ajax NS_ERROR_FAILURE : Firefox 失败
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NS_ERROR_FAILURE : Failure in Firefox
提问by ffledgling
I'm using javascript's XMLHttpRequest object to send a request to another page (not on the same server or domainname ) I get a ns_error_failure error in firefox, but the Javascript works in Google Chrome, after searching online it seems to be because of firefox's XSS policy. Cross-Domain requests are not allowed.
我正在使用 javascript 的 XMLHttpRequest 对象将请求发送到另一个页面（不在同一服务器或域名上）我在 firefox 中收到 ns_error_failure 错误，但 Javascript 在 Google Chrome 中工作，在网上搜索后似乎是因为 firefox 的 XSS政策。不允许跨域请求。
Is there anyway to work around this and make the JS run in both chrome and Firefox?
有没有办法解决这个问题并使 JS 在 chrome 和 Firefox 中运行？
Please feel free to ask for additional details you feel are needed!
请随时询问您认为需要的其他详细信息！
Here's the code that I was using.
这是我使用的代码。
"use strict";
function showFixed(username)
{
console.log("Entered script");
var url = 'https://api-dev.bugzilla.mozilla.org/latest/bug'
+ '?quicksearch='
+ encodeURIComponent('FIXED @'+username);
displayBug(url);
}
function showPending(username)
{
console.log("Entered script");
var url = 'https://api-dev.bugzilla.mozilla.org/latest/bug'
+ '?quicksearch='
+ encodeURIComponent('@'+username);
displayBug(url);
}
function showCC(username)
{
console.log("Entered script");
var url = 'https://api-dev.bugzilla.mozilla.org/latest/bug'
+ '?quicksearch='
+ encodeURIComponent('cc:'+username);
displayBug(url);
}
function displayBug(url)
{
var xmlhttp = new XMLHttpRequest();
xmlhttp.open("GET",url,false);
xmlhttp.send();
var text = xmlhttp.responseText;
var json = JSON.parse(text);
for(var i=0;i<json.bugs.length;i++)
{
var tempRow = document.createElement('tr');
var tempId = document.createElement('td');
tempId.innerHTML = '<a href=\'https://bugzilla.mozilla.org/show_bug.cgi?id=' + json.bugs[i].id + '\'>'+ json.bugs[i].id + '</a>';
var tempCreator = document.createElement('td');
tempCreator.innerHTML = json.bugs[i].creator.real_name;
var tempShortDesc = document.createElement('td');
tempShortDesc.innerHTML = json.bugs[i].summary;
var tempComponent = document.createElement('td');
tempComponent.innerHTML = json.bugs[i].component;
var tempAssignee = document.createElement('td');
tempAssignee.innerHTML = json.bugs[i].assigned_to.real_name;
var tempWhiteBoard = document.createElement('td');
tempWhiteBoard.innerHTML = json.bugs[i].whiteboard;
var tempBugStatus = document.createElement('td');
tempBugStatus.innerHTML = json.bugs[i].status;
var tempResolution = document.createElement('td');
tempResolution.innerHTML = json.bugs[i].resolution;
var tempLastChange = document.createElement('td');
tempLastChange.innerHTML = json.bugs[i].last_change_time;
tempRow.appendChild(tempId);
tempRow.appendChild(tempAssignee);
tempRow.appendChild(tempCreator);
tempRow.appendChild(tempBugStatus);
tempRow.appendChild(tempShortDesc);
tempRow.appendChild(tempLastChange);
document.getElementById('bugs-table-tbody').appendChild(tempRow);
}
document.getElementById('main').innerHTML = '';
}
function wrapper()
{
var waitString = "Please wait while bug list is loaded..."
document.getElementById('main').innerHTML = waitString;
回答by NullHappens
If you are able to use jQuery, I would suggest having a look at JSONP (http://www.jquery4u.com/json/jsonp-examples/) this effectively allows crossdomain ajax.
如果您能够使用 jQuery，我建议您查看 JSONP（http://www.jquery4u.com/json/jsonp-examples/），它有效地允许跨域 ajax。
