Rad*_*mko 3 javascript ajax xmlhttprequest
有没有办法在"onreadystatechange"方法中获取请求的URL?
我想运行多个XHR请求并知道它们中的哪一个回来:
xhr.open("GET", "https://" + url[i], true);
xhr.onreadystatechange = function(url) {
console.log("Recieved data from " + url);
};
xhr.send();
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有3种简单的方法可以做到这一点.
1:使用已经描述的闭包
2:在xhr对象上设置一个属性,稍后可以像这样引用:
xhr._url = url[i];
xhr.onreadystatechange = function(readystateEvent) {
//'this' is the xhr object
console.log("Recieved data from " + this._url);
};
xhr.open("GET", "https://" + url[i], true);
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3:将您需要的数据加入回调中(我的首选解决方案)
Function.prototype.curry = function curry() {
var fn = this, args = Array.prototype.slice.call(arguments);
return function curryed() {
return fn.apply(this, args.concat(Array.prototype.slice.call(arguments)));
};
};
function onReadystateChange(url, readystateEvent) {
console.log("Recieved data from " + url);
};
xhr.onreadystatechange = onReadystateChange.curry(url[i]);
xhr.open("GET", "https://" + url[i], true);
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