jQuery $.get or $.post to catch page load error (e

问题:

How do you catch Server Error or 404 page not found, when you use $.get or $.post ?

For example:

$.post("/myhandler", { value: 1 }, function(data) {
  alert(data);
});

That will do absolutely nothing if there is a Server Error loading "/myhandler", or if it is not found.

How do you make it notify you if there is an error?

回答1:

use error handler on $.ajax()

$.ajax({
    url: "/myhandler", 
    data: {value: 1},
    type: 'post',
    error: function(XMLHttpRequest, textStatus, errorThrown){
        alert('status:' + XMLHttpRequest.status + ', status text: ' + XMLHttpRequest.statusText);
    },
    success: function(data){}
});

demo

回答2:

you could do

$.post("/myhandler", { value: 1 }, function(data) {
  alert(data);
}).fail(function(){ 
  // Handle error here
});

fail will be called if theres an error

回答3:

The other answers are nice and all, but there's alternative solutions, namely .ajaxSetup, .ajaxError and other Ajax event handlers (check the ajaxSetup doc page for more info on the rest).

For example, with .ajaxError you can setup a global handler of all your ajax errors from .post, .get, .getJSON and .ajax for a specific set of elements.

$(selector).ajaxError(function(event, xhr, ajaxOptions, errorThrown) {
    // handle ajax error here
});

回答4:

Use $.ajax instead and use the error callback.

http://api.jquery.com/jQuery.ajax/

回答5:

Try using $.ajaxSetup() , stausCode option

$.ajaxSetup({
   statusCode : {
     // called on `$.get()` , `$.post()`, `$.ajax()`
     // when response status code is `404`
     404 : function (jqxhr, textStatus, errorThrown) {
             console.log(textStatus, errorThrown);
           }
     }
 });

// $.get("/path/to/url/");
// $.post("/path/to/url/");
// $.ajax({url:"/path/to/url/"})