javascript 查询。$.post request .done() .fail() 避免代码重复
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JQuery. $.post request .done() .fail() avoid code duplication
提问by Igor Slipko
I have a post request like
我有一个类似的帖子请求
$.post("test", {
ajax: "true",
action: ""
}).done(function(data){
if (data == "ok"){
//xxx
} else if (data == "err"){
//yyy
}
}).fail(function(){
//yyy
});
How to avoid code duplication in the post request if code in .done() method (comment 'yyy') the same in fail method (comment 'yyy') ??
如果 .done() 方法（注释 'yyy'）中的代码与失败方法（注释 'yyy'）中的代码相同，如何避免 post 请求中的代码重复？？
采纳答案by halilb
You can use alwayscallback method, and the request will always go in that block. As you know when data contains error and not, this method will work for server-side errors. And you can catch client-side errors by defining the final else block.
您可以使用始终回调方法，并且请求将始终进入该块。如您所知，数据何时包含错误而不是包含错误，此方法将适用于服务器端错误。您可以通过定义最后的 else 块来捕获客户端错误。
$.post("test", {
ajax: "true",
action: ""
}).always(function(data){
if (data == "ok"){
//xxx
} else if (data == "err"){
//handle server-side errors
} else {
//handle client-side errors like 404 errors
}
});
回答by Rudi Visser
The most obvious and simple solution would be to simply have a failure callback like so:
最明显和最简单的解决方案是简单地有一个失败回调，如下所示：
function ajaxFailed() {
// yyy
}
$.post("test", {
ajax: "true",
action: ""
}).done(function(data){
if (data == "ok"){
//xxx
} else if (data == "err"){
ajaxFailed();
}
}).fail(ajaxFailed);
回答by Jensen Ching
Have them call the same function, e.g.
让他们调用相同的函数，例如
function onErr() {
//yyy
}
$.post("test", {
ajax: "true",
action: ""
}).done(function(data){
if (data == "ok"){
//xxx
} else if (data == "err"){
onErr();
}
}).fail(onErr);
回答by John Dvorak
An alternative would be to change the protocol slightly, and make use of the HTTP status codesto indicate success or failure:
另一种方法是稍微更改协议，并使用HTTP 状态代码来指示成功或失败：
if($sqlError){
header("HTTP/1.1 503 Service unavailable");
}
...
...
.done(function(){
//xxx
}).fail(function(){
//yyy
});
