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我正在尝试将视频从 Android 代码上传到我的 PHP 网络服务器代码,但无法成功。我指的是以下 link执行上传任务,但我在我的 Android 代码中收到以下响应,但在 PHP 服务器上找不到任何文件。
安卓响应
DEBUG/ServerCode(29484): 200
DEBUG/serverResponseMessage(29484): OK
我检查了很多东西,比如在 php.ini
中设置值文件。尽管我可以从 Android 代码将图像上传到服务器,因为我发送的是 64 位编码的 64 位编码 ByteArray
并且在 php 中有一个现成的函数,它可以从编码的 ByteArray
创建图像
但是如果任何其他文件的类型不是图像,该代码也不起作用。
如果你们以前做过类似的事情,请指导我。
我正在使用的 PHP 代码:
<?php
$target_path = "./upload/";
$target_path = $target_path . basename( $_FILES['uploadedfile']['name']);
if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path))
{
echo "The file ".basename( $_FILES['uploadedfile']['name'])." has been uploaded";
}
else
{
echo "There was an error uploading the file, please try again!";
}
?>
我正在使用的安卓代码:
public void videoUpload()
{
HttpURLConnection connection = null;
DataOutputStream outputStream = null;
DataInputStream inputStream = null;
String pathToOurFile = "/sdcard/video-2010-03-07-15-40-57.3gp";
String urlServer = "http://10.0.0.15/sampleWeb/handle_upload.php";
String lineEnd = "\r\n";
String twoHyphens = "--";
String boundary = "*****";
int bytesRead, bytesAvailable, bufferSize;
byte[] buffer;
int maxBufferSize = 1*1024*1024;
try
{
FileInputStream fileInputStream = new FileInputStream(new File(pathToOurFile) );
URL url = new URL(urlServer);
connection = (HttpURLConnection) url.openConnection();
// Allow Inputs & Outputs
connection.setDoInput(true);
connection.setDoOutput(true);
connection.setUseCaches(false);
// Enable POST method
connection.setRequestMethod("POST");
connection.setRequestProperty("Connection", "Keep-Alive");
connection.setRequestProperty("Content-Type", "multipart/form-data;boundary");
outputStream = new DataOutputStream( connection.getOutputStream() );
outputStream.writeBytes(twoHyphens + boundary + lineEnd);
outputStream.writeBytes("Content-Disposition: form-data; name=\"uploadedfile\";filename=\"" + pathToOurFile );
outputStream.writeBytes(lineEnd);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
buffer = new byte[bufferSize];
// Read file
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
while (bytesRead > 0)
{
outputStream.write(buffer, 0, bufferSize);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
}
outputStream.writeBytes(lineEnd);
outputStream.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
// Responses from the server (code and message)
int serverResponseCode = connection.getResponseCode();
String serverResponseMessage = connection.getResponseMessage();
Log.d("ServerCode",""+serverResponseCode);
Log.d("serverResponseMessage",""+serverResponseMessage);
fileInputStream.close();
outputStream.flush();
outputStream.close();
}
catch (Exception ex)
{
ex.printStackTrace();
}
}
最佳答案
我的 java 有点笨拙但是......
看起来 connection.getResponseCode 仅返回 HTTP 状态代码,而 connection.getResponseMessage 仅返回 HTTP 状态消息 - 但您的 PHP 不执行任何操作这些值的操作。您可以尝试:
$target_path = "./upload/";
$src = $_FILES['uploadedfile']['name'];
$target_path .= basename($src);
if(file_exists($src)
&&
&& move_uploaded_file($src, $target_path)
) {
echo "The file ".basename( $_FILES['uploadedfile']['name'])." has been uploaded";
} else {
header("Server Error", true, 503);
echo "There was an error uploading the file, please try again!";
$msg = "src size ? "
. filesize($src) . "\n dest dir writable ?"
. is_writeable(dirname($target_path)) ? "Y\n" : "N\n"
. "FILES contains :\n";
. var_export($_FILES,true);
// now write $msg somwhere you can read it
}
这应该有助于缩小错误范围
关于php - 将视频从 Android 上传到 PHP 的代码,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/4944650/
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