Question

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了：

xhr.open("GET", "https://" + url[i], true);
xhr.onreadystatechange = function(url) {
    console.log("Recieved data from " + url);
};
xhr.send();

Answer 1

有三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);

3：将您需要的数据存入您的回调（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);

Answer 2

使用closures。

function doRequest(url) {
    // create the request here

    var requestUrl = "https://" + url;
    xhr.open("GET", requestUrl, true);
    xhr.onreadystatechange = function() {

        // the callback still has access to requestUrl
        console.log("Recieved data from " + requestUrl); 
    };
    xhr.send();
}

XHR在onreadystatechange中获取请求URL

2 个答案: