How do I catch an Ajax query post error?

2019-01-03 12:37发布

I would like to catch the error and show the appropriate message if the Ajax request fails.

My code is like the following, but I could not manage to catch the failing Ajax request.

function getAjaxData(id)
{
     $.post("status.ajax.php", {deviceId : id}, function(data){

        var tab1;

        if (data.length>0) {
            tab1 = data;
        }
        else {
            tab1 = "Error in Ajax";
        }

        return tab1;
    });
}

I found out that, "Error in Ajax" is never executed when the Ajax request failed.

How do I handle the Ajax error and show the appropriate message if it fails?

6条回答
叼着烟拽天下
2楼-- · 2019-01-03 13:18

Since jQuery 1.5 you can use the deferred objects mechanism:

$.post('some.php', {name: 'John'})
    .done(function(msg){  })
    .fail(function(xhr, status, error) {
        // error handling
    });

Another way is using .ajax:

$.ajax({
  type: "POST",
  url: "some.php",
  data: "name=John&location=Boston",
  success: function(msg){
        alert( "Data Saved: " + msg );
  },
  error: function(XMLHttpRequest, textStatus, errorThrown) {
     alert("some error");
  }
});
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戒情不戒烟
3楼-- · 2019-01-03 13:19

jQuery 1.5 added deferred objects that handle this nicely. Simply call $.post and attach any handlers you'd like after the call. Deferred objects even allow you to attach multiple success and error handlers.

Example:

$.post('status.ajax.php', {deviceId: id})
    .done( function(msg) { ... } )
    .fail( function(xhr, textStatus, errorThrown) {
        alert(xhr.responseText);
    });

Prior to jQuery 1.8, the function done was called success and fail was called error.

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Root(大扎)
4楼-- · 2019-01-03 13:30

A simple way is to implement ajaxError:

Whenever an Ajax request completes with an error, jQuery triggers the ajaxError event. Any and all handlers that have been registered with the .ajaxError() method are executed at this time.

For example:

$('.log').ajaxError(function() {
  $(this).text('Triggered ajaxError handler.');
});

I would suggest reading the ajaxError documentation. It does more than the simple use-case demonstrated above - mainly its callback accepts a number of parameters:

$('.log').ajaxError(function(e, xhr, settings, exception) {
  if (settings.url == 'ajax/missing.html') {
    $(this).text('Triggered ajaxError handler.');
  }
});
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唯我独甜
5楼-- · 2019-01-03 13:31
$.post('someUri', { }, 
  function(data){ doSomeStuff })
 .fail(function(error) { alert(error.responseJSON) });
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时光不老,我们不散
6楼-- · 2019-01-03 13:35
$.ajax({
  type: 'POST',
  url: 'status.ajax.php',
  data: {
     deviceId: id
  },
  success: function(data){
     // your code from above
  },
  error: function(xhr, textStatus, error){
      console.log(xhr.statusText);
      console.log(textStatus);
      console.log(error);
  }
});
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▲ chillily
7楼-- · 2019-01-03 13:39

I fixed this problem by declaring datatype: 'json' in my jQuery ajax call.

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