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我试图创建一个弹出窗口,将可见标签显示为小图像。从chrome.tabs.captureVisibleTab()函数返回的ImgSrc是“未定义”。我试过从各个地方运行。我可以验证从tabs.Query()返回的tab是否为null,因此tabs [0] .id不为空。Chrome Tabs.CaptureVisibleTab未定义
我这样做不正确吗?
这里是我的清单,popup.html和popup.js文件:
{
"manifest_version": 2,
"name": "SuperFave",
"description": "Saves favorites demo",
"version": "1.0",
"browser_action": {
"default_popup": "popup.html"
},
"permissions": [
"tabs",
"<all_urls>"
]
}
popup.html:
<html>
<head>
<script type="text/javascript" src="jquery-1.10.2.min.js">
</script>
<script type="text/javascript" src="popup.js">
</script>
</head>
<body>
</body>
</html>
popup.js:
$(document).ready(function() {
chrome.tabs.query({
// gets the window the user can currently see
active: true,
currentWindow: true
},
function (tabs) {
chrome.tabs.captureVisibleTab(
tabs[0].id,
function (src) {
// displays a link to the image. Can be replaced by an alert() to
// verify the result is 'undefined'
$('body').append("<a href='" + src + "'>" + tabs[0].url + "</a>");
}
);
}
);
});