Ajax导致alert（）[重复]

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                        <a href="/questions/14220321/how-do-i-return-the-response-from-an-asynchronous-call" dir="ltr">How do I return the response from an asynchronous call?</a>
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                                (38 answers)
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            <div class="grid--cell mb0 mt8">Closed <span title="2014-02-20 11:58:24Z" class="relativetime">6 years ago</span>.</div>
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How can I get imageX show it's value when used outside of $.ajax()?

var imageX;
$.ajax({
    type: 'GET',
    url: 'php/my1.php',
    dataType: 'json',
    async: 'false',
    success: function(response){
        imageX = response[0].studentName,
        groupX = response[0].subjectId;
    }
});
alert(imageX);

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Answer 1

Change

async: 'false',

to

async: false,

'false' is string which is cast to true by Jquery(JS).

Jquery Ajax Documentation says async accept boolean value.

async (default: true): Type: Boolean By default, all requests are sent asynchronously (i.e. this is set to true by default). If you need synchronous requests, set this option to false.

https://api.jquery.com/jQuery.ajax/

async:false will cause the jQuery.ajax() call to block until it returns. instead of this:

function your ajax_calling_function
    call your ajax 
    callback with results