Android - 如何将文件名添加到调用 PHP 文件的上传文件函数
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【中文标题】Android - 如何将文件名添加到调用 PHP 文件的上传文件函数【英文标题】：Android - How to add file name to an upload file function that calls a PHP file 【发布时间】：2020-06-06 01:48:38 【问题描述】：我的 android 应用中有以下 Java 代码 - 在扩展 Application 的 Java 类中：
public static void XSUploadFile(final String filePath, final String fileName, final Activity act, final XSFileHandler handler)
new Thread(new Runnable()
public void run()
upload(filePath, fileName, act, handler);
).start();
public interface XSFileHandler void done(String fileURL, String error);
public static void upload(String sourceFileUri, String fileName, Activity act, final XSFileHandler handler)
HttpURLConnection conn;
DataOutputStream dos;
String lineEnd = "\r\n";
String twoHyphens = "--";
String boundary = "*****";
final int code;
int bytesRead, bytesAvailable, bufferSize;
byte[] buffer;
int maxBufferSize = 1024*1024;
File sourceFile = new File(sourceFileUri);
try
FileInputStream fileInputStream = new FileInputStream(sourceFile);
URL url = new URL(DATABASE_PATH + "upload-file.php");
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("file", sourceFileUri);
dos = new DataOutputStream(conn.getOutputStream());
dos.writeBytes(twoHyphens + boundary + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name='file';fileName='" + sourceFileUri + "'" + 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);
code = conn.getResponseCode();
if (code == HttpURLConnection.HTTP_OK)
final HttpURLConnection finalConn = conn;
act.runOnUiThread(new Runnable()
public void run()
InputStream responseStream = null;
try
responseStream = new BufferedInputStream(finalConn.getInputStream());
BufferedReader responseStreamReader = new BufferedReader(new InputStreamReader(responseStream));
String line;
StringBuilder stringBuilder = new StringBuilder();
int i = 0;
while ((line = responseStreamReader.readLine()) != null)
if (i != 0) stringBuilder.append("\n");
stringBuilder.append(line);
i++;
responseStreamReader.close();
String response = stringBuilder.toString();
responseStream.close();
finalConn.disconnect();
// Log.i(TAG, "XSUploadFile -> " + response);
if (response != null) handler.done(DATABASE_PATH + response, null);
else handler.done(null, E_401);
// error
catch (IOException e) e.printStackTrace(); handler.done(null, XS_ERROR);
);
// Bad response from sever
else
act.runOnUiThread(new Runnable()
public void run() handler.done(null, XS_ERROR); );
fileInputStream.close();
dos.flush();
dos.close();
// No response from server
catch (final Exception ex)
act.runOnUiThread(new Runnable()
public void run() handler.done(null, XS_ERROR); );
我这样称呼它：
// Get demo image path
Bitmap bmp = BitmapFactory.decodeResource(getResources(), R.drawable.demo_img);
final String filePath = getRealPathFromURI(getImageUri(bmp, ctx), ctx);
XSUploadFile(filePath, "image.jpg", (Activity)ctx, new XServerSDK.XSFileHandler()
@Override
public void done(String fileURL, String e)
if (fileURL != null)
hideHUD();
Log.i("log-", "Uploaded FileURL: " + fileURL);
);
这是我的upload-file.php 脚本：
<?php include '_config.php';
if ($_FILES["file"]["error"] > 0)
echo "Error: " .$_FILES["file"]["error"]. "<br>";
else
// Check file size
if ($_FILES["file"]["size"] > 20485760) // 20 MB
echo "ERROR: Your file is larger than 20 MB. Please upload a smaller one.";
else uploadImage();
// ./ If
// UPLOAD IMAGE ------------------------------------------
function uploadImage()
// generate a unique random string
$randomStr = generateRandomString();
$filePath = "uploads/".$randomStr;
// upload image into the 'uploads' folder
move_uploaded_file($_FILES['file']['tmp_name'], $filePath);
// echo the link of the uploaded image
echo $filePath;
// GENERATE A RANDOM STRING ---------------------------------------
function generateRandomString()
$characters = '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ';
$charactersLength = strlen($characters);
$randomString = '';
for ($i = 0; $i<20; $i++)
$randomString .= $characters[rand(0, $charactersLength - 1)];
return $randomString."_".$_POST['fileName'];
?>
我在 Logcat 中得到的结果是这个：
Uploaded FileURL: https://xscoder.com/xserver/uploads/mocpWfxIvtRAacnk1lTV_
我需要做的是将“image.jpg”附加到该 URL 的末尾，所以最终结果应该是这样的：
Uploaded FileURL: https://xscoder.com/xserver/uploads/mocpWfxIvtRAacnk1lTV_image.jpg
我假设它可以在我的 java 类的 XSUploadFile() 函数中完成，也许通过编辑这一行：
