Is there a way in JavaScript to simulate POST request errors for testing purposes? I want to catch different responses other than the standard 200 status code.
$.ajax({
url: url,
type: 'POST',
// additional arguments
}).done(function (result) {
// Successful response
}).fail(function(result) {
// Handling errors here
});
I'm looking to fake various status codes as part of the response.
Extracted from the jQuery documentary page (Source):
jqXHR.fail(function(jqXHR, textStatus, errorThrown) {});
Providing an alternative approach to the error callback option, the .fail() method supersedes the deprecated .error() method. See deferred.fail() for detailed implementation guidelines.
Therefore, if you conclude your response of your url with a HTTP status code other than 200, it will result in '.fail'.
For instance, by adding a header to your PHP file as shown below, it should function correctly.
header("HTTP/1.0 404 Not Found");
An alternate technique to the header() function is the http_response_code(). Feel free to refer to this link for further information: http_response_code()
I trust that this explanation is beneficial.
A straightforward approach is to simply make a call to a URL that returns the desired code, like this:
$.ajax({
url: 'https://httpstat.us/400',
type: 'POST',
// additional arguments
}).done(function (result) {
console.log("Success");
}).fail(function(result) {
console.log("Fail");
console.log("Result", result);
});<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>Alternatively, as mentioned in this answer, you can override the $.ajax() function to always fail like this:
function ajax_fail(jqXHR, textStatus, errorThrown) {
return function (params) {
var deferred = $.Deferred().reject(jqXHR, textStatus, errorThrown);
return deferred.promise();
}
}
$.ajax = ajax_fail();
$.ajax({
url: '',
type: 'POST'
}).done(function (result) {
console.log("Success");
}).fail(function (result) {
console.log("Error");
});<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>The parameters passed to the ajax_fail() function will be included in your .fail() callback. You can customize the returned values as needed.
Keep in mind: This method should not be used in production or if you require other $.ajax calls to succeed. Every instance of $.ajax() will fail when implemented this way.
If you only want to apply this to a single AJAX call, you can follow these steps:
var ajaxFunc = $.ajax;
// utilize the above code
// restore original $.ajax function
$.ajax = ajaxFunc;
Using the autoResponder feature in Fiddler is more efficient than relying solely on JavaScript. By intercepting responses with a 200 code, you can modify the response and effectively test your error handling process.
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