从安卓设备上传多张图片到PHP服务器

Question

这是一段关于在服务器端使用 PHP 的代码的求助。我对 PHP 没有任何了解，但我需要从 Android 上传三张图片到这个 PHP 页面。

我已经尝试了很多方法并进行了搜索，但没有任何教程或其他东西能够帮助我。我的 Android 代码可以正常工作，DNS 也已经配置好了，但图片在服务器端没有显示出来。请帮我提供 Java 代码。

PHP:

<?php
if ($_FILES["file1"]["error"] > 0)
{
    header("HTTP/1.1 400 Bad Request");
    echo "Error: " . $_FILES["file1"]["error"] . "<br />";
}
else if ($_FILES["file2"]["error"] > 0)
{
    header("HTTP/1.1 400 Bad Request");
    echo "Error: " . $_FILES["file1"]["error"] . "<br />";
}
else if ($_FILES["file3"]["error"] > 0)
{
    header("HTTP/1.1 400 Bad Request");
    echo "Error: " . $_FILES["file1"]["error"] . "<br />";
}
else
{
    if ($_FILES["file1"]["error"] > 0)
    {
        echo "Error: " . $_FILES["file1"]["error"] . "<br />";
    }
    else
    {
        echo "Upload: " . $_FILES["file1"]["name"] . "<br />";
        echo "Type: " . $_FILES["file1"]["type"] . "<br />";
        echo "Size: " . ($_FILES["file1"]["size"] / 1024) . " Kb<br />";
        echo "Stored in: " . $_FILES["file1"]["tmp_name"]. "<br />";
    }

    //$target_path = "uploads/";
    $target_path = "elp/pendingimages/";

    $target_path = $target_path . basename( $_FILES['file1']['name']); 

    if(move_uploaded_file($_FILES['file1']['tmp_name'], $target_path)) {
        echo "The file ".  basename( $_FILES['file1']['name']). 
            " has been uploaded"; 
    } 
    else{
        echo "There was an error uploading the file, please try again!";
    }

    if ($_FILES["file2"]["error"] > 0)
    {
        echo "Error: " . $_FILES["file2"]["error"] . "<br />";
    }
    else
    {
        echo "Upload: " . $_FILES["file2"]["name"] . "<br />";
        echo "Type: " . $_FILES["file2"]["type"] . "<br />";
        echo "Size: " . ($_FILES["file2"]["size"] / 1024) . " Kb<br />";
        echo "Stored in: " . $_FILES["file2"]["tmp_name"]. "<br />";
    }

    //$target_path = "uploads/";
    $target_path = "elp/pendingimages/";

    $target_path = $target_path . basename( $_FILES['file2']['name']); 

    if(move_uploaded_file($_FILES['file2']['tmp_name'], $target_path)) {
        echo "The file ".  basename( $_FILES['file2']['name']). 
        " has been uploaded";
    } 
    else{
        echo "There was an error uploading the file, please try again!";
    }

    if ($_FILES["file3"]["error"] > 0)
    {
        echo "Error: " . $_FILES["file3"]["error"] . "<br />";
    }
    else
    {
        echo "Upload: " . $_FILES["file3"]["name"] . "<br />";
        echo "Type: " . $_FILES["file3"]["type"] . "<br />";
        echo "Size: " . ($_FILES["file3"]["size"] / 1024) . " Kb<br />";
        echo "Stored in: " . $_FILES["file3"]["tmp_name"]. "<br />";
    }


    //$target_path = "uploads/";
    $target_path = "elp/pendingimages/";

    $target_path = $target_path . basename( $_FILES['file3']['name']);  

    if(move_uploaded_file($_FILES['file3']['tmp_name'], $target_path)) {
        echo "The file ".  basename( $_FILES['file3']['name']). 
            " has been uploaded";
    } 
    else{
        echo "There was an error uploading the file, please try again!";
    }
}

?>

Java:

public class TryprojectActivity extends Activity {
    InputStream is;
    int pic_count = 0;
    Bitmap bitmap=null;
    FileInputStream in1,in2,in3;
    BufferedInputStream buf;

@Override
public void onCreate(Bundle icicle) {
    super.onCreate(icicle);
    setContentView(R.layout.main);

        try {
            in1 = new FileInputStream("/sdcard/1.jpg");
        } 
        catch (FileNotFoundException e2) {
        // TODO Auto-generated catch block
            e2.printStackTrace();
        }

        try {
            in2 = new FileInputStream("/sdcard/2.jpg");
        } catch (FileNotFoundException e1) {
        // TODO Auto-generated catch block
        e1.printStackTrace();
    } 

    try {
        in3 = new FileInputStream("/sdcard/3.jpg");
    } 
    catch (FileNotFoundException e1) {
        // TODO Auto-generated catch block
        e1.printStackTrace();
    } 

