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Thread: Problem Getting PHP Variables into Flash

  1. #1

    Problem Getting PHP Variables into Flash

    I've been following this tutorial on using Flash with PHP and mySQL:

    And I'm sort of stumbling at the first fence. I've got the dynamic text field set up with the variable myVar inside a movieclip with the actionScript:

    onClipEvent (load) {
    loadVariables("", this, "GET");
    And I've made and uploaded test.php:

    $x = "abc";
    print "myVar=$x";
    I've opened test.php, and it says 'myVar=abc' like it should.

    Now when I try to preview my flash movie I get this error message:

    Error opening URL ' %3E'
    As far as I can tell, its sending a bunch of data from the text field to the PHP which is mucking it up (I'm guessing this because when I change the text field's alignment from center to left the error changes to 'align%3D%22left').

    Does anyone know what I'm doing wrong?


  2. #2
    long time i did this but i think it should be:
    print "&myVar=$x";

  3. #3
    God, I'm such an idiot... It wasn't a problem with the script at all, my firewall was simply blocking Flash from connecting to the internet.

    Thanks anyway.

  4. #4
    Ok, new problem. I'm now having problems with it getting the variable out of a mySQL database. What I'm trying to do is search a table for an entry and then use that to set up my variable in Flash.

    So I have this for my php:


    include ("conninfo.php");

    $country = $_GET['country'];

    $query="SELECT * FROM PE_videos WHERE videocountry = $country";

    $result=mysql_query($query) or die ("Nope");
    $numrows = mysql_numrows($result);

    if ($numrows>0)
    print "myVar=$r[videox]";

    echo "None";

    So when I try going to '' in my browser's address bar it says 'myVar=30', as it should. This format worked fine in the first post where the variable was static.

    So I update my actionscript to:

    onClipEvent (load) {
    loadVariables("", this, "GET");
    And nothing. Going to this address should use the country=1 part to do the search query on the php page which should provide the myVar variable. I'm nearly 100% sure the php is correct (after all, it does work in the browser), so I'm guessing there's something wrong with that actionscript.

    Any ideas?


  5. #5
    this problem could be caused by caching. I'm not shure that it is but anyway i would advice you to put a random value in your value list tricking your browser to think there will be new info in processing the file

  6. #6
    prstudio's Avatar
    Formerly known as hybrd_dvlpr. This makes life simpler.
    I agree on the caching - what's the actionscript look like?

    Caching information:

    Prevent caching of loaded variables:

    Prevent caching of SWF files:

  7. #7
    Registered User


    Thx man, I was working on it 3 days and i could not get this tutorial running, thx of you it works now hooohhaaa thx alot!

    Note: changing "print "myVar=$x";" to "print "&myVar=$x";" works on my server PHP5.

    Quote Originally Posted by borrob View Post
    long time i did this but i think it should be:
    print "&myVar=$x";

  8. #8

    problem with same tutorial

    I've being trying to implement the same tutorial but I'm stuck.
    I've tried both locally and on my server with no success. I managed to obtain the print result from the php file but, when I go for the database, I get no results.
    I checked and double checked the table and all the scripting and all seems fine. The only problem I can remember of is the correct identification of the db hosting and user. How can I make sure that those are correct?
    thanks in advance

  9. #9


    I use this :

    but it dosent true ,

    here this code is :

    function lv(l, n, t, e, f) { 
                 if (l == undefined) { 
                     l = new LoadVars(); 
                     l.onLoad = function() {
                     var i; 
                     n.htmlText = "";
                     if (t == undefined) { 
                     n.htmlText += "<b>"+this["title"+e]+"</b><br>"; 
                    } else { 
                    for (i=0; i<this.n; i++) {
                    n.htmlText += "<a href='"+this["link"+i]+"'>
     lv(sites_txt, "cycle", null, "sites.php");

    with out change and next code :

    mysql_select_db ("THE DATABASE YOU WANT TO USE");
    $qResult = mysql_query ("SELECT * FROM sites ORDER BY id ASC");
    $nRows = mysql_num_rows($qResult);
    $rString ="&n=".$nRows;
    for ($i=0; $i< $nRows; $i++){
    $row = mysql_fetch_array($qResult);
    $rString .="&id".$i."=".$row['id']."&"."&title".$i."=".$row['title']."&".
    echo $rString."&";

    just change red words .and my sql :

