编程学习网 > 数据库 > laravel Validator ajax返回错误信息的方法
2019
09-29

laravel Validator ajax返回错误信息的方法

将return back()->withErrors($validator->errors())->withInput();替换为下面的代码

$validator = \Validator::make($request->input(), $rules, $messages);
      if ($validator->fails()) {
        //将返回错误循环组装成字符串
        $arr = [];
        foreach ($validator->getMessageBag()->toArray() as $k=>$error){
          array_push($arr, "<li>".$error[0]."</li>");
        }
        $str = implode(' ', $arr);
        return \Response::json([
          'success' => false,
          'errors' => $str
        ]);
      }
然后直接在ajax输出相应的错误就行了

<div class="alert alert-danger alert-dismissible fade in" style="display: none;" id="yc">
 <button type="button" class="close" data-dismiss="alert" aria-label="Close"><span aria-hidden="true">×</span></button>
 <ul class="margin-bottom-none padding-left-lg" id="li1">
 </ul>
</div>
$.ajax({
  method: 'post',
  data: form,
  dataType:'json',
  url: $('#url').val(),
  success: function (data) {
  //console.log(data);
  if (data.code == 0){
   $('#textarea').html(JSON.stringify(data.resp, null, "\t")).css("display","block");
  } else if(data.code == 1) {
   layer.msg("失败"+data.msg,{icon:2})
  }else if (data.success == false) {
   if (data.errors){
   $('#yc').css('display','block');
   $('#li1').html(data.errors);
   }
  }
  }
 })

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