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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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• Calling function and running it

  - by devs

  I am quite new to objects and OOP. I really don't know how to explain it well but I'll try. So I am trying to read though JSON with JS, the JSON is passed from PHP. This would be easy if all of the information was on the same html page, but I' am trying something that I am new too. So let me show my code... First is the JS which is in app.js var Donors = function(){ var api = this.list; $(document).ready(function(){ $.getJSON(api, function(){ var donorObj = api.payload; $.each(donorObj, function(i, donor){ //console.log(donor.ign); }); }); }); } What I want this part to do is read from the JSON I'm giving it and console.log each name (or donor.ign) when the document is ready. On the html page, or header.php <script> $(function(){ var list = <?php cbProxy(); ?>; var Dons = new Donors(); Dons.list = list; }); </script> the data that's in list is the below JSON. You already know what the rest does, it just passes the JSON to the Donors() function. JSON example: { "code": 0, "payload": [ { "time": 1349661897, "packages": [ "49381" ], "ign": "Notch", "price": "15.99", "currency": "USD" } I'm use to just making functions and calling it on the same page or file and this is my first doing this kind of function. How can I get the function to run with the data I sent it so it console.log() each name.

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• Pass Element value to $ajax->link in cakephp

  - by TwoThumbs

  I need to pass the value of an element to an $ajax-link without using a form/submit structure. (because I have a dynamically set number of clickable links through which I am triggering the action) I used to do this in Ruby using the Prototype javascript function $F like this: <%= link_to_remote "#{item.to_s}", :url => { :action => :add_mpc }, :with => "'category=' + $F('mpc_category')" -%> But this does not seem to work in Cakephp: <?php echo $ajax->link(substr($vehicle['vehicles']['year'], -2), array('action' => 'add_mpc', 'category' => '$F("mpc_category")'), array('update' => 'results', 'position' => 'top')); ?> PHP sees $F as a variable instead of a call to javascript. I'm not too familiar with Javascript, but is there another way to pass the value of the 'mpc_category' input element to the controller through this link? I have been looking for a couple days and can't find anyone dealing with this specific issue. Thanks for any assistance. Edit: fixed syntax in php statement.

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• JQuery, AJAX: Trying AJAX for the first time but cant get this to work...

  - by fwaokda

  Trying to get the code below to work but the success doesn't execute - the error does. How can I get more detailed information on what is exactly going wrong? I'll include the code for next.php in a pastebin link also. Thanks. [next.php: http://pastebin.com/Gnu2AfU8 ] $("a#next").click(function() { $.ajax({ type : 'POST', url : 'next.php', dataType : 'json', data : { nextID : $("a#next").attr("rel") }, success : function ( data ) { $("img#spotlight").attr("src",data.spotlightimage); $("div#showcase h1").text(data.title); $("div#showcase h2").text(data.subtitle); for(var i=0; i < data.size; i++) { $("ul#features").append("<li>").text(data.feature+i).append("</li>"); } $("div#showcase p").text(data.description); for(i=1; j < data.picsize; i++) { $("div.thumbnails ul").append("<li>").text(data.image+i).append("</li>"); } $("a#next").attr("rel", $a("a#next").attr("rel") + 1); }, error : function ( XMLHttpRequest, textStatus, errorThrown) { $("div#showcase h1").text("An error has occured."); } }); });

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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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• How to reslove mysql_fetch_assoc(): problems!

  - by sky

  When i use the code below, im getting this error: Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource when returning the data, anyone can fix it? Thanks! <?php $mysql_server_name="localhost"; $mysql_username=""; $mysql_password=""; $mysql_database=""; $conn=mysql_connect($mysql_server_name, $mysql_username, $mysql_password); ?> <?php $result = mysql_query("SELECT * FROM users"); $arrays = array(); while ($row = mysql_fetch_assoc($result)) { foreach ($row as $key => $val) { if (!array_contains_key($key)) { $arrays[$key] = array(); } $arrays[$key][] = $val; } } ?> <script type="text/javascript"> <?php foreach ($arrays as $key => $val) { print 'var ' . $key . ' = ' . json_encode($val) . ";\r\n"; } ?> </script>

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• Send form with jQuery using keypress()

  - by kexxcream

  What I have so far: I have a form in PHP that contain the following: <input type="submit" id="0" name="0" value="This button"> <input type="submit" id="1" name="1" value="This button"> <input type="hidden" id="response" name="response" value="150"> The problem: I would like to send the response value together with either name=1 or name=0, depending on which button the user submit. However, this should be done using the keyboard letters A and S. If a user press the letter A then the value 0 should be submitted, and if S is pressed then the value 1 should be sent. The jQuery code: $(document).ready(function() { // listens for any navigation keypress activity $(document).keypress(function(e) { switch(e.which) { // user presses the "a" case 97: submitViaKeypress("0"); break; // user presses the "s" key case 115: submitViaKeypress("1"); break; } }); }); // shows a given element and hides all others function submitViaKeypress(element_id) { var response = $('#response').attr('value'); $.ajax({ type: "POST", url: "initiate.php", data: "response=" + response + "&" + element_id + "=" + element_id }); } Goal: That "initiate.php receive two POST variables (response and either 0 or 1).

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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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• Strange mod_rewrite problem; Website works partially

  - by Camran

  I have Ubuntu 9.10 Server... I need to get mod_rewrite working... the mod_rewrite module IS LOADED. On my server, the httpd.conf is empty, instead everything (almost) is in a file called apache2.conf. Anyways, I have also read I have to change the AllowOverride None to AllowOverride All in some file... My httpd.conf is empty as you know, but I have a folder called sites-enabled which contains a 000-default file. This is where I have set: AllowOverride All Now my goal as I stated in the last Q is to turn this link: http://mydomain.com/ad.php?ad_id=Bmw_nice_M3_497379462 into this: http://mydomain.com/Bmw_nice_M3_497379462 So as I got an answer in the last Q i inserted this into the htaccess file: Options +FollowSymLinks Options +Indexes RewriteEngine On RewriteCond %{REQUEST_URI} !^/ad\.php RewriteRule ^(.*)$ ad.php?ad_id=$1 [L] Now, this works (no fully) when entering the url manually in the adress bar, but my website isn't working now for some reason. It is like the website is locked down or something, and unless I change AllowOverride to None it will act like that. Any ideas why? Also another note, the links inside the rewritten url doesn't work properly (images are not shown, while some are shown)...

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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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• 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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• Sort the $_POST variables

  - by Jerry

  Hello guys Might be an easy for you guys. I am trying to sort the $_POST variables that were sent by a form and update the sorted result in mysql. I am not sure how to do it and appreciate it anyone can help me about it. My main.php //I have a loop here. (omitted) //$k will be increased by 1 every time the loop starts, so I will know the total times of the loops //the form will be submitted to update.php echo "<input type='hidden' name='pickTotal' value='".$k."' />"; echo "<input type='hidden' id='point' name='earnedPoint".$k."' value='".$point."' />"; echo "<input type='hidden' id='users' name='userName".$k."' value='".$userPick['user']."' />"; //loop ends My update.php if(isset($_POST['submit'])){ $pickTotal=$_POST['pickTotal']; //get the total loop for ($p=0;$p<=$pickTotal;$p++){ $userToBeUpdated=$_POST['userName'.$p]; $userPoint=$_POST['earnedPoint'.$p]; //sort the $userPoint here. //I need to find out who got the most points //and list user's place. 1st, 2nd, 3rd...etc. //update my mysql } Thanks for any helps.

