load()不会加载我的脚本。

I have made a template that I would like to load in using jQuery .load() function. When testing I found out that it won't load any .

我已经创建了一个模板，我希望使用jQuery .load()函数加载它。测试时我发现它不会加载任何东西。

Here is my load code:

这是我的负载代码:

    function open() {

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

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

        }

Here is my template code:

这是我的模板代码:

<?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>

Can somebody help me with this? Thanks in advance.

有人能帮我吗?提前谢谢。

2 个解决方案

#1

Should be

应该是

 function open() {

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

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

    }

not

不

  function open() {

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

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

    }

#2

jQuery load() takes three parameters, URL, data and a function. These parameters have to be separated by a coma. Your load() has something behind your URL that is not part of the URL.

jQuery load()包含三个参数、URL、数据和一个函数。这些参数必须由昏迷来区分。您的load()在URL后面有一些不属于URL的内容。

Change

改变

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

来

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

#1