我已经找到一种方法。这是我的代码。我认为这会捕获所有错误。
// JavaScript Errors
window.onerror = function(messageOrEvent, source, lineno, colno, error) {
console.log("Captured: " + messageOrEvent)
}
// 404 FILES
window.addEventListener('error', function(e) {
console.log(e);
}, true);
// AJAX Errors
var open = window.XMLHttpRequest.prototype.open,
send = window.XMLHttpRequest.prototype.send;
function openReplacement(method, url, async, user, password) {
this._url = url;
return open.apply(this, arguments);
}
function sendReplacement(data) {
if(this.onreadystatechange) {
this._onreadystatechange = this.onreadystatechange;
}
this.onreadystatechange = onReadyStateChangeReplacement;
return send.apply(this, arguments);
}
function onReadyStateChangeReplacement() {
// CAPTURE HERE.
if(this.status != 200){
console.log(this.responseURL + " " + this.status + " " + this.statusText);
}
if(this._onreadystatechange) {
return this._onreadystatechange.apply(this, arguments);
}
}
window.XMLHttpRequest.prototype.open = openReplacement;
window.XMLHttpRequest.prototype.send = sendReplacement;
如果ajax调用jQuery可能会{错误:err =>抛出err} –
我必须做到全局。我无法编辑每一个功能。无论如何谢谢:) –
可能会覆盖一些XMLHttpRequest.prototype函数。 –