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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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• 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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• 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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• 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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• Page does update with details from the database after i hit a button

  - by swathi

  I have a code and the way it should work is,when they click on NEW CUSTOMER,it takes them to test1.php where in they enter the details and they hit submit.it saves all the details in properly in the database and when i go back and hit REFRESH ,it should come up with the customer details which they had entered in previously. But what happens is, when i click on the REFRESH,it refreshes the same old page which is empty.I wanted to find out where am i missing the logic.Thanks in advance. The sample code would be <tr> <td class="tdvisitbig" colspan="5">THIS IS A TEST</td> </tr> <tr> <td class='tdvisitbig' colspan="5"><input type="button" onClick="openVisit('test1.php?id=<?=$key?>&name=<?=$name?>');return false;" value="NEW CUSTOMER" class="submit">&nbsp;<input type="button" value="REFRESH" name="add_xyz" class="submit" onClick="document.add.target='_self';document.add.action='test3.php?redirect=visit&section=test page';document.add.submit();"></td> </tr> <? $q = "SELECT address,customernum,status FROM customer WHERE name='$name' ORDER BY customernum"; $r = mysql_query( $q , $Link ); while( $rw = mysql_fetch_assoc( $r ) ) { extract( $rw ); ?> <tr> <? } ?>

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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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• Apache mod_rewrite : How to REWRITE (or whatever) child directories to parent?

  - by ????

  Actually i am trying to make a PHP MVC like application. A basic one. The current milestone i am reaching already includes: Basic RESTful Routing Means, if i type: www.example.com/items/book/8888 .. it properly just stays there as it is and i can already slice out the URL by slashes / and loads the responsible Controllers .... etc from the top single index.php file. I mean, so it is OK for the backend PHP. But the only problem is, it still CAN NOT process the REWRITES properly. For example, the CSS & JS are BROKEN as if i VIEW PAGE SOURCE of the page www.example.com/items/book/8888, the asset files are being called as: www.example.com/items/book/8888/css/main.css www.example.com/items/book/8888/js/jquery.js .. which really are PROBLEMS because in the code is like: <link type="text/css" rel="stylesheet" media="all" href="css/main.css"> <script type="text/javascript" src="js/jquery.js"></script> So the question is: How can i use Apache REWRITE (or whatever approach) to make sure every ASSET FILES to be correctly being called from the DOCROOT. For example, if i am in the URL: www.example.com/items/book/8888 My ASSET FILES should still be called as: www.example.com/css/main.css www.example.com/js/jquery.js Or is there any other methods i need to follow? Please kindly help suggest. Thank you.

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• preg_match to find the current directory in a URL

  - by Ian

  I'm trying to detect the current section of a site that a user is viewing by checking for the final directory in the URL. I'm using a PHP and regex to do it and I think I'm close but unfortunately not quite there yet. Here's what I currently have: <?php $url = $_SERVER['REQUEST_URI_PATH'] = preg_replace('/\\?.*/', '', $_SERVER['REQUEST_URI']); $one = '/one/'; $two = '/three/'; $three = '/three/'; $four = '/four/'; $five = '/five/'; echo $url; if (substr($_SERVER['REQUEST_URI_PATH'], 0, strlen($one)) == $one) { // URI path starts with "/one/" echo "The section is one."; } elseif (substr($_SERVER['REQUEST_URI_PATH'], 0, strlen($two)) == $two) { // URI path starts with "/two/" echo "The section is two."; } elseif (substr($_SERVER['REQUEST_URI_PATH'], 0, strlen($three)) == $three) { // URI path starts with "/three/" echo "The section is three."; } elseif (substr($_SERVER['REQUEST_URI_PATH'], 0, strlen($four)) == $four) { // URI path starts with "/four/" echo "The section is four."; } elseif (substr($_SERVER['REQUEST_URI_PATH'], 0, strlen($five)) == $five) { // URI path starts with "/five/" echo "The section is five."; } ?> I've placed in the echo before the if statements just to get confirmation of the value of $url. This outputs /currentdirectory/file.php However the conditions themselves don't match anything and my individual echo for each section never displays. Also if there's a simpler way of doing it then I'm open to suggestions. Thanks

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• Scan HTML for values with a special character before them.

