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);
}
}
}
})
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。