jquery ajax http状态码
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jquery ajax http status code
提问by Zulakis
I am changing my scripts to jquery at the moment. This is my old javascript:
我现在正在将我的脚本更改为 jquery。这是我的旧 javascript：
var request = (window.XMLHttpRequest) ? new XMLHttpRequest() : (window.ActiveXObject ? new window.ActiveXObject("Microsoft.XMLHTTP") : false);
request.open("GET", url, true);
request.onreadystatechange = updatePage;
request.send(null);
function updatePage() {
if (request.readyState == 4) {
if (request.status == 200) {
var response = request.responseText;
doSomething();
} else if (request.status == 304) {
doSomethingElse();
} else {
}
}
}
I now want to change this to jquery-ajax:
我现在想将其更改为 jquery-ajax：
request = $.ajax({
type: "GET",
url: url,
data: data,
success: updatePage
});
How can I test for the status code (and the responseText) returned by the request, like I did in my old script?
我如何测试请求返回的状态代码（和 responseText），就像我在旧脚本中所做的那样？
回答by mgraph
try this:
尝试这个：
request = $.ajax({
type: "GET",
url: url,
data: data,
statusCode: {
200: function() {
doSomething();
},
304:function(){
doSomethingElse();
}
}
});
or:
或者：
request = $.ajax({
type: "GET",
url: url,
data: data,
complete: function(e, xhr, settings){
if(e.status === 200){
}else if(e.status === 304){
}else{
}
}
});
回答by techie_28
You need to just bind this request to an event like button click or something Like
你只需要将此请求绑定到一个事件，比如按钮点击或类似的东西
$('#btn').click({function(){
$.ajax({
type: "GET",
url: url,
data: data,
success: updatePage
});
});
You dont need to really keep this request against a variable as jQuery ajaxhas a success callbackwhich will help you to execute post success code
您不需要真正针对变量保留此请求，因为jQuery ajax有一个成功回调，这将帮助您执行成功后代码