    Bitmap bitmapOrg1 = BitmapFactory.decodeStream(in1);
    ByteArrayOutputStream bao1 = new ByteArrayOutputStream();
    bitmapOrg1.compress(Bitmap.CompressFormat.JPEG, 90, bao1);
    byte [] imagearray1 = bao1.toByteArray();
    String ba1=Base64.encode(imagearray1);

    Bitmap bitmapOrg2 = BitmapFactory.decodeStream(in2);
    ByteArrayOutputStream bao2 = new ByteArrayOutputStream();
    bitmapOrg2.compress(Bitmap.CompressFormat.JPEG, 90, bao2);
    byte [] imagearray2 = bao2.toByteArray();
    String ba2=Base64.encode(imagearray2);

    Bitmap bitmapOrg3 = BitmapFactory.decodeStream(in3);
    ByteArrayOutputStream bao3 = new ByteArrayOutputStream();
    bitmapOrg3.compress(Bitmap.CompressFormat.JPEG, 90, bao3);
    byte [] imagearray3 = bao3.toByteArray();
    String ba3=Base64.encode(imagearray3);

    List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(3);

    nameValuePairs.add(new BasicNameValuePair("image1",ba1));
    nameValuePairs.add(new BasicNameValuePair("image2",ba2));
    nameValuePairs.add(new BasicNameValuePair("image3",ba3));

    try{
        HttpClient httpclient = new DefaultHttpClient();
        HttpPost httppost = new HttpPost("http://helpdesk.cispl.com/upload_file.php");
        UrlEncodedFormEntity obj = new UrlEncodedFormEntity(nameValuePairs);
        obj.setChunked(true);
        httppost.setEntity(obj);
        HttpResponse response = httpclient.execute(httppost);
        HttpEntity entity = response.getEntity();
        //is = entity.getContent();
        httpclient.getConnectionManager().shutdown(); 
    }
    catch(Exception e){
        //CommonFunctions.writeLOG(ctx.getClass().toString(), e.toString());
        //CommonFunctions.showToast(ctx, "Unable to post captured image file: " +
        //e.toString());
    }
}

- user1160020

---

这些图片必须通过HTTP上传，你是如何进行上传的？（因为你提到了Java） - greut

我是指在Android应用程序中所需的Java Activity代码。我已经尝试使用HTTP和Multipart上传，但没有用。这可能是因为我在不了解PHP的情况下尝试编码。 - user1160020

@user1160020，PHP 代码是正确的。所以不要怪你的 PHP 知识。看起来你的多部分请求中有错误。 - Dmitry Zaytsev

我也尝试过使用multipart，但目前我正在使用这段代码，并且代码中没有错误或其他问题。：S - user1160020

@user1160020 你的问题解决了吗？能否请您告诉我解决方案？你提出的代码看起来和我的完全一样。 - Korhan

Answer 1

看起来你的PHP是正确的。

在你的设备上使用HTTP POST请求和MultipartEntity数据类型。在这里阅读更多信息。

编辑

来自我链接的示例：

您需要下载其他库才能运行MultipartEntity！

1）从http://james.apache.org/download.cgi#Apache_Mime4J下载httpcomponents-client-4.1.zip，并将apache-mime4j-0.6.1.jar添加到您的项目中。

2）从http://hc.apache.org/downloads.cgi下载httpcomponents-client-4.1-bin.zip，并将httpclient-4.1.jar、httpcore-4.1.jar和httpmime-4.1.jar添加到您的项目中。

3）使用下面的示例代码。

    private DefaultHttpClient mHttpClient;
 public ServerCommunication() {
        HttpParams params = new BasicHttpParams();
        params.setParameter(CoreProtocolPNames.PROTOCOL_VERSION, HttpVersion.HTTP_1_1);
        mHttpClient = new DefaultHttpClient(params);
    }
 public void uploadUserPhoto(File image1, File image2, File image3) {
try {
HttpPost httppost = new HttpPost("some url");
MultipartEntity multipartEntity = new MultipartEntity(HttpMultipartMode.BROWSER_COMPATIBLE);  
            multipartEntity.addPart("Title", new StringBody("Title"));
            multipartEntity.addPart("Nick", new StringBody("Nick"));
            multipartEntity.addPart("Email", new StringBody("Email"));
            multipartEntity.addPart("Description", new StringBody(Settings.SHARE.TEXT));
            multipartEntity.addPart("file1", new FileBody(image1));
            multipartEntity.addPart("file2", new FileBody(image2));
            multipartEntity.addPart("file3", new FileBody(image3));
            httppost.setEntity(multipartEntity);
mHttpClient.execute(httppost, new PhotoUploadResponseHandler());
} catch (Exception e) {
            Log.e(ServerCommunication.class.getName(), e.getLocalizedMessage(), e);
        }
    }
 private class PhotoUploadResponseHandler implements ResponseHandler {
@Override
        public Object handleResponse(HttpResponse response)
                throws ClientProtocolException, IOException {
HttpEntity r_entity = response.getEntity();
            String responseString = EntityUtils.toString(r_entity);
            Log.d("UPLOAD", responseString);
return null;
        }
}

