我没有收到弹出脚本收到消息时应该显示的警报。
manifest.json:
{
"manifest_version": 2,
"name": "Drill Sergeant",
"description": "Tracks time spent on 'watchlist' websites",
"version": "1.0.0",
"browser_action": {
"default_popup": "popup.html"
},
"background": {"scripts": ["background.js"]},
"web_accessible_resources": ["background.js"],
"permissions": ["activeTab", "webNavigation", "notifications", "tabs"]
}
在我的html文件中包含popup.js:
<body>
....
<script src = popup.js></script>
</body>
来自后台的消息通话:
chrome.tabs.onUpdated.addListener(function(tabId, changeInfo, tab)
{
chrome.runtime.sendMessage({
text: "please work"
});
checkMe();
});
popup.js代码:
chrome.runtime.onMessage.addListener(
function(request, sender, sendResponse) {
alert("PLEASE WORK");
console.log("PLEASE WORK");
}
);
每当我更改页面时都会执行checkMe()函数(我知道也是因为它也具有调试警报),但是我没有从popup.js收到警报...我看过其他线程,使用了runtime.onMessage而不是tabs.onMessage,但仍然无法正常工作。感谢您的帮助