I have in app-display.php: $query = 'SELECT num, title, thumb_url, url, type, cat, dated, details FROM app WHERE `num` = "$_POST[app]"'; but it is currently giving me a page without the contents of app-display.php. all the other fragments of the page are loading: header, footer etc. the PHP Response (in Firebug) I am getting is the normal html of the page. how should i do it?

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• Change flash src with jquery?

  - by Elliott

  Hi I have a flash menu showing a few links, but when the user is logged in I want to change the menu from menu1 to menu2 ... so that it will display "My Account" rather than "Signup" The code below is for my flash: <div id="menu"> <object classid="clsid:d27cdb6e-ae6d-11cf-96b8-444553540000" codebase="http://download.macromedia.com/pub/shockwave/cabs/flash/swflash.cab#version=9,0,0,0" width="825" height="69" id="menu1" align="middle"> <param name="allowScriptAccess" value="sameDomain" /> <param name="allowFullScreen" value="false" /> <param name="movie" value="menu1.swf" /><param name="quality" value="high" /><param name="wmode" value="transparent" /><param name="bgcolor" value="#ffffff" /> <embed src="menu1.swf" quality="high" wmode="transparent" bgcolor="#ffffff" width="825" height="69" name="menu1" align="middle" allowScriptAccess="sameDomain" allowFullScreen="false" type="application/x-shockwave-flash" pluginspage="http://www.macromedia.com/go/getflashplayer" /> </object> </div> Php: if (loggedin()) { echo '<script type="text/javascript"> CHANGE FLASH LINK HERE </script>'; } Could this be done without having to write all the above code out again? Thanks :)

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• help on ejb stateless datagram and message driven beans

  - by Kemmal

  i have a client thats sending a message to the ejbserver using UDP, i want the server(stateless bean) to echo back this message to the client but i cant seem to do this. or can i implement the same logic by using JMS? please help and enlighten. this is just a test, in the end i want a midp to be sending the message to the ejb using datagrams. here is my code. @Stateless public class SessionFacadeBean implements SessionFacadeRemote { public SessionFacadeBean() { } public static void main(String[] args) { DatagramSocket aSocket = null; byte[] buffer = null; try { while(true) { DatagramPacket request = new DatagramPacket(buffer, buffer.length); aSocket.receive(request); DatagramPacket reply = new DatagramPacket(request.getData(), request.getLength(), request.getAddress(), request.getPort()); aSocket.send(reply); } } catch (SocketException e) { System.out.println("Socket: " + e.getMessage()); } catch (IOException e) { System.out.println("IO: " + e.getMessage()); } finally { if(aSocket != null) aSocket.close(); } } } and the client: public static void main(String[] args) { DatagramSocket aSocket = null; try { aSocket = new DatagramSocket(); byte [] m = "Test message!".getBytes(); InetAddress aHost = InetAddress.getByName("localhost"); int serverPort = 6789; DatagramPacket request = new DatagramPacket(m, m.length, aHost, serverPort); aSocket.send(request); byte[] buffer = new byte[1000]; DatagramPacket reply = new DatagramPacket(buffer, buffer.length); aSocket.receive(reply); System.out.println("Reply: " + new String(reply.getData())); } catch (SocketException e) { System.out.println("Socket: " + e.getMessage()); } catch (IOException e) { System.out.println("IO: " + e.getMessage()); } finally { if(aSocket != null) aSocket.close(); } } please help.

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• PHP: Retrieving JSON via jQuery ajax help

  - by iamjonesy

  Hey I have a script that is creating and echoing a JSON encoded array of magento products. I have a script that calls this script using jQuery's ajax function but I'm not getting a proper response. This is the script that creates the array: $collection = Mage::getModel('catalog/product')->getCollection(); $collection->addAttributeToSelect('name'); $collection->addAttributeToSelect('price'); $products = array(); foreach ($collection as $product){ $products[$product->getPrice()] = $product->getName(); } header('Content-Type: application/x-json; charset=utf-8'); echo(json_encode($products)); Here is my jQuery: <select id="products"> <option value="#">Select</option> </select> <script type="text/javascript"> jQuery.noConflict(); jQuery(document).ready(function(){ jQuery.ajax({ type: "GET", url: "http://localhost.com/magento/modules/products/get.php", dataType: "json", success: function(products) { jQuery.each(products,function(price,name) { var opt = jQuery('<option />'); opt.val(name); opt.text(price); jQuery('#products').append(opt); }); } }); }); </script> I'm getting a response from this but I'm not seeing a any JSON. I'm using firebug. I can see there has been a JSON encoded response but the response tab is emtyp and my select boxes have no options. Can anyone see and problems with my code? Here is the response I should get: {"82.9230":"Dummy","177.0098":"Dummy 2","76.0208":"Dummy 3","470.6054":"Dummy 4","357.0083":"Dummy Product 5"} Thanks, Billy

