Question

从这里开始我就像从数据库中获取数据并以表格格式显示它仍然工作正常，点击division_edit()函数后我想传递值（id），怎么能这样做。

＆＃13;

<table id="dataTable" class="table table-bordered table-hover">
  <thead>
    <tr>
      <th>S.No</th>
      <th>Division</th>
      <th>Action</th>
    </tr>
  </thead>
  <tbody>
    <?php $sql=m ysql_query( "SELECT * FROM division WHERE status !='1'"); for($i=1;$row=m ysql_fetch_assoc($sql);$i++){ ?>
    <tr role="row" class="odd">
      <td>
        <?php echo $i;?>
      </td>
      <td>
        <?php echo $row[ 'division'];?>
      </td>
      <td><a class="btn btn-success btn-xs" href="state_edit.php?id=<?php echo base64_encode($row['id']);?>" onclick="division_edit()" id="division_edit"><i class="fa fa-pencil-square-o"></i> Edit</a> 
        <button class="btn btn-danger btn-xs" data-href="state_delete.php?id=<?php echo base64_encode($row['id']);?>"
        data-toggle="modal" data-target="#confirm-delete"><i class="fa fa-trash"></i> Delete</button>
      </td>
    </tr>
    <?php } ?>
  </tbody>

  <tfoot>
    <tr>
      <th>S.No</th>
      <th>Division</th>
      <th>Action</th>
    </tr>
  </tfoot>
</table>

＆＃13;

Answer 1

这可以帮到你：

<table id="dataTable" class="table table-bordered table-hover">
  <thead>
    <tr>
      <th>S.No</th>
      <th>Division</th>
      <th>Action</th>
    </tr>
  </thead>
  <tbody>
    <?php $sql=m ysql_query( "SELECT * FROM division WHERE status !='1'"); for($i=1;$row=m ysql_fetch_assoc($sql);$i++){ ?>
    <tr role="row" class="odd">
      <td>
        <?php echo $i;?>
      </td>
      <td>
        <?php echo $row[ 'division'];?>
      </td>
      <td><a class="btn btn-success btn-xs" href="state_edit.php?id=<?php echo base64_encode($row['id']);?>" onclick="division_edit(<?php echo $row['id']; ?>)" id="division_edit"><i class="fa fa-pencil-square-o"></i> Edit</a> 
        <button class="btn btn-danger btn-xs" data-href="state_delete.php?id=<?php echo base64_encode($row['id']);?>"
        data-toggle="modal" data-target="#confirm-delete"><i class="fa fa-trash"></i> Delete</button>
      </td>
    </tr>
    <?php } ?>
  </tbody>

  <tfoot>
    <tr>
      <th>S.No</th>
      <th>Division</th>
      <th>Action</th>
    </tr>
  </tfoot>
</table>

Answer 2

试试这个..

<a class="btn btn-success btn-xs" href="state_edit.php?id=<?php echo base64_encode($row['id']);?>" onclick="division_edit(<?php echo $row['id'] ?>)" id="division_edit"><i class="fa fa-pencil-square-o"></i> Edit</a>

如何在jquery中传递php参数

2 个答案: