php - 如何使用 php 响应 Json 中的错误消息？

Question

**var ajaxUrl='admin/includes/validation.php?ref=enquiry&name='+$("#name").val() + '&email1='+$("#email1").val() +  '&phone='+$("#phone").val()/*+ '&check='+a*/;
     $.ajax({

     type: "POST",
   url: ajaxUrl,
   data: "",
       dataType: 'json',
       success: function(msg){

    //alert(msg);
        if(msg!="")
           {
              $(".error").html("");
              $(".err").attr("style","");
               $.each(msg,function(key, value) { 

                    if(value!="") {
                    //alert(value);
                  $("#"+key).attr("style","border:2px solid #000000;");
                      $("#Enq_"+key).html(value);
                    }   


               });

           }

       else
       {



         $("#contact").submit();

       }

    }
    });**



Php file:

*if(!empty($check))
{

        $countema = "select id, email from contactus where email='".$email."' "; 
      $exe_query=$Obj->_query($countema);
$fetch_Mail= $Obj->_fetch_array($exe_query);

     $mail=($fetch_Mail['email']);  


      if(!empty($email) && ($email== $mail)){

      echo  'This Email id already subscribed '; 
      }
       /*$query=" insert into `news` set  `email`='".escapestr($email)."',`status`='1', createddate='now' ";  

$exe_query=$Obj->_query($query);

     }
     else
     {

    $query=" insert into `news` set  `email`='".escapestr($email)."',`status`='1', createddate='now' ";  

 $exe_query=$Obj->_query($query);
 }*/

Answer 1

json_encode在这里使用

$resp = array();
if(!empty($email) && ($email== $mail)){
  $resp['data'] = 'This Email id already subscribed '; 
 }
echo json_encode($resp);

Answer 2

您的JavaScript代码期望JSON，您返回的是text/plain. 字符串This Email id already subscribed不是有效JSON对象。有效JSON的是

{ "error" : true }

或类似的东西。在这种情况下，就知道如何告诉前端是否发生了错误。

你会像这样使用它：

success: function(msg){
    if (msg.error) { alert('An error has occurred! Everybody, get down!'); }