dataTable插件不工作它显示一个页面中的所有记录我想要每页只有10个

时间:2017-11-09 06:28:24

标签: javascript php jquery html css

我使用dataTable插件来格式化我的html表。但是它没有显示我想要的结果。它显示所有记录而不是每页10条记录。我使用了iDisplayLength:10但它也不起作用。

我的代码: -

<html>
<head>
<script src="https://code.jquery.com/jquery-1.11.1.min.js"></script>
<script src="https://cdn.datatables.net/1.10.4/js/jquery.dataTables.min.js"></script>
<script>
$(document).ready(function(){
    $('#count').DataTable({
           "pagingType": "full_numbers",
           "iDisplayLength": 10
         }
    );
});
</script>
<link rel="stylesheet" href="https://cdn.datatables.net/1.10.4/css/jquery.dataTables.min.css">
</head>
  <table id="count">
                <thead>
                 <tr>
                    <th>Quarter ID</th>
                    <th>Locality Name</th>
                    <th>Block</th>
                    <th>Floor</th>
                    <th>RB Type</th>
                    <th>Quarter Type</th>
                    <th>Quarter No</th>
                    <th>No of Rooms</th>
                    <th>Servant quarter</th>
                    <th>Status</th>
                 </tr>
                </thead>
                <?php
        require 'connection.php';
        $q8="select * from mstquarters";
        $q9=mysqli_query($con,$q8);
        if(mysqli_num_rows($q9)>=1)
        {
            while($row=mysqli_fetch_array($q9))
            {
                ?>
                <?php
                $y=$row['blockId'];
                $q5="select * from mstblocks where blockId='$y'";
                $r5=mysqli_query($con,$q5);
                $r6=mysqli_fetch_array($r5);
                $r7=$r6['block'];

                $z=$row['localityId'];
                $q6="select * from mstlocalities where localityId='$z'";
                $s5=mysqli_query($con,$q6);
                $s6=mysqli_fetch_array($s5);
                $s7=$s6['localitySName'];

                ?>
                <tbody>
                <tr>
                <td><?php echo $row['quarterId'];?></td>
                <td><?php echo $s7;?></td>
                <td><?php echo $r7;?></td>
                <td><?php echo $row['floor'];?></td>
                <td><?php echo $row['rbType'];?></td>
                <td><?php echo $row['quarterType'];?></td>
                <td><?php echo $row['quarterNo'];?></td>
                <td><?php echo $row['noOfRooms'];?></td>
                <td><?php echo $row['servantQuarter'];?></td>
                <td><?php echo $row['status'];?></td>
                </tr>
                </tbody>
                <?php
            }
        }
        ?>
    </table>
</html>

我已经包含了javascript文件,我应该包含更多文件以便正常工作吗?

4 个答案:

答案 0 :(得分:1)

问题是您在<tbody>内创建while(),这会导致错误的表格结构。

将它放在while()之外,你会很高兴

像这样: -

</thead>
    <tbody>
        <?php
        require 'connection.php';
        $q8="select * from mstquarters";
        $q9=mysqli_query($con,$q8);
        if(mysqli_num_rows($q9)>=1)
        {
            while($row=mysqli_fetch_array($q9))
            {
                ?>
                <?php
                $y=$row['blockId'];
                $q5="select * from mstblocks where blockId='$y'";
                $r5=mysqli_query($con,$q5);
                $r6=mysqli_fetch_array($r5);
                $r7=$r6['block'];

                $z=$row['localityId'];
                $q6="select * from mstlocalities where localityId='$z'";
                $s5=mysqli_query($con,$q6);
                $s6=mysqli_fetch_array($s5);
                $s7=$s6['localitySName'];

                ?>
                <tr>
                <td><?php echo $row['quarterId'];?></td>
                <td><?php echo $s7;?></td>
                <td><?php echo $r7;?></td>
                <td><?php echo $row['floor'];?></td>
                <td><?php echo $row['rbType'];?></td>
                <td><?php echo $row['quarterType'];?></td>
                <td><?php echo $row['quarterNo'];?></td>
                <td><?php echo $row['noOfRooms'];?></td>
                <td><?php echo $row['servantQuarter'];?></td>
                <td><?php echo $row['status'];?></td>
                </tr>

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

答案 1 :(得分:0)

尝试"_iDisplayLength": 10或将jquery.dataTables.js添加到您的项目中。

因为iDisplayLength位于 jquery.dataTables.js

答案 2 :(得分:0)

使用$ _REQUEST [“start”]和$ _REQUEST [“length”]  $ _REQUEST [“start”]默认以0开头,$ _REQUEST [“length”]在下面设置iDisplayLength。  $( '#计数')。数据表({            “pagingType”:“full_numbers”,            “iDisplayLength”:10          } 拿两个变量 $开始= $ _ REQUEST [ “开始”]; $长度= $ _ REQUEST [ “长度”];

并在此处使用----从mstquarters限制$ start,$ length

中选择*

答案 3 :(得分:0)

jquery代码中的任何变化

在php文件中添加代码。

取两个变量

$开始= $ _ REQUEST [ “开始”]; $长度= $ _ REQUEST [ “长度”];

并在此处使用----从mstquarters限制$ start,$ length

中选择*