我正在编写我的第一个程序以在 Google chrome 中进行扩展,我只是从这里以“hello world”教程为例
这是我的 html 文件源代码:
<!doctype html>
<html>
<head>
<title>Getting Started Extension's Popup</title>
<style>
body {
min-width:357px;
overflow-x:hidden;
}
img {
margin:5px;
border:2px solid black;
vertical-align:middle;
width:75px;
height:75px;
}
</style>
<!-- JavaScript and HTML must be in separate files for security. -->
<script src="popup.js"></script>
</head>
<body>
</body>
</html>
广告这是我的 javascript 文件源代码:
var req = new XMLHttpRequest();
req.open(
"GET",
"http://api.flickr.com/services/rest/?" +
"method=flickr.photos.search&" +
"api_key=90485e931f687a9b9c2a66bf58a3861a&" +
"text=hello%20world&" +
"safe_search=1&" + // 1 is "safe"
"content_type=1&" + // 1 is "photos only"
"sort=relevance&" + // another good one is "interestingness-desc"
"per_page=20",
true);
req.onload = showPhotos;
req.send(null);
function showPhotos() {
var photos = req.responseXML.getElementsByTagName("photo");
var element = document.createElement('h1');
element.appendChild(document.createTextNode
('tete '+document.location.href+'hgdfhgd'));
for (var i = 0, photo; photo = photos[i]; i++) {
var img = document.createElement("image");
img.src = constructImageURL(photo);
document.body.appendChild(img);
}
}
// See: http://www.flickr.com/services/api/misc.urls.html
function constructImageURL(photo) {
return "http://farm" + photo.getAttribute("farm") +
".static.flickr.com/" + photo.getAttribute("server") +
"/" + photo.getAttribute("id") +
"_" + photo.getAttribute("secret") +
"_s.jpg";
}
该示例非常简单并且工作正常,但是当添加我自己的 javascript 指令时,它不显示它,添加的指令在 showPhotos() 函数中,它是:
var element = document.createElement('h1');
element.appendChild(document.createTextNode
('tete '+document.location.href+'hgdfhgd'));
结果,我可以看到其他内容,但我的“h1”我看不到。我错过了什么?谁能帮帮我?
谢谢