为什么js全局变量没有被内部ajax函数改变？ [英] Why the js global variable not changed by internal ajax function ?

问题描述

var rcode = 0;

//微博认证函数
        function CNY_CheckWeiDo() {            
            

            $.post("do_chy_anthorize.aspx", function (data) {
                var result = eval('(' + data + ')');
                if (result.code == undefined) {                    
                    isLogin = false;
                    rcode = 2;
                    //alert("rcode is 2 :"+rcode);
                }
                if (result.code == 1) {                    
                    isLogin = true;
                    rcode = 1;
                    //alert("rcode is 1 :" + rcode);
                }
                if (result.code == 0) {
                    isLogin = false;
                    rcode = 0;
                    //alert("rcode is 0 :" + rcode);
                    document.location.href = "http://localhost:8870/docheck.aspx" ;
                    
                }
                if (result.code == -2) {                   
                    isLogin = false;
                    rcode = -2;
                    //alert("rcode is -2 :" + rcode);
                    document.location.href =  "http://localhost:8870/docheck.aspx";
                    
                }

                if (result.code == -1) {                    
                    isLogin = false;
                    rcode = -1;
                    //alert("rcode is -1 :" + rcode);
                    document.location.href = "http://localhost:8870/docheck.aspx";                   
                }


            });

            alert("return code :" + rcode);            
            //rcode = 1;

            return rcode;
        }

作为代码，当我提醒'rcode'的值时，为什么'rcode'总是返回0 ？



但是在功能中，当我发出警报时，它的价值已经改变了。不能在ajax函数中改变全局变量吗？

谁可以说实话？谢谢

as the code ,when i alert the value of 'rcode', why the 'rcode' always returns 0 ?

but in the function , when i alert , it's value has changed . can't the global variable be changed in the ajax function ?
who can tell the truth ? thanks

推荐答案

.post（ do_chy_anthorize.aspx ， function （data）{
var result = eval （' （' + data + ' ）'）;
if （结果。 code == undefined ）{
isLogin = false ;
rcode = 2 ;
// alert（rcode is 2 ：+ rcode）;
}
if （result.code == 1 ）{
isLogin = true ;
rcode = 1 ;
// alert（rcode is 1：+ rcode）;
}
if （result.code == 0 ）{
isLogin = < span class =code-keyword> false ;
rcode = 0 ;
// alert（rcode is 0：+ rcode）;
document 。 location .href = < span class =code-string> http：// localhost：8870 / docheck.aspx;

}
if （result.code == -2）{
isLogin = 假;
rcode = -2;
// alert（rcode is -2：+ rcode）;
document 。 location .href = http：// localhost：8870 / docheck.aspx;

}

if （result.code == -1）{
isLogin = 假;
rcode = -1;
// alert（rcode is -1：+ rcode）;
document 。 location .href = http：// localhost：8870 / docheck.aspx;
}


}）;

alert（ return code： + rcode）;
// rcode = 1;

返回 rcode;
}

.post("do_chy_anthorize.aspx", function (data) { var result = eval('(' + data + ')'); if (result.code == undefined) { isLogin = false; rcode = 2; //alert("rcode is 2 :"+rcode); } if (result.code == 1) { isLogin = true; rcode = 1; //alert("rcode is 1 :" + rcode); } if (result.code == 0) { isLogin = false; rcode = 0; //alert("rcode is 0 :" + rcode); document.location.href = "http://localhost:8870/docheck.aspx" ; } if (result.code == -2) { isLogin = false; rcode = -2; //alert("rcode is -2 :" + rcode); document.location.href = "http://localhost:8870/docheck.aspx"; } if (result.code == -1) { isLogin = false; rcode = -1; //alert("rcode is -1 :" + rcode); document.location.href = "http://localhost:8870/docheck.aspx"; } }); alert("return code :" + rcode); //rcode = 1; return rcode; }

作为代码，当我提醒'rcode'的值时，为什么'rcode'总是返回0 ？



但是在功能中，当我发出警报时，它的价值已经改变了。不能在ajax函数中改变全局变量吗？

谁可以说实话？谢谢

as the code ,when i alert the value of 'rcode', why the 'rcode' always returns 0 ?

but in the function , when i alert , it's value has changed . can't the global variable be changed in the ajax function ?
who can tell the truth ? thanks

原因是ajax post需要时间，之前你从CNY_CheckWeiDo方法返回并且初始值为rcode

the reason is when ajax post take time and before that you are returning from the CNY_CheckWeiDo method and having initial value as rcode

replace alert（ 返回代码：+ rcode）;



with

setTimeout（function（）{alert（rcode）}，5000）;





这里5000毫秒（5秒）延迟运行警报。



实际上我在这里延迟5秒。因此，在AJAX帖子之后执行警报可以完成其工作。只是减少或增加时间延迟，看看发生了什么？

replace alert("return code :" + rcode);

with
setTimeout (function (){ alert (rcode)}, 5000);


here 5000 milliseconds( 5 Seconds) is delay to run alert.

Actually here i am giving 5 seconds delay. So that alert execute after AJAX post can finish its work. Just decrease or increase time delay and see whats happens?

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