我有一个chrome扩展程序。以下是清单。
{
"name": "example",
"description": "sample",
"version": "1.0",
"permissions": [
"tabs",
"activeTab"
],
"options_ui": {
"page": "options.html",
"chrome_style": true
},
"background": {
"scripts": ["background.js"]
},
"content_scripts": [{
"matches": ["https://www.google.co.in/*"],
"js": ["jquery-1.8.0.min.js", "content_script.js"]
}],
"browser_action": {
"default_title": "Title.",
"default_icon": "abc.png",
"default_popup": "popup.html"
},
"manifest_version": 2
}
popup.html
<html >
<head>
<title>ABC</title>
<script src="./popup.js"></script>
</head>
<body id="body" style="width:400px;max-height:80%;">
</body>
</html>
popup.js
document.addEventListener('DOMContentLoaded', function () {
<my code here>
});
当单击图标时,弹出窗口打开,一切正常。双击时,弹出窗口会打开并关闭。
我需要禁用双击或双击工作方式与单击工作方式相同。请建议如何做到这一点。