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• Getting Started with Ruby & Ruby on Rails

  - by JakeTheSnake

  Some background: I'm a jack-of-all traits, one of which is programming. I learned VB6 through Excel and PHP for creating websites and so far it's worked out just fine for me. I'm not CS major or even mathematically inclined - logic is what interests me. Current status: I'm willing to learn new and more powerful languages; my first foray into such a route is learning Ruby. I went to the main Ruby website and did the interactive intro. (by the way, I'm currently getting redirected to google.com when I try the link...it's happening to other websites as well...is my computer infected?) I liked what I learned and wanted to get started using Ruby to create websites. I downloaded InstantRails and installed it; everything so far has been fine - the program starts up just fine, and I can test some Ruby code in the console. However my troubles begin when I try and view a web page with Ruby code present. Lastly, my problem: As in PHP, I can browse to the .php file directly and through using PHP tags and some simple 'echo' statements I can be on my way in making dynamic web pages. However with the InstantRails app working, accessing a .rb or .rhtml page doesn't produce similar results. I made a simple text file named 'test.rb' and put basic HTML tags in there (html, head, body) and the Ruby tags <%= and % with some ruby code inside. The web page actually shows the tags and the code - as if it's all just plain HTML. I take it Ruby isn't parsing the page before it is displayed to the user, but this is where my lack of understanding of the Ruby environment stops me short. Where do I go from here?

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• OOP 101 - quick question.

  - by R Bennett

  I've used procedural for sometime now and trying to get a better understanding of OOP in Php. Starting at square 1 and I have a quick question ot get this to gel. Many of the basic examples show static values ( $bob-name = "Robert";) when assigning a value, but I want to pass values... say from a form ( $name = $_POST['name']; ) class Person { // define properties public $name; public $weight; public $age; public function title() { echo $this->name . " has submitted a request "; } } $bob = new Person; // want to plug the value in here $bob->name = $name; $bob->title(); I guess I'm getting a little hung up in some areas as far as accessing variables from within the class, encapsulation & "rules", etc., can $name = $_POST['name']; reside anywhere outside of the class or am I missing an important point? Thanks

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• save html-formatted text to database

  - by yozhik

  Hi all! I want to save html-formatted text to database, but when I do that it is don't save html-symbols like < / ' and others This is how I read article from database for editing: <p class="Title">??????????? ???????:</p> <textarea name="EN" cols="90" rows="20" value="<?php echo $articleArr['EN']; ?>" ></textarea> And this is how I save it to database: function UpdateArticle($ArticleID, $ParentName, $Title, $RU, $EN, $UKR) { //fetch data from database for dropdown lists //connect to db or die) $db = mysql_connect($GLOBALS["gl_dbName"], $GLOBALS["gl_UserName"], $GLOBALS["gl_Password"] ) or die ("Unable to connect"); //to prevenr ????? symbols in unicode - utf-8 coding mysql_query("SET NAMES 'UTF8'"); mysql_set_charset('utf8'); mysql_query("SET NAMES 'utf8' COLLATE 'utf8_general_ci'"); //select database mysql_select_db($GLOBALS["gl_adminDatabase"], $db); $sql = "UPDATE Articles SET AParentName='".$ParentName."', ATitle='".$Title."', RU='".$RU."', EN='".$EN."', UKR='".$UKR."' WHERE ArticleID='".$ArticleID."';"; //execute SQL-query $result = mysql_query($sql, $db); if (!$result) { die('Invalid query: ' . mysql_error()); } //close database = very inportant mysql_close($db); } Help me please, how can I save such texts properly, Thanx!

