我正在使用phoengap捕获视频插件 我使用过这段代码
<!DOCTYPE html>
<html>
<head>
<title>Capture Video</title>
<script type="text/javascript" charset="utf-8" src="cordova.js"> </script>
<script type="text/javascript" charset="utf-8" src="json2.js"></script>
<script type="text/javascript" charset="utf-8">
// Called when capture operation is finished
//
function captureSuccess(mediaFiles) {
var i, len;
for (i = 0, len = mediaFiles.length; i < len; i += 1) {
uploadFile(mediaFiles[i]);
}
}
// Called if something bad happens.
//
function captureError(error) {
var msg = 'An error occurred during capture: ' + error.code;
navigator.notification.alert(msg, null, 'Uh oh!');
}
// A button will call this function
//
function captureVideo() {
// Launch device video recording application,
// allowing user to capture up to 2 video clips
navigator.device.capture.captureVideo(captureSuccess, captureError, {limit: 2});
}
// Upload files to server
function uploadFile(mediaFile) {
var ft = new FileTransfer(),
path = mediaFile.fullPath,
name = mediaFile.name;
ft.upload(path,
"http://my.domain.com/upload.php",
function(result) {
console.log('Upload success: ' + result.responseCode);
console.log(result.bytesSent + ' bytes sent');
},
function(error) {
console.log('Error uploading file ' + path + ': ' + error.code);
},
{ fileName: name });
}
</script>
</head>
<body>
<button onclick="captureVideo();">Capture Video</button> <br>
</body>
</html>
它工作正常..但我想要
如何在触摸选择时显示我的视频图像或播放视频?
我确实使用了这个
path = mediaFile.fullPath
$("div").html("<img src="+path+">");
不显示视频图像。你能救我吗?
答案 0 :(得分:5)
@DavidC非常接近,你只需要知道要使用的MediaResult ob的属性是fullPath。这是一个完整的例子。
document.addEventListener("deviceready", init, false);
function init() {
document.querySelector("#takeVideo").addEventListener("touchend", function() {
console.log("Take video");
navigator.device.capture.captureVideo(captureSuccess, captureError, {limit: 1});
}, false);
}
function captureError(e) {
console.log("capture error: "+JSON.stringify(e));
}
function captureSuccess(s) {
console.log("Success");
console.dir(s[0]);
var v = "<video controls='controls'>";
v += "<source src='" + s[0].fullPath + "' type='video/mp4'>";
v += "</video>";
console.log(v);
document.querySelector("#videoArea").innerHTML = v;
}
答案 1 :(得分:2)
因为它是一个视频,我想你应该使用标签视频而不是img。我现在无法验证这一点
$("div").html(' <video controls="controls">
<source src="'+path+'" type="video/mp4" />
</video>');