我想要那样的东西=> $(本)('#标识&#39)。CSS。('显示''直列&#39);

时间:2017-05-18 16:35:16

标签: javascript

我一直试图解决这个问题很长一段时间。 这就是我想要做的事情。

单击缺席选项时,应弹出一个弹出框。 我发生的是,同样的弹出框显示我想要与表名相关的弹出框。

以下是我的代码:

    while($row_students = mysqli_fetch_assoc($result_students))
    {
    $student_name = $row_students['first_name'] .' '. $row_students['middle_name']  .' '. $row_students['last_name'];
    $student_id = $row_students['id'];
    echo 
    "<tr> 
    <td>{$student_id}</td>
    <td>{$student_name}</td>
    <td>{$row_students['batch_id']}</td>
    <td>
    <select class='form-control' name='reason{$student_id}' id='selection' >
    <option value='NA' style='display:none';>Type of Absence</option>
    <option value='absent'>Absent</option>
    <option value='late'>Late</option>
    <option value='sick'>Sick</option>
    </select>
    </td>
    <td>

    <select class='form-control' name='popupreason{$student_id}' 
    id='popup-text' style='display:none;'>

    <option value='NA' style='display:none';>Please Select Reason</option>
    <option value='death'>حالة وفاه</option>
    <option value='tardy'>لم يتم اللحاق بحافلة المدرسة </option>
    <option value='force'>مانع خارج عن الإرادة </option>
    <option value='accident'>حادث في الطريق  </option>
    <option value='travel'>السفر </option>
    <option value='expelled'>رفد مؤقت  </option>
    <option value='financial'>مانع مادي</option>
    <option value='unknown'>غير معروف</option>                    
    </select>     
    </div>
    </td>
    </tr>";   
    } 
    }
    ?>
And for Jquery:

    <script>


  $("select[id='selection']").change(function(){
    if ($(this).val() === 'absent'){
         $('#popup-text').css('display','inline');
        }
      });



 /*I BELIEVE THE ISSUE IS HERE IN THE ABOVE AS I NEED TO USE SOMETHING LIKE 


$(this).('#popup-text').css('display','inline');
  BUT I COULDN'T GET THE SYNTAX CORRECT  */

  $('#popup-text').change(function(){
    $("#popup-text").hide();
  })

   </script>

2 个答案:

答案 0 :(得分:0)

我认为这样可行:

$("select[id='selection']").change(function() {
  var studentId = $(this).attr('name').match(/^reason(\d+)$/i)[1];

  $('#popup-text[name=popupreason' + studentId + ']')
    .css('display','inline')
    .focus()
  ;
});

或者只是:

$("select[id='selection']").change(function(){
  $("#popup-text").hide(); // hide others

  if ($(this).val() === 'absent'){
      $('[name=popup' + $(this).attr('name') + ']')
        .css('display','inline')
      ;
  }
});

答案 1 :(得分:0)

每次迭代都有相同的ID,因为每页ID必须是唯一的 - 它无效。使用类。

比你能做的

var p = 'my family ‍‍‍‍❤️‍‍ here';
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var res = p.split(rx).filter(Boolean);
document.body.innerHTML = res;