处理 JSON (jQuery) 中的 500 个错误
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Handle 500 errors in JSON (jQuery)
提问by Fraser
This JSON request:
这个 JSON 请求：
$.ajax({
url:jSONurl+'?orderID='+thisOrderID+'&variationID='+thisVariationID+'&quantity='+thisQuantity+'&callback=?',
async: false,
type: 'POST',
dataType: 'json',
success: function(data) {
if (data.response == 'success'){
//show the tick. allow the booking to go through
$('#loadingSML'+thisVariationID).hide();
$('#tick'+thisVariationID).show();
}else{
//show the cross. Do not allow the booking to be made
$('#loadingSML'+thisVariationID).hide();
$('#cross'+thisVariationID).hide();
$('#unableToReserveError').slideDown();
//disable the form
$('#OrderForm_OrderForm input').attr('disabled','disabled');
}
},
error: function(data){
alert('error');
}
})
In certain circumstances will bring back a 500 error in the form of:
在某些情况下会以以下形式返回 500 错误：
jQuery17205593111887289146_1338951277057({"message":"Availability exhausted","status":500});
This however is still useful to me and I need to be able to be able to handle this correctly.
然而，这对我来说仍然有用，我需要能够正确处理这个问题。
For some reason though, when this 500 error is returned, my error function is not called and I just get a "NetworkError: 500 Internal Server Error" error in firebug.
但是由于某种原因，当返回这个 500 错误时，我的错误函数没有被调用，我只是在 firebug 中得到一个“NetworkError: 500 Internal Server Error”错误。
How can I handle this?
我该如何处理？
回答by Prasenjit Kumar Nag
Did you try statuscodecallback like
你试过statuscode回调吗
$.ajax({
statusCode: {
500: function() {
alert("Script exhausted");
}
}
});
回答by emispowder
Check out the jqXHR Objectdocs. You could use the fail method to capture any errors.
查看jqXHR 对象文档。您可以使用 fail 方法来捕获任何错误。
Something like the following for your case:
您的情况类似于以下内容：
$.post(jSONurl+'?orderID='+thisOrderID+'&variationID='+thisVariationID+'&quantity='+thisQuantity+'&callback=?')
.done(function(data){
if (data.response == 'success'){
//show the tick. allow the booking to go through
$('#loadingSML'+thisVariationID).hide();
$('#tick'+thisVariationID).show();
}else{
//show the cross. Do not allow the booking to be made
$('#loadingSML'+thisVariationID).hide();
$('#cross'+thisVariationID).hide();
$('#unableToReserveError').slideDown();
//disable the form
$('#OrderForm_OrderForm input').attr('disabled','disabled');
}
}, "json")
.fail(function(jqXHR, textStatus, errorThrown){
alert("Got some error: " + errorThrown);
});
I would also look into passing a json data string via post instead of attaching query variables:
我还将研究通过 post 传递 json 数据字符串而不是附加查询变量：
$.post(jSONurl, $.toJSON({orderID: thisOrderID, variationID: thisVariationID, quantity: thisQuantity, callback: false}))
回答by Mladen Janjetovic
If you are using POST you can use something like this:
如果您正在使用 POST，您可以使用以下内容：
$.post('account/check-notifications')
.done(function(data) {
// success function
})
.fail(function(jqXHR){
if(jqXHR.status==500 || jqXHR.status==0){
// internal server error or internet connection broke
}
});
回答by Vince V.
I think you could catch it by adding this:
我认为您可以通过添加以下内容来捕捉它：
$.ajax({
statusCode: {
500: function() {
alert("error");
}
},
url:jSONurl+'?orderID='+thisOrderID+'&variationID='+thisVariationID+'&quantity='+thisQuantity+'&callback=?',
async: false,
type: 'POST',
dataType: 'json',
success: function(data) {
if (data.response == 'success'){
//show the tick. allow the booking to go through
$('#loadingSML'+thisVariationID).hide();
$('#tick'+thisVariationID).show();
}else{
//show the cross. Do not allow the booking to be made
$('#loadingSML'+thisVariationID).hide();
$('#cross'+thisVariationID).hide();
$('#unableToReserveError').slideDown();
//disable the form
$('#OrderForm_OrderForm input').attr('disabled','disabled');
}
},
error: function(data){
alert('error');
}
})
回答by Fraser
I removed the dataType:json from the ajax call and I was able to catch the error. In these situations I do not need the content of the returned jSON luckily; only to know that there is an error being returned so this will suffice for now. Firebug still has a hissy fit but I am at least able to perform some actions when there is an error
我从 ajax 调用中删除了 dataType:json 并且我能够捕捉到错误。在这些情况下，幸运的是我不需要返回的 json 的内容；只知道返回了一个错误，所以现在就足够了。Firebug 仍然有嘶嘶声，但我至少能够在出现错误时执行一些操作
$.ajax({
url:'http://example.com/jsonservice/LiftieWeb/reserve?token=62e52d30e1aa70831c3f09780e8593f8&orderID='+thisOrderID+'&variationID='+reserveList+'&quantity='+thisQuantity+'&callback=?',
type: 'POST',
success: function(data) {
if (data.response == 'Success'){
//show the tick. allow the booking to go through
$('#loadingSML'+thisVariationID).hide();
$('#tick'+thisVariationID).show();
}else{
//show the cross. Do not allow the booking to be made
$('#loadingSML'+thisVariationID).hide();
$('#cross'+thisVariationID).hide();
$('#unableToReserveError').slideDown();
//disable the form
$('#OrderForm_OrderForm input').attr('disabled','disabled');
}
},
error: function(data){
alert('error');
}
})
