Question

我正在制作一个上传一个文件从SD卡到PHP服务器的应用程序。但是，当我尝试这样做时，我收到了一个错误。

我的Android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：

upload1.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;
 }
 ?>

任何人都可以告诉我哪里错了吗？它始终显示文件的上传失败。我的服务器正在运行。

Answer 1

我正在使用此处描述的方法：

Uploading files to HTTP server using POST on Android.

它就像一个魅力。

Answer 2

尝试使用move_uploaded_file()，功能copy()。它们使用相同的输入参数，如下所示：

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

成功上传后，PHP进程很可能无权将上传的文件移动到不同的文件夹，因为移动包含本质上的复制和删除操作。

还有一件事 - 尽量不要将上传文件的权限更改为0644，因为这也可以限制在PHP进程中，即在Linux上处理文件系统操作时（我假设你使用Linux机器）服务器）工作过程（在你的情况下是PHP和apache）具有特定的权限设置，也许他们没有能力删除/移动工作文件夹之外的文件。您还应将上传文件夹权限更改为755或777。

Answer 3

我按照这个链接......像魅力一样工作..

Upload File to server

希望这能帮到你： - ）

Answer 4

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上传图片的AsyncTask“Helpher”类。要调用此类，请使用下面的语法。

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

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

无法将文件从Android上传到PHP服务器

4 个答案: