First time click:

Screenshot-110

Second Round Mouse click:

Screenshot11

The above images shows screen short of a popup that is shown by clicking on a link from a html table i created using Jquery DataTables plugin and data from mysql database. in essense, when i click the linkable column of a table for the first time, it shows the popup but under the main table.closing the popup and click on the same link again now displays the popup clearly IF I DONT REFRESH THE browser. restarting machine or firefox doesnt help. below is part of css code:

  <link href="css/dropdown/dropdown.css" media="screen" rel="stylesheet" type="text/css" />
<link href="css/dropdown/themes/default/default.ultimate.css" media="screen" rel="stylesheet" type="text/css" />

        <style type="text/css">
            @import "media/css/demo_table_jui.css";
            @import "media/themes/smoothness/jquery-ui-1.8.4.custom.css";

        </style>
 <script type="text/javascript" charset="utf-8">
            $(document).ready(function(){
                $('#custlist').dataTable({
                    "sPaginationType":"full_numbers",
                    "aaSorting":[[2, "desc"]],
                     "iDisplayLength":5,
                    "bJQueryUI":true
                });
            })

        </script>

  <table width="100%" border="0" cellpadding="0" class="body" cellspacing="0" align="center">
  <tr>
    <td align="center" valign="top">
      <table width="100%" border="0" cellspacing="0" cellpadding="0" class="body">
        <tr>
          <td valign="top" align="center" width="35%">
            <table width="100%" border="0" cellspacing="0" cellpadding="0">
              <tr>
                <td>
                  <table width="70%" border="0" cellspacing="0" cellpadding="0" class="table_background">
                      <tr>
                        <td>
<!-- gap between tables>-->
                          <table width="70%" border="0" cellspacing="1" cellpadding="0">
                            <tr valign="top">
                              <td width="70%" class="table_heading">
                                <div align="center">
                                Customers List
                                </div>
                              </td>

                            </tr>
                            <tr valign="top">
                              <td width="65%" class="row1">
                                <table width="100%" border="0" cellspacing="5" cellpadding="1" class="row1">
                                  <tr>
                                    <td width="70%">

                                <div></div><div id="grid">
                    <table id="custlist" class="display">
                <thead>
                    <tr>
                        <th>Customer Name</th>
                        <th>Category</th>
                        <th>Contact</th>
                        <th>email</th>
                    </tr>
                </thead>
                <tbody>
                    <?php
                    while ($row = mysql_fetch_array($result)) {
                        ?>
                        <tr>
                            <td><?= "<a href=? onclick=\"xajax_getContact('".$row['mobile']."','".$row['id']."');return false;\">".$row['customer']." (".$row['mobile'].")</a>"?></td>
                            <td><?=$row['account_type']?></td>
                            <td><?=$row['mobile']?></td>
                            <td><?=$row['email']?></td>

                        </tr>
                        <?php
                    }
                    ?>
                </tbody>
            </table>
        </div>



    So what can be the cause ?
Member Avatar for LastMitch

Jquery DataTables popup error

What error? I'm having a hard time understand what you are having?

You are having trouble making another window open a website?

Where is your ajax code?

I mean this:

<?= "<a href=? onclick=\"xajax_getContact('".$row['mobile']."','".$row['id']."');return false;\">".$row['customer']." (".$row['mobile'].")</a>"?>
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