Jquery Ajax返回数组——如何处理Javascript。

时间:2022-08-25 18:13:42

If someone could assist me please. I'm doing a jquery Ajax post, for some reason the Json object isn't working so just returning a php array instead

如果有人能帮助我,请。我正在做jquery Ajax贴子,由于某种原因Json对象不能工作,所以只返回一个php数组

$.post
(  
  "classes/RegisterUser.php",  
  $("#frmRegistration").serialize(),  
  function(data)
  {  
   alert(data);  
  } 
);

The data is returned to Javascript 100% as

数据返回到Javascript 100%。

array
(
   [key]=>value
   [Name] => SomeOneName
   [Surname] => SomeOneSurName
)

How would i go about getting the value of Surname in Javascript?

在Javascript中如何获得姓氏的价值呢?

Thanks for your assistance? Regards

谢谢你的援助?问候

2 个解决方案

#1


3  

Expanding on The MYYN's answer, after you get your script to return JSON, you must specify that you're receiving JSON and act accordingly. You can do this with .ajax():

扩展MYYN的答案,在得到返回JSON的脚本之后,必须指定正在接收JSON并相应地执行。您可以使用.ajax():

$.ajax({
    type: 'post',
    url: 'classes/RegisterUser.php',
    data: $("#frmRegistration").serialize(),
    dataType: 'json',
    success: function(obj) {
        // This alerts SomeOneSurName
        alert(obj.Surname);
    }
});

#2


0  

Maybe your PHP script should return json (right now it seem to return something like var_dump($some_ary);. A proper way to do this is via php's json_encode.

也许您的PHP脚本应该返回json(现在它似乎返回了var_dump($some_ary);正确的方法是通过php的json_encode。

#1


3  

Expanding on The MYYN's answer, after you get your script to return JSON, you must specify that you're receiving JSON and act accordingly. You can do this with .ajax():

扩展MYYN的答案,在得到返回JSON的脚本之后,必须指定正在接收JSON并相应地执行。您可以使用.ajax():

$.ajax({
    type: 'post',
    url: 'classes/RegisterUser.php',
    data: $("#frmRegistration").serialize(),
    dataType: 'json',
    success: function(obj) {
        // This alerts SomeOneSurName
        alert(obj.Surname);
    }
});

#2


0  

Maybe your PHP script should return json (right now it seem to return something like var_dump($some_ary);. A proper way to do this is via php's json_encode.

也许您的PHP脚本应该返回json(现在它似乎返回了var_dump($some_ary);正确的方法是通过php的json_encode。