我正在制作一个Chrome扩展,它将打开新标签页中的所有链接.
的manifest.json
{
"name": "A browser action which changes its icon when clicked.","version": "1.1","permissions": [
"tabs","<all_urls>"
],"browser_action": {
"default_title": "links",// optional; shown in tooltip
"default_popup": "popup.html" // optional
},"content_scripts": [
{
"matches": [ "<all_urls>" ],"js": ["background.js"]
}
],"manifest_version": 2
}
popup.html
<!doctype html>
<html>
<head>
<title>My Awesome Popup!</title>
<script>
function getPageandSelectedTextIndex()
{
chrome.tabs.getSelected(null,function(tab) {
chrome.tabs.sendRequest(tab.id,{greeting: "hello"},function (response)
{
console.log(response.farewell);
});
});
}
chrome.browserAction.onClicked.addListener(function(tab) {
getPageandSelectedTextIndex();
});
</script>
</head>
<body>
<button onclick="getPageandSelectedTextIndex()">
</button>
</body>
</html>
background.js
chrome.extension.onRequest.addListener(
function(request,sender,sendResponse) {
console.log(sender.tab ?
"from a content script:" + sender.tab.url :
"from the extension");
if (request.greeting == "hello")
updateIcon();
});
function updateIcon() {
var allLinks = document.links;
for (var i=0; i<allLinks.length; i++) {
alllinks[i].style.backgroundColor='#ffff00';
}
}
最初我想突出显示页面上的所有链接或以某种方式标记它们;但是我收到错误“拒绝执行内联脚本,因为Content-Security-Policy”.
当我按弹出窗口内的按钮时,我得到这个错误：拒绝执行内联事件处理程序,因为Content-Security-Policy.
解决方法
“manifest_version”的后果之一是2,默认情况下启用
Content Security Policy. Chrome开发人员选择严格执行此操作,并始终禁止使用内置JavaScript代码 – 只允许在外部JavaScript文件中执行代码(以防止扩展中的
Cross-Site Scripting vulnerabilities).因此,不要在popup.html中定义getPageandSelectedTextIndex()函数,您应该将其放入popup.js文件并将其包含在popup.html中：
<script type="text/javascript" src="popup.js"></script>
< button onclick =“getPageandSelectedTextIndex()”>必须改变,onclick属性也是一个内联脚本.您应该分配ID属性：< button id =“button”> ;.然后在popup.js中,您可以将该事件处理程序附加到该按钮：
window.addEventListener("load",function()
{
document.getElementById("button")
.addEventListener("click",getPageandSelectedTextIndex,false);
},false);
