问在chrome扩展中正确链接JavaScript
EN

Stack Overflow用户

提问于 2016-04-26 16:40:50

回答 1查看 125关注 0票数 0

因此，我有一个铬扩展，但JavaScript没有正确加载。我试图链接一个文件，但不幸地失败了。对于如何使按钮在单击时运行函数，有什么想法吗？标记在这种情况下不起作用，按钮什么也不做，但是初始的JavaScript确实运行.这是我的密码

Manifest.json(我认为它是正确的)：

{
"manifest_version": 2,

"name": "Getting started example",
"description": "This extension shows a Google Image search result for the current page",
"version": "1.0",

"browser_action": {
"default_icon": "icon.png",
"default_popup": "popup.html"
},
"permissions": [
"activeTab",
"https://ajax.googleapis.com/"
 ]

  }

Popup.html( JavaScript中的函数不链接到此)：

<body style = "margin:10px">

<!-- Latest compiled and minified CSS -->
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/css/bootstrap.min.css" integrity="sha384-1q8mTJOASx8j1Au+a5WDVnPi2lkFfwwEAa8hDDdjZlpLegxhjVME1fgjWPGmkzs7" crossorigin="anonymous">

<!-- Optional theme -->
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/css/bootstrap-theme.min.css" integrity="sha384-fLW2N01lMqjakBkx3l/M9EahuwpSfeNvV63J5ezn3uZzapT0u7EYsXMjQV+0En5r" crossorigin="anonymous">

<!-- Latest compiled and minified JavaScript -->
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/js/bootstrap.min.js" integrity="sha384-0mSbJDEHialfmuBBQP6A4Qrprq5OVfW37PRR3j5ELqxss1yVqOtnepnHVP9aJ7xS" crossorigin="anonymous"></script>
<script src="popup.js"></script>
<style>
//Fancy styles
large{font-size:30px}
span{}
.input-group{width:400px}
</style>
<!-- title -->
<large>
SpeedyURLs
</large>
<hr/>
<!-- Code that won't link with JavaScript -->
<div id = "stuff">

</div>
<div class = "input-group" id = "content"><input class = "form-control" placeholder = "Enter a URL here. Ex: https://www.google.com"/><span class="input-group-addon btn btn-default" style = "font-size:20px;width:50px;color:black" id="basic-addon1" onclick = "addNew()">+        </span></div>
<button onclick = 'document.write("HELLO")'>YAS</button>
</body>

Popup.js(不链接的函数)：

function open(url){
window.open(url)}
function addNew(){
document.getElementById("stuff").innerHTML = document.getElementById("stuff").innerHTML+'<div class = "input-group" id = "content"><input class = "form-control" placeholder = "Enter a URL here. Ex: https://www.google.com"/><span  style = "font-size:20px;width:50px;color:black" class="input-group-addon btn btn-danger" id="basic-addon1">x</span></div>'}

Icon.png(默认图像中没有代码，工作正常)

因此，JS中的函数不会链接到HTML。有什么想法吗？

google-chrome-extension

回答 1

Stack Overflow用户

回答已采纳

发布于 2016-04-26 17:00:16

默认情况下，将不会执行内联JavaScript。

你可以

放松默认策略。有关更多详细信息，请参阅内联脚本。
删除onclick中的popup.html并将该逻辑移动到您的popup.js中，代码将类似于(假设您的按钮id是buttonId0) document.getElementById("buttonId0").addEventListener("click"，函数(){ //您的逻辑在这里}，false)；