我是Laravel新手,因此这是Interview Schedule表单的脚本,该脚本基于使用onChange选择的Program出现:

function changeProgram(input){
      var intake_id = $("[name='intake_id']").val();
      var campus_input = $(input);
      var campus_id = $('[name="program[' + campus_input.data('id') + '][campus_id]"]').val();
      $('[name="program[' + campus_input.data('id') + '][program_id]"').empty();

      if( campus_id != '' ){

        $.ajax({
          type: 'GET',
          dataType: 'json',
          url: '{{ url("campus_program/programs") }}',
          data: {
            intake_id: intake_id,
            campus_id: campus_id
          },
          success: function (data) {
            var select_data = [];

            $.each(data, function(i, object) {
              select_data.push({id:i, text:object});
            });

            $('[name="program[' + campus_input.data('id') + '][program_id]"').select2({
              data: select_data,
              placeholder: "Please Select",
              allowClear: true
            });
            $('[name="program[' + campus_input.data('id') + '][program_id]"]').val('').trigger('change');

            @if(!empty($intake_select))
            if (intake_id == {{ $intake_select }})
            {
                @for ($i = 0; $i < $form_setting['total_program']['show']; $i++)
                @if (!empty($student_program[$i]->program_id))
                <?php $program_id = $student_program[$i]->program_id; ?>
                $('[name="program['+{{$i}}+'][program_id]"]').val({{$program_id}}).trigger('change');
                @else
                $('[name="program['+{{$i}}+'][program_id]"]').val('').trigger('change');
                @endif
                @endfor
            }
            @endif


          }
        });

            $('[name="program[' + campus_input.data('id') + '][program_id]"]').attr('onchange','getInterviewSchedule(this)');
      } else {
        $('[name="program[' + campus_input.data('id') + '][program_id]"').select2({
          data: [],
          placeholder: "Please Select",
          allowClear: true
        });
      }
}


我希望仅在有数据时显示Interview Schedule表单。

如果下拉列表中没有表单,如何隐藏表单?提前致谢。

javascript - Laravel使用onchange隐藏表格-LMLPHP

也是刀片的一部分:

<div class="form-group">
      <label class="col-lg-2 control-label">{{ msg('lbl_program') }}</label>
      <div class="col-lg-6">
        {!! Form::dropdown('program['.$i.'][program_id]', $program, @$student_program[$i]->program_id, [
          'data-id' => $i,
          'class' => 'select2-form program-input'
        ]) !!}
      </div>
    </div>

    <div class="form-group interview_schedule-div{{$i}} hide">
      <label class="col-lg-2 control-label">{{ msg('lbl_interview_schedule') }}</label>
      <div class="col-lg-6">
          {!! Form::dropdown('program['.$i.'][interview_schedule_id]', ['' => ''], @$student_program[$i]->interview_schedule_id, [
            'data-id' => $i,
            'class' => 'select2-form program-input'
          ]) !!}
      </div>
    </div>

最佳答案

默认情况下,隐藏表单,为您的表单显示提供类和样式:
上课下来尝试一下

$(document).on('change','.dropdown-class-name', funictiion(){
    if($(this).val() != "")
{
$('.form-class').show();
}
else
{
$('.form-class').hide();
}
});


谢谢

09-19 08:25