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Why is my AJAX-call failing in Google Chrome?

开发者 https://www.devze.com 2023-04-07 12:53 出处:网络
I am curious as to why my AJAX-call is failing in Google Chrome, it works perfectly fine in Firefox. Before anyone asks, no I\'m not using JQuery because I need to have access to readyState == 3 which

I am curious as to why my AJAX-call is failing in Google Chrome, it works perfectly fine in Firefox. Before anyone asks, no I'm not using JQuery because I need to have access to readyState == 3 which JQuery doesn't seem to have.

My script currently looks like this(with large unneccessary parts stripped out):

function fetch()
{
    main = new XMLHttpRequest();
    main.open("GET", "<?php echo anchor("thescript"); ?>", true);

    var lastResponse = '';
    var statusString = 'Step 1(of 3), please wait... ';

    main.onreadystatechange = function()
    {
        if( main开发者_如何学运维.readyState == 1 ) 
        {
            alert('Fetch!');
            $("#ajax-status").html( statusString );
        }

        // If there's been an update
        if( main.readyState == 3 )
        {
        }
        if( main.readyState == 4 )
        {
        }
    };
    main.send(null);
}

It works perfectly in Firefox but in Chrome it doesn't even alert anything so it doesn't even get into readyState 1(which is when you send it) -- that seems rather odd..

Any ideas??


As noted above:

does it any difference to put the .open() after you set .onreadystatechange

And yes Ein~, it actually does a difference! The readystate's are now working properly, I think! I recieve the alert when it sends the request and I also tried an alert in readyState == 3 and it alerts that too. However, the response seems to be empty for some reason

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