jQuery .load() lädt meine Skripte nicht

Question

Ich habe eine Vorlage erstellt, die ich mit jQuery .load() - Funktion laden möchte. Beim Testen habe ich herausgefunden, dass es keine laden wird.

Hier ist meine Last Code:

function open() { 

      history.pushState(null, null, "artiesten.php?u=Chato De Veirman"); 

      $('.content').load('artiest_template.php .content'); 

     } 

Hier mein Template Code:

<?php include('includes/connect.php') ?> 
    <span class="content"> 

    <div class="right_col" role="main"> 

    <div class=""> 
    <script>alert("nothing")</script> 
<div class="row" id="groot"> 
     <div class="col-md-4 col-sm-2 col-xs-12 profile_details"> 
     </div> 
     <div class="col-md-4 col-sm-8 col-xs-12 profile_details"> 
        <div class="well profile_view"> 
         <div class="col-sm-12"> 
         <h4 class="brief a-titel"><i>Panda 2.0</i></h4> 
         <div class="left col-xs-7"> 
          <h2 class="a-naam">Laurent Meersman</h2> 
          <p><strong>Over: </strong><span class="a-over"> Te weinig tijd/Grafische artiest/Gek </span></p> 
          <ul class="list-unstyled"> 
          <li><i class="fa fa-user"></i> Kernmerk: <span class="a-kernmerk">Gek</span></li> 
          <li><i class="fa fa-envelope-o"></i> Email: <span class="a-email"><a href= "mailto:laurent.meersman@oc.be">laurent.meersman@oc.be</a></span></li> 
          </ul> 
         </div> 
         <div class="right col-xs-5 text-center"> 
          <img src="images/Laurent Meersman.jpg" alt="" class="img-circle img-responsive a-foto"> 
         </div> 
         </div> 
         <div class="col-xs-12 bottom text-center"> 
          <button type="button" class="btn btn-info btn-xs"> 
          <i class="fa fa-video-camera"> </i> Bekijk meer video's. 
          </button> 
        </div> 
        </div> 

     </div> 
     </div> 
     <div class="row" id="klein"> 
     <div class="col-md-2 col-sm-2 col-xs-12 profile_details"> 
     </div> 
     <div class="col-md-8 col-sm-8 col-xs-12 profile_details"> 
        <div class="well profile_view"> 
         <div class="col-sm-12"> 
         <h4 class="brief a-titel"><i>Panda</i></h4> 
         <div class="left col-xs-7"> 
          <h2 class="a-naam">Laurent Meersman</h2> 
          <p><strong>Over: </strong><span class="a-over"> Te weinig tijd/Grafische artiest/Gek </span></p> 
          <ul class="list-unstyled"> 
          <li><i class="fa fa-user"></i> Kernmerk: <span class="a-kernmerk">Gek</span></li> 
          <li><i class="fa fa-envelope-o"></i> Email: <span class="a-email"><a href= "mailto:laurent.meersman@oc.be">laurent.meersman@oc.be</a></span></li> 
          </ul> 
         </div> 
         <div class="right col-xs-5 text-center"> 
          <img src="images/Laurent Meersman.jpg" alt="" class="img-circle img-responsive a-foto"> 
         </div> 
         </div> 
         <div class="col-xs-12 bottom text-center"> 
          <button type="button" class="btn btn-info btn-xs"> 
          <i class="fa fa-video-camera"> </i> Bekijk meer video's. 
          </button> 
        </div> 
        </div> 

     </div> 
     </div> 

     <?php 
     echo '<script>alert("' . $_GET['u'] . ' or nothing")</script>'; 
if($_GET['u']){ 
$t = mysqli_real_escape_string($connect,$_GET['u']); 
$res = mysqli_query($connect, "SELECT * FROM artiesten WHERE Naam='" . $t . "'"); 
$i = mysqli_fetch_assoc($res); 
echo ' 
    <script src="vendors/jquery/dist/jquery.min.js"></script> 
    <script type="text/javascript"> 
    $(document).ready(function() { 
     $(".a-titel").html("' . $i['Titel'] . '"); 
     $(".a-naam").html("' . $i['Naam'] . '"); 
     $(".a-over").html("' . $i['Over'] . '"); 
     $(".a-kernmerk").html("' . $i['Kernmerk'] . '"); 
     $(".a-email").html("' . $i['Email'] . '"); 
     $(".a-foto").attr("src", "images/' . $i['Naam'] . '.jpg"); 
     history.pushState(null, null, "?u=' . $i['Naam'] . '"); 
    }); 
    </script>'; 
} 
?> 



    </div> 

    </div> 

    </span> 

Kann mir jemand dabei helfen? Vielen Dank im Voraus.

Answer 1

sein sollte

function open() { 

     history.pushState(null, null, "artiesten.php?u=Chato De Veirman"); 

     $('.content').load('artiest_template.php'); 

    } 

nicht

function open() { 

     history.pushState(null, null, "artiesten.php?u=Chato De Veirman"); 

     $('.content').load('artiest_template.php .content'); 

    } 

Answer 2

jQuery load() nimmt drei Parameter, URL, Daten und eine Funktion. Diese Parameter müssen durch ein Koma getrennt sein. Ihr load() hat etwas hinter Ihrer URL, die nicht Teil der URL ist.

ändern

$('.content').load('artiest_template.php .content'); 

zu

$('.content').load('artiest_template.php');