从jquery获取返回的值吗？ [英] get returned values from jquery get?

问题描述

我让jquery调用get和getJSON，但不能访问回调函数以外的返回值。如何访问返回的数据？

  var firstid; 
 var nextid; 
 $ b $ .get（callUrl，function（data）{//调用添加节点
 var n = data.indexOf（id-）; 
 var m = data .indexOf（id-）; 
 firstid = data.substr（n + 3，m  - （n + 3））; 
 nextid = data.substr（m + 3）; 
 
 alert（firstid：+ firstid）; //返回正确的值
}）; 
 
 alert（firstid：+ firstid）; //返回undefined for firstid 
  

如何在函数外获得firstid？

解决方案

所有AJAX调用都是异步的

>需要使用回调。除此之外的任何内容都会返回 undefined 。

  $。get（callUrl，函数（数据）{//调用添加节点
 var n = data.indexOf（id-）; 
 var m = data.indexOf（id-）; 
 firstid = data.substr（n + 3，m  - （n + 3））; 
 nextid = data.substr（m + 3）; 
 
 doSomethingWithFirst（firstid）; 
 }）; 
 
函数doSomethingWithFirst（f）{
 //现在做某事
} 
  

I am making jquery calls to get and getJSON, but cannot access returned values outside of callback function. How can I access the returned data?

        var firstid;
        var nextid;

    $.get(callUrl, function(data) {  // call to add node
         var n = data.indexOf("id-");
         var m = data.indexOf("id-");
         firstid = data.substr(n+3,m - (n+3));
         nextid = data.substr(m+3);

             alert("firstid:" + firstid);  // returns correct value
    });

        alert("firstid:" + firstid);  // returns undefined for firstid

how can I get firstid outside the function?

解决方案

ALL AJAX CALLS ARE ASYNCHRONOUS

SO you need to use callbacks. anything outside that will return undefined.

$.get(callUrl, function(data) {  // call to add node
     var n = data.indexOf("id-");
     var m = data.indexOf("id-");
     firstid = data.substr(n+3,m - (n+3));
     nextid = data.substr(m+3);

     doSomethingWithFirst(firstid);
});

function doSomethingWithFirst(f)   {
     //NOW do something
}

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