dos.writeBytes("Content-Disposition: form-data; name='file';fileName='" + sourceFileUri + "'" + fileName + "\"" + lineEnd);
我尝试了一些编辑，但完全没有成功。
请注意，我无法编辑 PHP 脚本，我有一个 PHP 和 ios SDK，它们都调用该脚本并且工作正常，所以我必须只编辑 Android Java 代码。
【问题讨论】：
请在下面找到答案。 谢谢，我已经编辑了我的问题 没有扩展的ios如何显示图片？ 【参考方案1】：您需要从上传的文件中获取扩展名。请看下面的代码。
$filename = $_FILES["file"]["name"];
$file_extension = end((explode(".", $filename)));
$filePath = 'uploads/' . $randomStr.$file_extension;
move_uploaded_file($_FILES["file"]["tmp_name"], $filePath);
编辑
@xscoder 无法编辑 php，所以我在 java 中给出了一个函数，当你将文件名传递给它时，它会返回文件扩展名。
private static String getFileExtension(String fileName)
if(fileName.lastIndexOf(".") != -1 && fileName.lastIndexOf(".") != 0)
return fileName.substring(fileName.lastIndexOf(".")+1);
else return "";
您可以将这一行编辑为
dos.writeBytes("Content-Disposition: form-data; name='file';fileName='" + sourceFileUri + "'" + fileName +"."+ getFileExtension(fileName) +"\"" + lineEnd);
注意：在 java 中使用文件时，您需要包含这一行 import java.io.File;
【讨论】：
谢谢，但我无法编辑我的 php 脚本，我忘了在我的问题中提及它，因为我还有一个 PHP 和 iOS SDK 可以按原样调用该文件工作正常，所以我必须只编辑我的 Java 代码 @xscoder 没有扩展 ios如何显示图片？ 因为在一个类似的 Swift 函数中，这就是我使用的：body.append("\(fileName)\r\n".data(using: String.Encoding.utf8)!) AND if fileName.hasSuffix(".jpg") body.append("Content-Type:image/png\r\n\r\n".data(using: String.Encoding.utf8)!)。所以我的http请求发送的是文件数据+文件名String，完整的文件名就变成了randomChars_image.jpg。我也必须对我的 Android 代码做类似的工作
@xscoder 您在 php 文件中上传的所有扩展名是什么？
主要是 jpg、png 和 mp4，但我可以上传任何文件，因为我在代码中附加了文件名和扩展名。【参考方案2】：
我从我的 Swift 代码中获得灵感找到了解决方案，我必须将此代码添加到我的 upload 函数中：
dos.writeBytes(twoHyphens + boundary + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name=\"fileName\"\r\n\r\n" +
fileName + "\r\n" +
"--" + boundary + "\r\n" +
"Content-Disposition: form-data; name=\"file\"; filename=\"file\"\r\n"
);
所以，最终的 Java 函数现在看起来像这样：
public static void upload(String sourceFileUri, String fileName, Activity act, final XSFileHandler handler)
HttpURLConnection conn;
DataOutputStream dos;
String lineEnd = "\r\n";
String twoHyphens = "--";
String boundary = "*****";
final int code;
int bytesRead, bytesAvailable, bufferSize;
byte[] buffer;
int maxBufferSize = 1024*1024;
File sourceFile = new File(sourceFileUri);
StrictMode.ThreadPolicy policy = new StrictMode.ThreadPolicy.Builder().permitAll().build();
StrictMode.setThreadPolicy(policy);
try
FileInputStream fileInputStream = new FileInputStream(sourceFile);
URL url = new URL(DATABASE_PATH + "upload-file.php");
conn = (HttpURLConnection) url.openConnection();
conn.setDoInput(true);
conn.setDoOutput(true);
conn.setUseCaches(false);
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("file", sourceFileUri);
dos = new DataOutputStream(conn.getOutputStream());
dos.writeBytes(twoHyphens + boundary + lineEnd);
/* HERE'S THE MAGIC CODE :) */
dos.writeBytes(twoHyphens + boundary + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name=\"fileName\"\r\n\r\n" +
fileName + "\r\n" +
"--" + boundary + "\r\n" +
"Content-Disposition: form-data; name=\"file\"; filename=\"file\"\r\n"
);
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);
code = conn.getResponseCode();
if (code == HttpURLConnection.HTTP_OK)
final HttpURLConnection finalConn = conn;
act.runOnUiThread(new Runnable()
public void run()
InputStream responseStream = null;
try
responseStream = new BufferedInputStream(finalConn.getInputStream());
BufferedReader responseStreamReader = new BufferedReader(new InputStreamReader(responseStream));
String line;
StringBuilder stringBuilder = new StringBuilder();
int i = 0;
while ((line = responseStreamReader.readLine()) != null)
if (i != 0) stringBuilder.append("\n");
stringBuilder.append(line);
i++;
responseStreamReader.close();
String response = stringBuilder.toString();
responseStream.close();
finalConn.disconnect();
// Log.i(TAG, "XSUploadFile -> " + response);
if (response != null) handler.done(DATABASE_PATH + response, null);
else handler.done(null, E_401);
// error
catch (IOException e) e.printStackTrace(); handler.done(null, XS_ERROR);
);
// Bad response from sever
else act.runOnUiThread(new Runnable()
public void run() handler.done(null, XS_ERROR); );
fileInputStream.close();
dos.flush();
dos.close();
// No response from server
catch (final Exception ex)
act.runOnUiThread(new Runnable()
public void run() handler.done(null, XS_ERROR); );
【讨论】：
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