1

我正在制作一个将文件从卡上传到服务器的应用程序。但是,当我尝试执行此操作时,出现错误。SDPHP

我的安卓代码如下:

package de.fileupload;

import android.app.Activity;
import android.os.Bundle;
import android.util.Log;
import android.view.View;
import android.view.View.OnClickListener;
import android.widget.Button;
import android.widget.TextView;
import org.apache.http.client.ClientProtocolException;
import org.apache.http.client.HttpClient;
import org.apache.http.client.methods.HttpEntityEnclosingRequestBase;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.client.methods.HttpUriRequest;
import org.apache.http.entity.mime.MultipartEntity;
import org.apache.http.entity.mime.content.FileBody;
import org.apache.http.entity.mime.content.StringBody;
import org.apache.http.impl.client.DefaultHttpClient;
import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.Date;
import android.os.Bundle;

@SuppressWarnings("unused")
public class FileUpload extends Activity {
/** Called when the activity is first created. */
@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);

    final TextView tmp = (TextView) findViewById(R.id.textView1);
    tmp.setText("Hi! Click the button!");

    Button b = (Button) findViewById(R.id.button1);
    b.setOnClickListener(new OnClickListener() {
    public void onClick(View v) {
    File f = new File("mnt/sdcard/SMSBackup.txt");
    try {
    f.createNewFile();
    Date d = new Date();
    PrintWriter writer = new PrintWriter(f);
    writer.println(d.toString());
    writer.close();
    HttpClient client = new DefaultHttpClient();
    httpPostFileUpload(client, "mnt/sdcard/SMSBackup.txt", "http://10.0.2.2:8080/admin/admin/upload1.php", "uploadedfile");
    } catch (Exception e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
    }
    }
    });
    }
    public void httpPostFileUpload(
    HttpClient client,
    String filePath,
    String uploadUri,
    String inputNameAttr) throws ClientProtocolException,
    IOException {
    HttpUriRequest request = new HttpPost(uploadUri);
    MultipartEntity form = new MultipartEntity();

    client.getParams().setBooleanParameter("http.protocol.expect-continue", false);
    form.addPart(inputNameAttr, new FileBody(new File(filePath)));
    ((HttpEntityEnclosingRequestBase) request).setEntity(form);
    try {
    client.execute(request);
    } catch (ClientProtocolException e) {
    throw e;
    } catch (IOException ee) {
    throw ee;
    }
    }
    }

我的PHP文件如下:

上传1.php

<meta name="generator" content="Namo WebEditor(Trial)">
<form enctype="multipart/form-data" action="upload.php" method="POST"> 
<input type="hidden" name="MAX_FILE_SIZE" value="100000" /> 
 Choose a file to upload: <input name="uploadedfile" type="file" /><br /><input 
 type="submit" value="Upload File" /> 
 </form> 
 <?php 
 $to_file = "tmp/" . basename($_FILES['uploadedfile']['name']); 
 $from_file = $_FILES['uploadedfile']['tmp_name']; 
 if (move_uploaded_file($from_file, $to_file)) { 
  echo "Successful upload"; 
 ?> 
 <a href="<?php echo $to_file;?>"><?php echo $to_file;?></a> 
 <?php 
 } else { 
 echo "Unsuccessful upload"; 
 } 
 ?> 

和 upload.php 是:

<?php
// Where the file is going to be placed 
$target_path = "localhost/admin/admin/uploads/";

/* Add the original filename to our target path.  
 Result is "uploads/filename.extension" */
 $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";
 chmod ("uploads/".basename( $_FILES['uploadedfile']['name']), 0644);
 } else{
 echo "There was an error uploading the file, please try again!";
 echo "filename: " .  basename( $_FILES['uploadedfile']['name']);
 echo "target_path: " .$target_path;
 }
 ?>

谁能告诉我哪里错了?它总是显示文件上传不成功。当我的服务器正在运行时。

4

2 回答 2

1

Try to use instead move_uploaded_file(), function copy(). They use the same input parameters, like this:

if(copy($_FILES['uploadedfile']['tmp_name'], $target_path)) { ...

Most likely the PHP process has no permissions to move the uploaded file to different folder after successful upload, because the moving contains in essence copy and delete operation.

One more thing - try not to change the permission of the uploaded file to 0644, because this also can be restricted to the PHP process, i.e. when you deal with file system operations on Linux (I assume you use Linux machine for your server) the working process (in your case PHP and apache) has particular permissions set and maybe they do not have ability to delete/move files outside their working folder. You should also change the uploading folder permission to 755 or 777.

于 2012-05-29T13:26:47.417 回答
0

此类允许您直接上传文件。无需解码您的文件。

public class Helpher extends AsyncTask<String, Void, String> {
    Context context;
    JSONObject json;
    ProgressDialog dialog;
    int serverResponseCode = 0;
    DataOutputStream dos = null;
    FileInputStream fis = null;
    BufferedReader br = null;


    public Helpher(Context context) {
        this.context = context;
    }

    protected void onPreExecute() {

        dialog = ProgressDialog.show(Main2Activity.this, "ProgressDialog", "Wait!");
    }

    @Override
    protected String doInBackground(String... arg0) {

        try {
            File f = new File(arg0[0]);
            URL url = new URL("http://localhost:8888/imageupload.php");
            int bytesRead;
            HttpURLConnection conn = (HttpURLConnection) url.openConnection();
            conn.setDoOutput(true);
            conn.setDoInput(true);
            conn.setUseCaches(false);
            conn.setRequestProperty("Content-Type", "multipart/form-data; boundary=" + BOUNDARY);

            String contentDisposition = "Content-Disposition: form-data; name=\"keyValueForFile\"; filename=\""
                    + f.getName() + "\"";
            String contentType = "Content-Type: application/octet-stream";


            dos = new DataOutputStream(conn.getOutputStream());
            fis = new FileInputStream(f);


            dos.writeBytes(SPACER + BOUNDARY + NEW_LINE);
            dos.writeBytes(contentDisposition + NEW_LINE);
            dos.writeBytes(contentType + NEW_LINE);
            dos.writeBytes(NEW_LINE);
            byte[] buffer = new byte[MAX_BUFFER_SIZE];
            while ((bytesRead = fis.read(buffer)) != -1) {
                dos.write(buffer, 0, bytesRead);
            }
            dos.writeBytes(NEW_LINE);
            dos.writeBytes(SPACER + BOUNDARY + SPACER);
            dos.flush();

            int responseCode = conn.getResponseCode();
            if (responseCode != 200) {
                Log.w(TAG,
                        responseCode + " Error: " + conn.getResponseMessage());
                return null;
            }

            br = new BufferedReader(
                    new InputStreamReader(conn.getInputStream()));
            StringBuilder sb = new StringBuilder();
            String line = null;
            while ((line = br.readLine()) != null) {
                sb.append(line + "\n");
            }
            Log.d(TAG, "Sucessfully uploaded " + f.getName());

        } catch (MalformedURLException e) {
        } catch (IOException e) {
        } finally {
            try {
                dos.close();
                if (fis != null)
                    fis.close();
                if (br != null)
                    br.close();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return String.valueOf(serverResponseCode);
    }


    @Override
    protected void onPostExecute(String result) {
        dialog.dismiss();

    }

}

这是用于从 Android 上传图像的 AsyncTask “Helpher”类。要调用此类,请使用以下语法。

new Main2Activity.Helpher(this).execute(fileUri.getPath());

这里 fileUri.getPath() 本地图像位置。如果您想在“StringBuilder sb”中查看服务器响应值,您可以打印 sb 值

于 2016-09-08T01:17:32.383 回答