2010-05-17 67 views
0

要从$ .ajax调用中返回一个错误,有一个比在ajax.php文件中回显错误然后修剪它更好的方法!jquery ajax sucess - 可能返回JSON而不是字符串?

这似乎很笨拙和稳健:

success: function(e){ 
    var e = trim(e); 
    if(e == 'SUCCESS') 
     {alert('your password has been changed!');} 
    if(e == 'ERROR1') 
     {alert('please fill in all inputs!');} 
    if(e == 'ERROR2') 
     {alert('password incorrect!');} 
    if(e == 'ERROR3') 
     {alert('change failed!');} 
} 

我应该怎么来代替干什么?

回答

2

返回JSON:

{ success: false, errorMessage: 'please fill in all inputs!' } 

然后:

success: function(e) { 
    if(e.success) { 
     alert('your password has been changed!'); 
    } 
    else { 
     alert(e.errorMessage); 
    } 
} 
+0

啊凉,这样我就可以回声出/ json_encode在一些脚本标签上ajax.php文件? – Haroldo 2010-05-17 16:46:13

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