本文介绍了从jquery获取返回的值吗?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
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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