我正在科尔多瓦开发一个应用程序。此应用程序应该能够在您按下捕获按钮时捕获视频,然后将其上载到特定服务器上。
但是我遇到了捕获API的问题。当我在模拟器或物理设备上运行它时,没有任何反应,并且在纹波上,会返回错误。
我的HTML代码在这里
<div class="button">
<div class="button1">
<button id="captureVideo" class="btn btn-lg btn-primary">launch a capture</button>
</div>
</div>
我的js
document.addEventListener("DOMContentLoaded", function () {
document.getElementById("captureVideo").addEventListener("click", function () {
document.addEventListener("deviceready", onDeviceReady, false);
});
})
function onDeviceReady() {
console.log(navigator.device.capture);
navigator.device.capture.captureVideo(captureSuccess, captureError, { limit: 1 });
console.log("Record launched !");
}
// Called when capture operation is finished
function captureSuccess(mediaFiles) {
navigator.notification.alert("Capture Success");
/*var i, len;
for (i = 0, len = mediaFiles.length; i < len; i += 1) {
uploadFile(mediaFiles[i]);
}*/
}
// Called if something bad happens
function captureError(error) {
navigator.notification.alert("Capture failed");
var msg = 'An error occurred during capture: ' + error.code;
navigator.notification.alert(msg, null, 'Uh oh!');
}
// 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 });
}*/
非常感谢您的回答:)
根据要求,我将代码直接放在这里而不是pastebin。
答案 0 :(得分:0)