    CREATE TABLE `sites` ( `id` int(11) NOT NULL auto_increment, `link` varchar(100) NOT NULL default '', `title` varchar(100) NOT NULL default '', PRIMARY KEY (`id`) ) TYPE=MyISAM AUTO_INCREMENT=19 ;

    INSERT INTO `sites` VALUES (5, '', 'Kirupa');
    INSERT INTO `sites` VALUES (4, '', 'Voetsjoeba');
    INSERT INTO `sites` VALUES (3, '', 'Canned Laughter');
    INSERT INTO `sites` VALUES (6, '', 'Spoono');
    INSERT INTO `sites` VALUES (7, '', 'ReadyMadeMag');
    INSERT INTO `sites` VALUES (9, '', 'Weebl and Bob');
    INSERT INTO `sites` VALUES (10, ''); ?
    INSERT INTO `sites` VALUES (12, '', 'Flip Flop Flyin'''); INSERT INTO `sites` VALUES (15, '', 'KirupaForum');
    INSERT INTO `sites` VALUES (16, '', 'Razyr');
    INSERT INTO `sites` VALUES (17, '', 'Senocular');
    INSERT INTO `sites` VALUES (18, '', 'May 1st Reboot');

    the red line in the top code have problem but it s dosent matter .
    and in the end it s dosent work . why ?
    Last edited by danielelvito; November 5th, 2008 at 09:55 PM.

  10. #10

    variable vanishes in PHP

    Hi all,

    I've overcome most of my problems with this project except for one issue.

    I have a simple one table MySQL database with 4 fields: numero, nome, email, resultado. Sorry about the Portuguese but I think its preferable, for the sake of better understanding my code, which follows:

    $nome = $_POST["nomeVar"];
    //$nome = "rf teste";
    $email = '';
    $results = mysql_query("SELECT email FROM colaboradores WHERE nome = '$nome'")or die(mysql_error());
    while($row = mysql_fetch_assoc($results)) {
        if($email != '') { $email .= ', '; }
        $email .= $row['email'];
    if($email != '') { $email .= ''; }
    echo "<?xml version=\"1.0\" encoding=\"ISO-8859-1\"?>\n";
    echo "<colaboradores>\n";
    echo "<email>" . $email . "</email>\n";
    echo "</colaboradores>\n";
    $sendTo = "";
    $headers = "From: " . $nome;
    $subject = "qemail.php - Resultado de " . $email;
    $message = "O nosso colaborador, " . $nome . ";
    mail($sendTo, $subject, $message, $headers);
    The e-mail is just to check if the "nome" variable is reaching the PHP, which it is. "rf teste" is the name I have inserted in the table to perform all the testing.

    I have a flash quizz game which populates a combobox with the names from the database (through PHP and XML). After answering the questions, each user selects his name and, by pressing a confirmation button, the DB should be queried for the corresponding e-mail address and then return it to flash. The result is also inserted in the DB (no problem with that).

    I just can't understand what's wrong with this. If I assign the "nome" in the PHP, using the commented 2nd line of code, the query is executed and I get the intended e-mail address back, in XML, ready to be read by flash.

    When I send the variable from flash, it reaches PHP and is sent back to me via e-mail but, it fails to reach MySQL as the query isn't executed. Consequently, I get a null result in my flash application.

    Somehow, the variable "nomeVar" sent from flash to php is lost between querying for the corresponding e-mail address and sending the "email" variable back to flash.

    I hope I've clearly put it down for you. Thanks in advance



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