jQuery如何用json建立链接？，错误

发布于 2024-10-31 19:21:12 字数 761 浏览 0 评论 0原文

我有这个：

    $.get('xxx.php', { username: userName },
    function(data){
        var get_back = data;
        alert(get_back);
});

这将返回 get_back=12345

并且我正在尝试构建这个：

url: "http://www.test.com/users/" + get_back,

结果为 http://www.test.com/users/12345

由于某种原因它不想工作。如果我在链接中硬编码 12345 ，它将起作用。我还尝试过 url: "http://www.test.com/users/" + get_back + "", 和 url: 'http://www.test.com /users/' + get_back，

有什么想法吗？

编辑：

$.ajax({
type: "POST",
data: JSON.stringify(formData),
dataType: "json",
url: "http://www.test.com/users/" + get_back + "",
success: function(t){    alert(t);  }
});

原文

i have this:

    $.get('xxx.php', { username: userName },
    function(data){
        var get_back = data;
        alert(get_back);
});

this will return get_back=12345

and i'm trying to build this:

url: "http://www.test.com/users/" + get_back,

the result to be http://www.test.com/users/12345

for some reason it doesn't want to work. If i hard-code the 12345 in the link it will work. i've also tryed url: "http://www.test.com/users/" + get_back + "", and url: 'http://www.test.com/users/' + get_back,

any ideas?

edit:

$.ajax({
type: "POST",
data: JSON.stringify(formData),
dataType: "json",
url: "http://www.test.com/users/" + get_back + "",
success: function(t){    alert(t);  }
});

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埋葬我深情 2024-11-07 19:21:12

因为 get_back 可能在其他所有内容的范围内。您可以在 $.get 调用之前调用 get_back ，它将是全局的，或者您可以将其全部放入对象中，如下所示：

var get = {
   get_back: null,
   init: function(username){
      var self = this;
      $.get('xxx.php', { username: userName },
          function(data){
              self.get_back = data;
              self.runAjax(); //run the ajax when get_back is instantiated
          }
      });

   },
   runAjax: function(){
      var self = this;
      $.ajax({
         type: "POST",
         data: JSON.stringify(formData),
         dataType: "json",
         url: "http://www.test.com/users/" + self.get_back + "",
         success: function(t){    alert(t);  }
         });
   }
}

//to use it:
get.init(username);
//to use get_back
alert(get.get_back);

使用 AJAX 更新

你可能必须以某种方式将formdata添加到对象中，我不知道它是在哪里创建的

that because get_back might be in the scope of everything else. either you can call get_back before the $.get call and it will be global or you can put it all into on object like this:

var get = {
   get_back: null,
   init: function(username){
      var self = this;
      $.get('xxx.php', { username: userName },
          function(data){
              self.get_back = data;
              self.runAjax(); //run the ajax when get_back is instantiated
          }
      });

   },
   runAjax: function(){
      var self = this;
      $.ajax({
         type: "POST",
         data: JSON.stringify(formData),
         dataType: "json",
         url: "http://www.test.com/users/" + self.get_back + "",
         success: function(t){    alert(t);  }
         });
   }
}

//to use it:
get.init(username);
//to use get_back
alert(get.get_back);

UPDATED WITH AJAX

you might have to add formdata to the object somehow, i dont know where that was created

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