Question

The spec for the jQuery ajax.error function is:

error(XMLHttpRequest, textStatus, errorThrown)Function

I’m trying to catch the error and display the textStatus, but I can’t figure out how to specify only the textStatus without having to put in a variable name for XMLHttpRequest and errorThrown. My code currently looks like this:

$.ajax({
    type: "POST",
    contentType: "application/json; charset=utf-8",
    url: hbAddressValidation.webServiceUrl,
    data: this.jsonRequest,
    dataType: "json",
    timeout: 5,
    success: function (msgd) {
         //...
    },
    error: function (a,textStatus,b) {
        $("#txtAjaxError").val("There was an error in the AJAX call: " 
                                + textStatus);
    }
});

You can see in my code that I’m putting variables a and b as placeholders for the first and last variables in the error function. I know that in my success function, I’m only providing one parameter and it works fine, but in that case data is the first parameter. In the case of error, textStatus is the second parameter, but that’s the only one I want to specify. Is this possible?

Answer 1

No. The closest that you can get is:

error: function (a,textStatus) {
    $("#txtAjaxError").val("There was an error in the AJAX call: " 
                            + textStatus);
}

Only the last parameters can be omitted