Question

这可能是一个奇怪的问题，但我一直试图在codeigniter中反映db中的切换按钮，但它不起作用。在另一个页面中选择切换按钮。这是其在第一页上的选择方式：

<td>
    <label class="switch m-r-40">
    <input type="checkbox" class="switch-input" id="<?php echo $rows['applicationKey']?>" data-mail="<?php echo $rows['email']; ?>" data-confirm="<?php echo $rows['applicationId']; ?>" id="check" name="check" <?php echo $rows['status'] ? 'checked':''; ?> >
    <span class="switch-label" data-on="yes" data-off="no"></span>
    <span class="switch-handle"></span>
    </label>
  </td>

js

<script type="application/javascript">
$(function()
{




  $('input[type="checkbox"]').change(function(){
    //alert('shoe');
    //var id=$(this).attr('id');
    var appid = $(this).data('confirm');
    var value=$(this).prop('checked');

    if ($(this).prop('checked')){
        // do something
    $.post("<?php echo base_url(); ?>employer/view/accept", {check: appid, checked: 1}, function(result){
      //$("span").html(result);
    });
    }
  else
  {
    // 
    $.post("<?php echo base_url(); ?>employer/view/accept", {check: appid, checked: 0}, function(result){
      //$("span").html(result);
    });
  }
  });


});
</script>

我尝试在此处从db中提取数据并将其反映在其他页面中：

$page['applications'] = $this->db->query("SELECT a.*, b.postingName, b.`postDescription`, b.`state` FROM applications AS a INNER JOIN jobposting AS b ON a.`jobPostingKey`=b.`jobPostingKey` WHERE a.applicantKey = '$user'")->result_array();

html

<td>
  <label class="switch m-r-40">
  <input type="checkbox" class="switch-input" id="check" name="check" <?php echo $rows['status'] ? 'checked':''; ?> disabled="true">
  <span class="switch-label" data-on="yes" data-off="no"></span>
  <span class="switch-handle"></span>
  </label>
</td>

Answer 1

确保您具有正确的数据类型。字符串的真值与布尔真值不同！！！

    <?php
       $rows['status'] = false;
    ?>
   <td>
     <label class="switch m-r-40">
     <input type="checkbox" class="switch-input" id="check" name="check" <?php echo 
     $rows['status'] ? 'checked':''; ?> disabled="true">
      <span class="switch-label" data-on="yes" data-off="no"></span>
      <span class="switch-handle"></span>
     </label>
   </td>

   <hr />

   <?php
     $rows['status'] = "false";
    ?>
    <td>
      <label class="switch m-r-40">
      <input type="checkbox" class="switch-input" id="check" name="check" <?php echo 
      $rows['status'] ? 'checked':''; ?> disabled="true">
      <span class="switch-label" data-on="yes" data-off="no"></span>
      <span class="switch-handle"></span>
      </label>
    </td>

在另一页中反映数据库中存储的单选切换按钮的选择

1 个答案: