变量无法按预期方式工作

Question

I have a function:

function name(type)
{
    $.i="";
    $.post("name.php", type, function(data, status){
        $(".pic").attr("src", data);
        $.i=data;

    })
    alert($.i); //this is not working
}

In the above code the alert is displaying empty alert box. But when i have

function name(type)
{
    $.i="";
    $.post("name.php", type, function(data, status){
        $(".pic").attr("src", data);
        $.i=data;
        alert($.i); //now this is working
    })

}

I want to return the value returned by name.php file . name.php contains echo "string" . It is for testing only. In the second given code string is displaying but in first it is not working.

What can i do to return the value from the function and assign the received value to variable declared outside $.post() or in the function.

Thanks in advance.

Answer 1

The async call back of post is call after the execution of post. You can call a function from post to perform operation on value return in callback.

function name(type)
{
    $.i="";
    $.post("name.php", type, function(data, status){
        $(".pic").attr("src", data);
        $.i=data;
        //alert($.i); //now this is working
        callToFunctionPassingData(data);
    })    
}

You can try using deffered execution this post will guid you how to achieve it.