Question

我有以下伪代码例如：我想使用相同的ajax脚本来选择每条记录的phone_type和方法，我尝试使用.attr（）

<script>
   $('#phone_type').on('change', function (e) {
     console.log(e);
   
     var phone_type = e.target.value;
   
     //ajax
     $.get('../get_conversion_method_by_phone_type?phone_type=' + phone_type, function (data) {
       //success data
       $('#method').empty();
       $('#method').append('<option value="0" selected>Choose Method</option>');
       $.each(data, function (index, conMethObj) {
         $('#method').append('<option value="' + conMethObj.id + '">' + conMethObj.method + '</option>');
       });
     });      
   });
</script>

foreach($foo as $bar){
<tr>
   <select id="phone_type" name="phone_type">
      <option>Choose</option>
   </select>
   <select id="method" name="method">
     
   </select>
</tr>
}

Answer 1

我希望我能正确理解你想做什么，试试这段代码

<script>
   $('.phone_type').on('change', function (e) {
       console.log(e);
       var $self = $(this);
       var phone_type = e.target.value;
   
       //ajax
       $.get('../get_conversion_method_by_phone_type?phone_type=' + phone_type, function (data) {
           //success data
           $method = $self.next('.method');
           $method.empty();
           $method.append('<option value="0" selected>Choose Method</option>');
           $.each(data, function (index, conMethObj) {
              $method.append('<option value="' + conMethObj.id + '">' + conMethObj.method + '</option>');
   
           });
       });
   
   });
</script>

和html

foreach($foo as $bar){
<tr>
   <select class="phone_type" name="phone_type">
      <option>Choose</option>
   </select>
   <select class="method" name="method">
     
   </select>
</tr>
}

每行jQuery AJAX多选

1 个答案: