从函数中抛出和捕获异常

Question

function connectTo(url) {
    var xhr = new XMLHttpRequest();
    xhr.open("GET", url, false);
    xhr.onreadystatechange = function () {
        if (xhr.readyState == xhr.DONE) {
            throw "Troubles.";
        }
    };
    xhr.send();
}

try {
    connectTo("http://www.google.com");
} catch (e) {
    console.log('Exception happend.');
}

也许"catch"部分将执行(在控制台中显示消息),但异常保持未捕获状态(=在控制台中显示"Uncaught Troubles".).为什么？

Answer 1

投掷不会像这样回调.传入错误处理回调并手动处理.

让我来说明你的堆栈跟踪

onreadystatechange函数和connectTo函数之间没有堆栈跟踪连接.所以当你抛出一个错误时,它永远不会冒泡到connectTo周围的try catch块.

firefox正在做的是说 "Oh you did something that doesn't work. let me fix that for you and do what you think it does"

function connectTo(url, err) {
    var xhr = new XMLHttpRequest();
    xhr.open("GET", url, false);
    xhr.onreadystatechange = function () {
        if (xhr.readyState == xhr.DONE) {
            err.call(this, new Error("troubles"));
        }
    };
    xhr.send();
}

connectTo("http://www.google.com", function(e) {
     console.log(e);
});