我在java中使用下面的代碼通過網絡在php服務器上傳圖像。在android中將文件和圖像上傳到php服務器
import java.io.DataInputStream;
import java.io.DataOutputStream;
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.net.HttpURLConnection;
import java.net.MalformedURLException;
import java.net.URL;
import android.util.Log;
public class UploadFiles {
public void upload(String selectedPath) throws IOException {
HttpURLConnection conn = null;
DataOutputStream dos = null;
DataInputStream inStream = null;
String lineEnd = "\r\n";
String twoHyphens = "--";
String boundary = "AaB03x87yxdkjnxvi7";
int bytesRead, bytesAvailable, bufferSize;
byte[] buffer;
int maxBufferSize = 1 * 1024 * 1024;
String urlString = "http://android.1mohammadi.ir/nightly/upload_files.php";
try {
// ------------------ CLIENT REQUEST
FileInputStream fileInputStream = new FileInputStream(new File(
selectedPath));
// open a URL connection to the Servlet
URL url = new URL(urlString);
// Open a HTTP connection to the URL
conn = (HttpURLConnection) url.openConnection();
// Allow Inputs
conn.setDoInput(true);
// Allow Outputs
conn.setDoOutput(true);
// Don't use a cached copy.
conn.setUseCaches(false);
// Use a post method.
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", selectedPath);
dos = new DataOutputStream(conn.getOutputStream());
dos.writeBytes(twoHyphens + boundary + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name=\"uploaded_file\";filename=\""
+ selectedPath + "\"" + lineEnd);
dos.writeBytes(lineEnd);
dos.writeBytes(lineEnd);
// create a buffer of maximum size
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
buffer = new byte[bufferSize];
// read file and <span id="IL_AD4" class="IL_AD">write</span> it
// into form...
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);
}
// send multipart form data necesssary after file data...
dos.writeBytes(lineEnd);
dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
// close <span id="IL_AD3" class="IL_AD">streams</span>
Log.e("Debug", "File is written");
fileInputStream.close();
dos.flush();
dos.close();
} catch (MalformedURLException ex) {
Log.e("Debug", "error: " + ex.getMessage(), ex);
} catch (IOException ioe) {
Log.e("Debug", "error: " + ioe.getMessage(), ioe);
}
// ------------------ read the SERVER RESPONSE
try {
inStream = new DataInputStream(conn.getInputStream());
String str;
while ((str = inStream.readLine()) != null) {
Log.e("Debug", "Server Response " + str);
}
inStream.close();
} catch (IOException ioex) {
Log.e("Debug", "error: " + ioex.getMessage(), ioex);
}
}
}
而且在PHP服務器使用這樣的:
<?php
// Where the file is going to be placed
$target_path = "/uploads/";
/* Add the original filename to our target path.
Result is "uploads/filename.<span id="IL_AD5" class="IL_AD">extension</span>" */
$target_path = $target_path . basename($_FILES['uploadedfile']['name']);
error_log("Upload File >>" . $target_path . "\r\n", 3, "Log.log");
if (move_uploaded_file($_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!";
echo "filename: " . basename($_FILES['uploadedfile']['name']);
echo "target_path: " . $target_path;
}
?>
android系統中的代碼
不發生任何錯誤,並且在PHP代碼也。當文件上傳不能移動文件時。我該如何解決這個問題? 謝謝。
你檢查過上傳是否成功的PHP端? $ _FILES中的'['error']'參數應該**總是被檢查。 –
@MarcB沒有錯誤。 –
@MarcB我使用'error_log(「上傳文件>>」.basename($ _ FILES ['uploadedfile'] ['name'])。「\ r \ n」,3, 「Log.log」);'和輸出爲:'上傳文件>>上傳/ 上傳文件>> ' –