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

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$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

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< 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 >

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$.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);     }    }    }   })

以上这篇laravel Validator ajax返回错误信息的方法就是小编分享给大家的全部内容了，希望能给大家一个参考，也希望大家多多支持服务器之家。

原文链接：https://blog.csdn.net/qq_39864544/article/details/89239117