从jquery获取返回的值吗

从jquery获取返回的值吗

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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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08-29 07:04