[英]Upload file to server Android
我想将一些mp3文件上传到我的本地主机服务器,但是我不确定我要去哪里。 我认为这是php
文件,但这是我的php
文件:
<?php
$file_path = "/storage/emulated/0/sample.txt";
$file_path = $file_path . basename( $_FILES['uploaded_file']['name']);
if(move_uploaded_file($_FILES['uploaded_file']['tmp_name'], $file_path)) {
echo "success";
} else{
echo "fail";
}
?>
这是我的java
代码:
public class UploadFile extends Activity {
int serverResponseCode = 0;
String serverResponseMessage;
String uploadFileName;
String uploadFilePath;
String upLoadServerUri = null;
ProgressBar loading;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.uploads);
uploadFilePath = "/storage/emulated/0/";
uploadFileName = "sample.txt";
upLoadServerUri = "http://10.0.3.2/files/uploadfile.php";
new Thread(new Runnable() {
public void run() {
uploadFile(uploadFilePath + "" + uploadFileName);
}
}).start();
}
public void uploadFile(String sourceFileUri) {
String fileName = sourceFileUri;
HttpURLConnection conn = null;
DataOutputStream dos = null;
String lineEnd = "\r\n";
String twoHyphens = "--";
String boundary = "*****";
int code;
int bytesRead, bytesAvailable, bufferSize;
byte[] buffer;
int maxBufferSize = 1 * 1024 * 1024;
File sourceFile = new File(sourceFileUri);
if (!sourceFile.isFile()) {
code = 0;
this.runOnUiThread(new Runnable() {
public void run() {
}
});
}
else {
try {
FileInputStream fileInputStream = new FileInputStream(sourceFile);
URL url = new URL(upLoadServerUri);
conn = (HttpURLConnection) url.openConnection();
conn.setDoInput(true); // Allow Inputs
conn.setDoOutput(true); // Allow Outputs
conn.setUseCaches(false); // Don't use a Cached Copy
conn.setRequestMethod("POST");
conn.setRequestProperty("Connection", "Keep-Alive");
conn.setRequestProperty("ENCTYPE", "multipart/form-data");
conn.setRequestProperty("Content-Type", "multipart/form-data;boundary=" + boundary);
conn.setRequestProperty("uploaded_file", fileName);
dos = new DataOutputStream(conn.getOutputStream());
dos.writeBytes(twoHyphens + boundary + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name='uploaded_file';filename='"+fileName+"'"+lineEnd);
dos.writeBytes(lineEnd);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
buffer = new byte[bufferSize];
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
while (bytesRead > 0) {
dos.write(buffer, 0, bufferSize);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
}
dos.writeBytes(lineEnd);
dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
serverResponseCode = conn.getResponseCode();
code = serverResponseCode;
if (code==200) {
this.runOnUiThread(new Runnable() {
public void run() {
}
});
}
else {
this.runOnUiThread(new Runnable() {
public void run() {
}
});
}
serverResponseMessage = conn.getResponseMessage();
fileInputStream.close();
dos.flush();
dos.close();
}
catch (MalformedURLException ex) {
this.runOnUiThread(new Runnable() {
public void run() {
}
});
}
catch (Exception e) {
this.runOnUiThread(new Runnable() {
public void run() {
}
});
}
}
}
}
我将此代码在线获取并出于我的目的对其进行了修改-但它无法正常工作,但正如我所提到的,它可能是php
文件
任何帮助都将真正起到很大的作用(即使对此进行修复,也可以让我正常饮食和睡眠)
谢谢
使用以下php代码更改您的代码,即可正常运行。
<?php
$file_path = "uploads/";
$file_path = $file_path . basename( $_FILES['uploaded_file']['name']);
if(move_uploaded_file($_FILES['uploaded_file']['tmp_name'], $file_path)) {
echo "success";
} else{
echo "fail";
}
?>
我已经测试过它工作完美
我认为您应该为此尝试使用json解析器,将您的PHP脚本设置为输出json,以获取成功和失败的信息。 您需要将Java代码中的图像文件名字符串解析为Web服务,并获取它来执行操作。 您还可以使用$ _GET从Java代码获取文件
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