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• security deleting a mysql row with jQuery $.post

  - by FFish

  I want to delete a row in my database and found an example on how to do this with jQuery's $.post() Now I am wondering about security though.. Can someone send a POST request to my delete-row.php script from another website? JS function deleterow(id) { // alert(typeof(id)); // number if (confirm('Are you sure want to delete?')) { $.post('delete-row.php', {album_id:+id, ajax:'true'}, function() { $("#row_"+id).fadeOut("slow"); }); } } PHP: delete-row.php <?php require_once("../db.php"); mysql_connect(DB_SERVER, DB_USER, DB_PASSWORD) or die("could not connect to database " . mysql_error()); mysql_select_db(DB_NAME) or die("could not select database " . mysql_error()); if (isset($_POST['album_id'])) { $query = "DELETE FROM albums WHERE album_id = " . $_POST['album_id']; $result = mysql_query($query); if (!$result) die('Invalid query: ' . mysql_error()); echo "album deleted!"; } ?>

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• Access/Download server files, not in site root, with PHP

  - by user271619

  Usually I save documents (images, mpegs, excel, word docs, etc...) for my friends or family on my website's root, inside a directory called /files/ or something similar. Nothing too uncommon. But, I have been playing with user session control, and allowing users to upload files to the dedicated /files/ directory. (the file names are saved in a db, with that user's ID) But, that means other people could try to guess and locate other people's files. I do randomize the file names, upon upload. And I stop the apache from displaying the /files/ directory content. However, I'd like to start saving the files outside of the website's root. This way it can't be accessible via the browser. I don't have any code to show, but I didn't want to even start on this endeavor if it's not able to be accomplished. I did find this snippet that shows how to display an image, from outside your website root: $file = $_GET['file']; $fileDir = '/path/to/files/'; if (file_exists($fileDir . $file)) { // Note: You should probably do some more checks // on the filetype, size, etc. $contents = file_get_contents($fileDir . $file); // Note: You should probably implement some kind // of check on filetype header('Content-type: image/jpeg'); echo $contents; } ? Maybe I can use this for any file type, but has anyone heard of a better way to allow users (logged in) to access their files from online, but not letting other users has similar access?

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• highlighting search results in php/mysql

  - by fusion

  how do i highlight search results from mysql query using php? this is my code: $search_result = ""; $search_result = $_GET["q"]; $result = mysql_query('SELECT cQuotes, vAuthor, cArabic, vReference FROM thquotes WHERE cQuotes LIKE "%' . $search_result .'%" ORDER BY idQuotes DESC', $conn) or die ('Error: '.mysql_error()); function h($s) { echo htmlspecialchars($s, ENT_QUOTES); } ?> <div class="center_div"> <table> <caption>Search Results</caption> <?php while ($row= mysql_fetch_array($result)) { ?> <tr> <td style="text-align:right; font-size:15px;"><?php h($row['cArabic']) ?></td> <td style="font-size:16px;"><?php h($row['cQuotes']) ?></td> <td style="font-size:12px;"><?php h($row['vAuthor']) ?></td> <td style="font-size:12px; font-style:italic; text-align:right;"><?php h($row['vReference']) ?></td> </tr> <?php } ?> </table> </div>

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• servlet resopnse data for autocomplete

  - by shams haque

  Hello experts, Following code is in php. i want to do same in java. Please tell me how do i generate this type of array or collection in java. I need this to response to json autocomplete. <?php $q = strtolower($_GET["q"]); if (!$q) return; $items = array( "Peter Pan"=>"[email protected]", "Molly"=>"[email protected]", "Forneria Marconi"=>"[email protected]", "Master Sync"=>"[email protected]", "Dr. Tech de Log"=>"[email protected]", "Don Corleone"=>"[email protected]", "Mc Chick"=>"[email protected]", "Donnie Darko"=>"[email protected]", "Quake The Net"=>"[email protected]", "Dr. Write"=>"[email protected]" ); $result = array(); foreach ($items as $key=>$value) { if (strpos(strtolower($key), $q) !== false) { array_push($result, array( "name" => $key, "to" => $value )); } } echo json_encode($result); ?>

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• Problem changing Java version using alternatives

  - by Brian Lewis

  I'm not quite sure how I got into this mess, but for some reason I'm not able to change the current version of Java using alternatives. I can run alternatives --config java and type my selection but when I echo the version number for either java or javac, it spits back out 1.5 every time (despite alternatives showing the current version is 1.6). The server I'm working with is running RHEL5, by the way. I have verified that the paths used in alternatives are pointing to the correct directories. Here's some output from my session: [brilewis@myserver]$ sudo /usr/sbin/update-alternatives --config java There are 3 programs which provide 'java'. Selection Command ** 1 /usr/lib/jvm/jre-1.4.2-gcj/bin/java + 2 /usr/java/jdk1.5.0_10/bin/java 3 /usr/java/jdk1.6.0_16/bin/java Enter to keep the current selection[+], or type selection number: 3 [brilewis@myserver]$ java -version java version "1.5.0_10" Java(TM) 2 Runtime Environment, Standard Edition (build 1.5.0_10-b03) Java HotSpot(TM) Server VM (build 1.5.0_10-b03, mixed mode) [brilewis@myserver]$ sudo /usr/sbin/update-alternatives --config java There are 3 programs which provide 'java'. Selection Command ** 1 /usr/lib/jvm/jre-1.4.2-gcj/bin/java 2 /usr/java/jdk1.5.0_10/bin/java + 3 /usr/java/jdk1.6.0_16/bin/java Enter to keep the current selection[+], or type selection number:

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• I need some help cropping an image in PHP (GD)

  - by evan

  http://i.imgur.com/foT9u.jpg Using that image as an example, here's what I need to do: Crop the blue square to have the same proportional ratio as that of the black square From doing that, I should then be able to resize the blue square to fit into the black square without losing stretching it - It'll retain its proportions. Note: The blue square must be cropped 'from the center'. The original center should remain the center after the crop (it can't be cropped from the top left, for example). Here's what I'm thinking needs to be done (using the, landscape, blue square as the example): Figure out the difference between the black squares width and height Figure out the difference between the blue squares width and height This should tell me how much to crop the blue square by and with how much of a 'top offset' Once it's cropped to fit the black squares proportions, it can then be resized I've been messing around with code similar to: if (BLACK_WIDTH > BLACK_HEIGHT) { $diffHeight = BLACK_WIDTH - BLACK_HEIGHT; $newHeight = $blue_Height - $blue_Height; echo $newHeight; } And using Photoshop to try and get a feel for how this should be done, but it continues to fail .< How should I go about doing this? How can I figure out how much to crop by (depending on if the blue square is landscape or portrait)? How do I then get the offset to retain the blue squares center?

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• PHP script not automatically updating when moved to another server

  - by user32007

  A friend built a ranking system on his site and I am trying to host in on mine via WordPress and Go Daddy. It updates for him but when I load it to my site, it works for 6 hours, but as soon as the reload is supposed to occur, it errors and I get a 500 timeout error. His page is at: jeremynoeljohnson .com/yakezieclub My page is currently at http://sweatingthebigstuff.com/yakezieclub but when you ?reload=1 it will give the error. Any idea why this might be happening? Any settings that I might need to change? Here is the top of the index.php file. I'm not sure which part of any of it is messing up. I literally uploaded the same code as him. Here's the reload part: $cachefile = "rankings.html"; $daycachefile = "rankings_history.xml"; $cachetime = (60 * 60) * 6; // every 6 hours, the cache refreshes $daycachetime = (60 * 60) * 24; // every 24 hours, the history will be written to // - or whenever the page is requested after 24 hours has passed $writenewdata = false; if (!empty($_GET['reload'])) { if ($_GET['reload']== 1) { $cachetime = 1; } } if (!empty($_GET['reloadhistory'])) { if ($_GET['reloadhistory'] == 1) { $daycachetime = 1; $cachetime = 1; } } if (file_exists($daycachefile) && (time() - $daycachetime < filemtime($daycachefile))) { // Do nothing } else { $writenewdata = true; $cachetime = 1; } // Serve from the cache if it is younger than $cachetime if (file_exists($cachefile) && (time() - $cachetime < filemtime($cachefile))) { include($cachefile); echo "<!-- Cached ".date('jS F Y H:i', filemtime($cachefile))." -->"; exit; } ob_start(); // start the output buffer ?>

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• PHP - Select from database the same query

  - by How to PHP

  I created a table that contains the name of the user and his job, and created PHP page that shows me all the users that works doctor, I entered doctor into a variable then I selected from the table where Jobs equal to $doctor, that is great, but I need it to get the same Jobs into a table in the page and the other same jobs into a table in the same page. this is my code that shows only the users works doctor in one table, <html> <h1>Doctors</h1> </html> <?php mysql_connect('localhost','root',''); mysql_select_db('data'); $doctor='doctor'; $query= mysql_query("SELECT * FROM `users` WHERE `job` = '$doctor'")or die(mysql_error()); while ($arr = mysql_fetch_array($query)) $name= $arr['name']; echo $name; } ?> That shows me doctors when I put doctor in a variable I want to show all same Jobs in a table. Is there is a way to do this? Thanks :)

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