Ajax, PHP and MySql

Question

I'm having a problem with Ajax. Nothing happen when I change my select value.

I have a div with the id textHint in order to print the result.

Here is my select :

<form>              
    <select id="choix" name="choix" onchange="showUser(this.value)">
        <div class="tutorial_list">
            <?php 
                $db = mysql_connect('localhost', 'root', 'root'); 
                mysql_select_db('Projet',$db); 

                $sql = 'select NomPromo, NumPromo from Promo';
                $req = mysql_query($sql) or die('Erreur SQL !<br>'.$sql.'<br>'.mysql_error());

                while ($data = mysql_fetch_array($req)){
                     echo'<option value="'.$data['NumPromo'].'">'.$data['NomPromo'].'</option>';
                }
            ?>
        </div>
    </select>
</form>

Here's my script :

<script>
    function showUser(str) {
        if (str == "") {
           document.getElementById("txtHint").innerHTML = "";
           return;
           } else { 
                if (window.XMLHttpRequest) {
                    // code for IE7+, Firefox, Chrome, Opera, Safari
                    xmlhttp = new XMLHttpRequest();
                } else {
                    // code for IE6, IE5
                    xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
                }
                xmlhttp.onreadystatechange = function() {
                    if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
                        document.getElementById("txtHint").innerHTML = xmlhttp.responseText;
                    }
                };
                xmlhttp.open("GET","data.php?q="+str,true);
                xmlhttp.send();
        }
    }
</script>

And here's my data.php :

<?php 

    $q = intval($_GET['q']);

    $db = mysql_connect('localhost', 'root', 'root'); 
    mysql_select_db('Projet',$db);

    $sql = "select Nom, Prenom from User where Groupe ='".$q."'";
    $req = mysql_query($sql) or die('Erreur SQL !<br>'.$sql.'<br>'.mysql_error());

    while ($data = mysql_fetch_array($req)){
         echo $data['Nom'].' '.$data['Prenom'];
    }
?>

mysql_* functions are deprecated since PHP 5.5 (and removed entirely in PHP 7) and you should stop using them if you can. You should choose another API, like mysqli_* or PDO instead - see choosing an API. — Qirel
– Qirel, Commented Feb 1, 2016 at 13:30
Check whether showUser method is getting invoked or not on changing value in selectbox — Hari Darshan
– Hari Darshan, Commented Feb 1, 2016 at 13:31
Oh yes I will change that when this things work. I'll use mysqli — FlorianSL
– FlorianSL, Commented Feb 1, 2016 at 13:33
Does it generate any errors? In PHP, check the error_log (error_reporting(E_ALL);, ini_set('display_errors', 1);) or in JavaScript (see Console in your browser). — Qirel
– Qirel, Commented Feb 1, 2016 at 13:41
There is an error : [Error] TypeError: null is not an object (evaluating 'document.getElementById("txtHint").innerHTML = xmlhttp.responseText') onreadystatechange on the line document.getElementById("txtHint").innerHTML = xmlhttp.responseText; — FlorianSL
– FlorianSL, Commented Feb 1, 2016 at 13:47

Community · Accepted Answer · 2017-05-23 10:27:58Z

2

I don't understand what <div class='tutorial_list'></div> doing inside <select></select>

This error

[Error] ReferenceError: Can't find variable: $ (fonction anonyme)prof.php:75

may be because of few reasons. a) jquery library is not loaded correctly b) path could be not correct. Check this link

I've done little changes, you can try this out.

<form>              
  <select id="choix" name="choix">
    <div class="tutorial_list">
      <?php 
      $db = mysql_connect('localhost', 'root', 'root'); 
      mysql_select_db('Projet',$db); 

      $sql = 'select NomPromo, NumPromo from Promo';
      $req = mysql_query($sql) or die('Erreur SQL !<br>'.$sql.'<br>'.mysql_error());

      while ($data = mysql_fetch_array($req)){
           echo'<option value="'.$data['NumPromo'].'">'.$data['NomPromo'].'</option>';
      }
      ?>
    </div>
  </select>
</form>

JS

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.0/jquery.min.js"></script>

<script>
  $(document).ready(function(){
    $('#choix').change(function(){
      var q= $('#choix').val();
      $.ajax({url:"data.php?q="+q,cache:false,success:function(result){
        $('#txtHint').html(result);
      }});
    });
  });
</script>

[NOTE: mysql_* functions are deprecated since PHP 5.5. Use mysqli_* or PDO]

edited May 23, 2017 at 10:27

CommunityBot

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answered Feb 1, 2016 at 13:42

Nana Partykar

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12 Comments

FlorianSL Over a year ago

It don't works there is the error $('#choix').change(function(){ he can't find the variable $

FlorianSL Over a year ago

"[Error] ReferenceError: Can't find variable: $ (fonction anonyme)prof.php:75"

Nana Partykar Over a year ago

I have updated my answer. Please go through it @FlorianSL

Nana Partykar Over a year ago

Can you do one thing for me. For a time being in data.php page. Remove all code. and just <?php echo "asd";?> and see what happens. You are close to it. @FlorianSL

FlorianSL Over a year ago

When i go in the development tool and I press on data.php, it tells me that there is the value that I want to print. So my data.php works, there is an error for print this

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