Des*_*901 2 javascript ajax jquery
我正在尝试为 AJAX 调用实现一个 jQuery 默认 HTTP 错误处理程序。我已经为通用应用程序错误实现了默认错误处理程序,如下所示:
$(document).ajaxError(function(event, jqXHR, settings, thrownError) {
switch (jqXHR.status) {
case 403:
console.log('Forbidden, go away!');
break;
case 404:
console.log('Resource not found :O');
break;
case 422:
console.log('Unprocessable entity.');
break;
case 500:
console.log('Someone have to pay for this!');
break;
default:
console.log('Anyone knows what is going on here?');
}
});
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现在我想要实现的是在单个 AJAX 调用中覆盖特定的状态代码。例如
$.ajax({
...
error: function(jqXHR, textStatus, errorThrown) {
if(jqXHR.status === 404){
console.log('The requested unicorn is actually missing');
}
}
})
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现在,如果我如上所示执行,将显示两条消息
>The requested unicorn is actually missing
>Resource not found :O
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虽然我只是想获取The requested unicorn is actually missing消息。
global: false在单个 AJAX 调用设置中设置标志意味着忽略该 AJAX 调用中的所有全局 AJAX 函数,不幸的是,这对我的应用程序来说不是一个可行的解决方案。
任何的想法?谢谢!
您可以执行以下操作
$.ajax({
handleErrors: [404],
...
error: function(jqXHR, textStatus, errorThrown) {
if(jqXHR.status === 404){
console.log('The requested unicorn is actually missing');
}
}
});
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然后在您的全局处理程序中检查是否处理了错误
$(document).ajaxError(function(event, jqXHR, settings, thrownError) {
if (settings.handleErrors && settings.handleErrors.indexOf(jqXHR.status) !== -1) {
return;
}
switch (jqXHR.status) {
case 403:
console.log('Forbidden, go away!');
break;
case 404:
console.log('Resource not found :O');
break;
case 422:
console.log('Unprocessable entity.');
break;
case 500:
console.log('Someone have to pay for this!');
break;
default:
console.log('Anyone knows what is going on here?');
}
});
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