- Dmitry Zaytsev

---

我正在将我的三张图片转换为字节数组，然后将其添加到列表中。之后我该怎么做？还是说我走错了路线？ - user1160020

in1 = new FileInputStream("/sdcard/1.jpg"); Bitmap bitmapOrg1 = BitmapFactory.decodeStream(in1); ByteArrayOutputStream bao1 = new ByteArrayOutputStream(); bitmapOrg1.compress(Bitmap.CompressFormat.JPEG, 90, bao1); byte [] imagearray1 = bao1.toByteArray(); String ba1=Base64.encode(imagearray1);我对所有三个图像都执行这些操作。 - user1160020

看看我的例子。采用这种方法，MultipartEntity将为您完成一切。您只需要从文件创建FileBody（并从图像创建文件）。 - Dmitry Zaytsev

reqEntity.addpart(); 命令出错，配置路径... 即使我已经提供了 Apache 和 Http Jar 路径。 - user1160020

类型 org.apache.james.mime4j.message.SingleBody 无法解析。它是从所需的 .class 文件中间接引用的。 - user1160020

@DmitryZaitsev能否回答一下我关于类似问题的提问（http://stackoverflow.com/questions/14567050/uploading-multiple-images-to-server）？但我的图像数据是位图。 - Korhan

Answer 2

首先，您的代码可以优化为以下内容：

$files = array('file1', 'file2', 'file3');
$path = 'elp/pendingimages/';

foreach ($files as $file) {
    if ($_FILES[$file]['error'] > 0) {
        echo 'Error: '. $_FILES[$file]['error'] .'<br />';
    }
    else {
        echo 'Upload: '. $_FILES[$file]['name'] .'<br />';
        echo 'Type: '. $_FILES[$file]['type'] .'<br />';
        echo 'Size: '. ($_FILES[$file]['size'] / 1024) .' Kb<br />';
        echo 'Stored in: '. $_FILES[$file]['tmp_name'] .'<br />';
    }

    $basename = basename($_FILES[$file]['name']);

    if (move_uploaded_file($_FILES[$file]['tmp_name'], $path . $basename) {
        echo "The file {$basename} has been uploaded";
    }
    else {
        echo 'There was an error uploading the file, please try again!';
    }
}

如果您为每个文件使用不同的字段，则没问题。

接下来，您可以看到当进行多文件上传时，$_FILES数组存储了什么：

$_FILES = array(
    ['files'] => array(
        ['name'] => array(
            [0] => 'WALL_video.jpg'
            [1] => 'WALLc.jpg'
        )
        ['type'] => array(
            [0] => 'image/jpeg'
            [1] => 'image/jpeg'
        )
        ['tmp_name'] => array(
            [0] => '/tmp/phpnbKcdM'
            [1] => '/tmp/phpnrHSN1'
        )
        ['error'] => array(
            [0] => 0
            [1] => 0
        )
        ['size'] => array(
            [0] => 885968
            [1] => 839713
        )
    )
)

如果您正在使用名称为files[]的文件数组字段，则以下代码将适用于您。

$target_path = 'elp/pendingimages/';

foreach ($_FILES['files']['name'] as $index => $name) {
    if ($_FILES['files']['error'][$index] > 0) {
        echo 'Error: ' . $_FILES['files']['error'][$index] . '<br />';
    }
    else {
        echo 'Upload: '. $_FILES['files']['name'][$index] .'<br />';
        echo 'Type: '. $_FILES['files']['type'][$index] .'<br />';
        echo 'Size: '. ($_FILES['files']['size'][$index] / 1024) .' Kb<br />';
        echo 'Stored in: '. $_FILES['files']['tmp_name'][$index] .'<br />';
    }

    $path = $target_path . basename($name);

    if (move_uploaded_file($_FILES['files']['tmp_name'][$index], $path) {
        echo "The file {$name} has been uploaded";
    }
    else {
        echo 'There was an error uploading the file, please try again!';
    }
}

- Vlad Stratulat

Answer 3

请查看我的关于向服务器发送多张图片的帖子 点击这里

- Prashant Maheshwari Andro