我正在使用此phonegap（js）代码将录制的视频上传到php服务器。
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<meta name="viewport" content="initial-scale=1, maximum-scale=1, user-scalable=no, width=device-width">
<link rel="stylesheet" href="http://code.jquery.com/mobile/1.3.2/jquery.mobile-1.3.2.min.css" />
<script src="http://code.jquery.com/jquery-1.9.1.min.js"></script>
<script src="http://code.jquery.com/mobile/1.3.2/jquery.mobile-1.3.2.min.js"></script>
<title>Mobile_Insurance</title>
</head>
<body>
<script type="text/javascript">
$(document).ready(function(){
$('input[name="visit"]').click(function(){
var inputValue = $(this).attr("value");
var targetBox = $("." + inputValue);
$(".box").not(targetBox).hide();
$(targetBox).show();
});
});
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: 1});
}
// Upload files to server
function uploadFile(mediaFile) {
var ft = new FileTransfer(),
path = mediaFile.fullPath,
name = mediaFile.name;
var options = new FileUploadOptions();
options.chunkedMode = true;
options.fileKey = "file";
options.fileName = name;
options.mimeType = "video/mp4";
var params = new Object();
params.value1 = "test";
params.value2 = "param";
options.params = params;
ft.upload(path, "http://192.168.0.46/upload/upload.php",
function(result) {
console.log('Upload success: ' + result.responseCode);
console.log(result.bytesSent + ' bytes sent');
console.log("Response = " + r.response);
alert("Response = " + r.response);
},
function(error) {
console.log('Error uploading file ' + path + ': ' + error.code);
alert('Error uploading file ' + path + ': ' + error.code);
},
options);
alert(mediaFile.fullPath);
}
</script>
<script type="text/javascript" src="cordova.js"></script>
<div data-role="page">
<div data-role="header">
<h3>Welcome </h3>
</div>
<div data-role="main" class="ui-content">
<h3 style="text-align: center;">Input Your IMEI:</h3>
<input type="number"/>
<h3 style="text-align: center;"> yes?</h3>
<input type="radio" name="visit" value="YES" id="Video"> YES
<input type="radio" name="visit" value="NO" id="self"> NO
<br>
<h3 style="text-align: center;"> damage.</h3>
<input type="radio" name="damage" value="Physical"> Physical
<input type="radio" name="damage" value="Water"> Water <br><br>
<h3 style="text-align: center;">Please give a breig description about the damage</h3><br>
<textarea rows="5" cols="10" style="resize:none"></textarea>
<div class="YES box"><input type="button" value="self analysis" hidden="true"></div>
<div class="NO box"> <button onclick="captureVideo();">Capture Video</button></div>
</div>
</div>
</body>
</html>
这是我的PHP代码..
<?php
print_r($_FILES);
$new_image_name = "r.mp4";
move_uploaded_file($_FILES["file"]["tmp_name"], $new_image_name);
?>
uploadFile函数应该将文件上传到指定的php文件。但在我的情况下，phonegap文件传输给出错误代码1，找不到文件。我在捕获后提醒文件路径，这是要上传的文件。如何抛出错误代码1？
答案 0 :(得分：1)
尝试一下，您也许可以使用
http://findnerd.com/list/view/Capturing-and-Uploading-Video-to-PHP-Server-in-Cordova/9398/
从网站上
如果要将图像作为base64字符串发送，则可以更改目标 输入Camera.DestinationType.DATA_URL，然后可以将imageData发送到 通过ajax调用服务器。或者，如果您要作为文件数组发送，则保留 与camera.DestinationType.FILE_URI相同的目的地类型，并使用 cordova文件插件以在服务器中发送文件数据：
var options = new FileUploadOptions();
options.fileKey="tickitFile";
options.fileName=imageData.substr(imageData.lastIndexOf('/')+1);
options.contentType = "multipart/form-data";
options.chunkedMode = false;
options.mimeType="image/jpeg";
options.httpMethod="POST";
options.headers = {
Connection: "close"
};
var ft = new FileTransfer();
ft.upload(imageData, PHP_URL, win, fail, options);
function win(r) {
console.log("Response = " + r.response);
console.log("Sent = " + r.bytesSent);
}
function fail(error) {
console.log(JSON.stringify(error));
}
答案 1 :(得分：0)
你可以尝试以下方法：
function upload(file){
var fd = new FormData();
fd.append("dir", dir);
fd.append("file", file);
var xhr = new XMLHttpRequest();
xhr.open('POST', 'upload.php', true);
xhr.send(fd);
xhr.onreadystatechange = function() {
if (xhr.readyState == 4 && (xhr.status == 200 || xhr.status == 0)) {
//alert(xhr.responseText);
var message = xhr.responseText;
message=message.trim();
if ( message != 0)
{
//alert(message);
}
}
};
}
和php文件：
<?php
if (isset($_FILES["file"]["name"])) {
$destination = $_POST["dir"];
$name = $_FILES["file"]["name"];
$tmp_name = $_FILES['file']['tmp_name'];
$error = $_FILES['file']['error'];
//echo $name;
//echo $tmp_name;
//echo $error;
move_uploaded_file($_FILES['file']['tmp_name'], $destination.$name);
}
echo "File transfer completed";
?>
XHR POST没有大小限制，但您正在向具有大小限制的PHP发送数据;）创建以下php文件并在浏览器中打开它：
现在搜索变量“post_max_size”，这个变量限制了可以发送到PHP的最大数据（但可以在php.ini中更改）
我的上传功能和我的php文件完美适用于输入文件，如：
var obj=document.getElementById("inputfile");
var len = obj.files.length;
for (i=0; i<=len; i++){
upload( obj.files[i] );
}
对我来说，问题是你的capturevideo（）函数的输出类型或captureSuccess（mediaFiles）中的错误：尝试更改如下所示的内容：
function captureSuccess(mediaFiles) {
var i, len;
for (i = 0, len = mediaFiles.length; i < len; i += 1) {
uploadFile(mediaFiles[i].fullPath);
}
}