How can I catch jQuery AJAX errors?

2019-02-05 20:40发布

问题:

When an AJAX request is submitted to a site, server-side errors are easily handled with the jQuery promise approach. .done(), .fail(), etc. However for some requests (e.g. to an invalid site or one that doesn't accept cross-origin requests), an exception occurs immediately as the call is made. Here's an example of one error in the console:

XMLHttpRequest cannot load http://someotherserver/api/blahblah. Origin http://localhost:52625 is not allowed by Access-Control-Allow-Origin.

Yes, I know about CORS...that's not the issue. What I'm actually doing is trying a web api call to test if the server IP/name is correct

I'm aware of the error option in the jQuery request syntax:

$.ajax({
    url: remoteURL,
    type: 'GET',
    error: function (err) {
        console.log("AJAX error in request: " + JSON.stringify(err, null, 2));
    }
}).etc.etc.

The error is handled here, but exceptions are still logged in the console. It seemed reasonable to wrap the above in a try-catch block, but that doesn't seem to help.

I've found this question, but the solution involves hacking the jQuery code. Surely there's a better way to catch these errors and not clog up the console logs??

回答1:

try this:

$.ajax({
    url: remoteURL,
    type: 'GET',
    error: function (err) {
        console.log("AJAX error in request: " + JSON.stringify(err, null, 2));
    }
}).always(function(jqXHR, textStatus) {
    if (textStatus != "success") {
        alert("Error: " + jqXHR.statusText);
    }
});

XHR Listener:

$.ajax({
    url: remoteURL,
    type: 'GET',
    xhr: function(){
        var xhr = new window.XMLHttpRequest();
        xhr.addEventListener("error", function(evt){
            alert("an error occured");
        }, false);
        xhr.addEventListener("abort", function(){
            alert("cancelled");
        }, false);

        return xhr;
    },
    error: function (err) {
        console.log("AJAX error in request: " + JSON.stringify(err, null, 2));
    }
});


回答2:

You can use web developer console in google chrome. Press F12. And use Networks tab for checking response, And for JavaSvript and jQuery and Ajax errors you can use Console tab. :)

Try this by adding to your ajax function :

error: function(XMLHttpRequest, textStatus, errorThrown) {
 alert(errorThrown);