从XMLHttpRequest send()捕获异步网络错误

Question

我使用XMLHttpRequest以下方式异步发出http请求:

xhr.open(method, uri, true);

当我发送something:

xhr.send(something)

当服务器关闭时,它会引发以下错误:

net::ERR_CONNECTION_REFUSED

如何捕获并处理此错误？标准try..catch块不起作用,因为请求是异步的.

提前致谢.

Answer 1

使用以下onerror事件XMLHttpRequest:

function aGet(url, cb) {
    var x = new XMLHttpRequest();
    x.onload = function(e) {
        cb(x.responseText)
    };
    x.onerror= function(e) {
        alert("Error fetching " + url);
    };
    x.open("GET", url, true);
    x.send();
}

var dmp = console.log.bind(console); // Dummy callback to dump to console
aGet("/", dmp) // Ok, uses onload to trigger callback
aGet("http://dgfgdf.com/sdfsdf", dmp); // Fails, uses onerror to trigger alert