  - by DogPooOnYourShoe

  Say I have values on my page, like #100 #246, What I want to do is scan the page for values with a # before them and then alter them to put a hyperlink on it $MooringNumbers = '#!' . $MooringNumbers . ' | ' . '#!' . $row1["Number"]; } $viewedResult = '<tr><td>' .$Surname.'</td><td>'.$Title.'</td><td>'.$MooringNumbers . '</td><td>'.$Telephone.'</td><td>' . '<a href="rlayCustomerUpdtForm.php?id='.$id.'">[EDIT]</a></td>'.'<td>'. '<a href="deleteCustomer.php?id='.$id.'">[x]</a></td>'. '</tr>'; preg_replace('/#!(\d\d\d)/', '<a href="update.php?id=${1}">${1}</a>', $viewedResult); echo $viewedResult; This is the broken code which doesnt work.

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• Overload the behavior of count() when called on certain objects

  - by Tom

  In PHP 5, you can use magic methods, overload some classes, etc. In C++, you can implement functions that exist is STL as long as the argument types are different. Is there a way to do this in PHP? An example of what I'd like to do is this: class a { function a() { $this->list = array("1", "2"); } } $blah = new a(); count($blah); I would like blah to return 2. IE count the values of a specific array in the class. So in C++, the way I would do this might look like this: int count(a varName) { return count(varName->list); } Basically, I am trying to simplify data calls for a large application so I can call do this: count($object); rather than count($object->list); The list is going to be potentially a list of objects so depending on how it's used, it could be really nasty statement if someone has to do it the current way: count($object->list[0]->list[0]->list); So, can I make something similar to this: function count(a $object) { count($object->list); } I know PHP's count accepts a mixed var, so I don't know if I can override an individual type.

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• Cannot Display Data from MySQL table

  - by MxmastaMills

  I've got a pretty standard call to a MySQL database and for some reason I can't get the code to work. Here's what I have: $mysqli = mysqli_connect("localhost","username","password"); if (!$mysqli) { die('Could not connect: ' . mysqli_error($mysqli)); } session_start(); $sql = "SELECT * FROM jobs ORDER BY id DESC"; $result = $mysqli->query($sql); $num_rows = mysqli_num_rows($result); Now, first, I know that it is connecting properly because I'm not getting the die method plus I added an else conditional in there previously and it checked out. Then the page displays but I get the errors: Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in blablabla/index.php on line 11 Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in blablabla/index.php on line 12 I've double-checked my database and there is a table called jobs with a row of "id" (it's the primary row). The thing that confuses me is this is code that I literally copied and pasted from another site I built and for some reason the code doesn't work on this one (I obviously copy and pasted it and then just changed the table name and rows accordingly). I saw the error and tried: $num_rows = $mysqli_result->num_rows; $row_array = $mysqli_result->fetch_array; and that fixed the errors but resulted in no data being passed (because obviously $mysqli_result has no value). I don't know why the error is calling for that (is it a difference in version of MySQL or PHP from the other site)? Can someone help me track down the problem? Thanks so much. Sorry if it's something super simple that I'm overlooking, I've been at it for a while.

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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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• Infrastructure for high transactional system (language & hosting suggestion help)

  - by RPS

  Some of our friends (University students) are trying to develop a twitter type application, I want to plan for at least 1000 transactions per second (I know it's wishful thinking) for initial launch. This involves several people connecting and getting updates and posting (text + images) to site. In the back end db will server the data and also calculates rankings of what to push to user based on complex algorithm on the fly real-time. Our group is familiar with Java and Tomcat/MySQL. We can also easily learn/code in PHP/MySQL. What is the best suited platform for our purpose ? Though Java seem to be easy to implement for us I am afraid that hosting will be a bit difficult. I could find cloud based php hosting services (like rackspace cloudsites) at reasonable cost. Amazon EC2 is a bit over our heads to manage on day-to-day. Also any recommendation on hosting ? (PHP or Java) We don't have millions in seed money but about $20K to start with. Any advice on above or any thing in general approach is much appreciated.

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