Ajax简单的例子不起作用

Question

I haven't used ajax in a while and now I can't even get this simple program to work, am I doing something wrong ?

        <script type = "text/javascript" src = "jquery-1.9.1.min.js"></script>
    <script type = "text/javascript">
        alert("");
        var count = 0;
        $.ajax({
               url: 'get.php',
               dataType: 'json',
               success: function () 
               {       
                alert("");
               } 
        });
        alert("");
    </script>

<?php
echo "yay";
?>

Thanks

Answer 1

Setting the dataType as "json" means the response from get.php is parsed as JSON. If it's not valid JSON or the response is empty, the request will fail.

If the URL is incorrect (can't be found...HTTP 404 error), then the request will fail.

The default type of request is "GET", so if get.php doesn't allow "GET" (for some reason), it will return an HTTP error, and the request will fail.

If there's an error on the server, it will likely return an HTTP 500 error, and the request will fail.

Something to help debug would be to add the error option to the $.ajax call and see if that's called. Instead, I use the .fail() method...it does the same thing.

Of course, the more direct way of debugging is opening your browser's console and viewing the AJAX request. It should show multiple details about it, that can help you determine any problems.

It might seem as if the AJAX request was never executed/sent, because you don't see the alert in the middle. Well, just because the request wasn't successful, doesn't mean it was skipped. There are plenty of reasons (I named several above) why the request may fail. And the .fail() method will help you determine the cause.

Also, the universal convention for handling deferred objects in jQuery is to use the done and fail methods, so that is an option. Of course, $.ajax has specific options you can specify ( success , error , and complete - which is for something else), so that is also an option. You can also use special methods ( .success() , .error() , .complete() ) that are part of the object returned from $.ajax , but those are deprecated as of version 1.8 - take a look at the .ajax docs towards the bottom - http://api.jquery.com/jQuery.ajax/ . But here's how I'd set it up, which shouldn't be any different from yours, but does catch errors:

$.ajax({
    url: 'get.php',
    dataType: 'json'
}).done(data) {
    console.log("successful response");
}).fail(jqXHR, textStatus, errorThrown) {
    console.log("error: " + textStatus + ", " + errorThrown);
});

Answer 2

Are you sure jQuery has loaded by the time your ajax call is hit?

Try wrapping it in a doc ready to make sure you are actually loaded

$(document).ready(function () {
    alert("");
    var count = 0;
    $.ajax({
        url: 'get.php',
        dataType: 'json',
        }).alwaysfunction () {
           alert("call was hit");
        });
    });

});

Answer 3

If there's a problem with get.php, your code will not indicate any problems, jquery.ajax() just swallow the error.

Even before adding the .failure callback routine to your ajax call, you should be able to check the error log on web server executing get.php, any errors related to executing get.php will be logged.