使用 Android SDK 发布多部分请求
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Post multipart request with Android SDK
提问by jpoz
I'm trying to do something I thought would be relatively simple: Upload an image to a server with the Android SDK. I'm found a lot of example code:
我正在尝试做一些我认为相对简单的事情：使用 Android SDK 将图像上传到服务器。我发现了很多示例代码：
http://linklens.blogspot.com/2009/06/android-multipart-upload.html
http://linklens.blogspot.com/2009/06/android-multipart-upload.html
But neither work for me. The confusion I keep running into is what is really needed to make a multipart request. What is the simplestway to have a multipart upload (with an image) for Android?
但两者都不适合我。我一直遇到的困惑是发出多部分请求真正需要的东西。为 Android 进行分段上传（带有图像）的最简单方法是什么？
Any help or advice would be greatly appreciated!
任何帮助或建议将不胜感激！
回答by vinzenzweber
Update April 29th 2014:
2014 年 4 月 29 日更新：
My answer is kind of old by now and I guess you rather want to use some kind of high level library such as Retrofit.
我的答案现在有点老了，我猜你更愿意使用某种高级库，例如Retrofit。
Based on this blog I came up with the following solution: http://blog.tacticalnuclearstrike.com/2010/01/using-multipartentity-in-android-applications/
基于这个博客，我想出了以下解决方案：http: //blog.tacticalnuclearstrike.com/2010/01/using-multipartentity-in-android-applications/
You will have to download additional libraries to get MultipartEntityrunning!
您必须下载其他库才能MultipartEntity运行！
1) Download httpcomponents-client-4.1.zip from http://james.apache.org/download.cgi#Apache_Mime4Jand add apache-mime4j-0.6.1.jar to your project.
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) Download httpcomponents-client-4.1-bin.zip from http://hc.apache.org/downloads.cgiand add httpclient-4.1.jar, httpcore-4.1.jar and httpmime-4.1.jar to your project.
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) Use the example code below.
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 image) {
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("Image", new FileBody(image));
httppost.setEntity(multipartEntity);
mHttpClient.execute(httppost, new PhotoUploadResponseHandler());
} catch (Exception e) {
Log.e(ServerCommunication.class.getName(), e.getLocalizedMessage(), e);
}
}
private class PhotoUploadResponseHandler implements ResponseHandler<Object> {
@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;
}
}
回答by Muhammad Babar
As MultiPartEntityis deprecated. So here is the new way to do it! And you only need httpcore.jar(latest)and httpmime.jar(latest)download them from Apache site.
由于MultiPartEntity已弃用。所以这里是新的方法来做到这一点！而你只需要httpcore.jar(latest)和httpmime.jar(latest)Apache的网站下载。
try
{
HttpClient client = new DefaultHttpClient();
HttpPost post = new HttpPost(URL);
MultipartEntityBuilder entityBuilder = MultipartEntityBuilder.create();
entityBuilder.setMode(HttpMultipartMode.BROWSER_COMPATIBLE);
entityBuilder.addTextBody(USER_ID, userId);
entityBuilder.addTextBody(NAME, name);
entityBuilder.addTextBody(TYPE, type);
entityBuilder.addTextBody(COMMENT, comment);
entityBuilder.addTextBody(LATITUDE, String.valueOf(User.Latitude));
entityBuilder.addTextBody(LONGITUDE, String.valueOf(User.Longitude));
if(file != null)
{
entityBuilder.addBinaryBody(IMAGE, file);
}
HttpEntity entity = entityBuilder.build();
post.setEntity(entity);
HttpResponse response = client.execute(post);
HttpEntity httpEntity = response.getEntity();
result = EntityUtils.toString(httpEntity);
Log.v("result", result);
}
catch(Exception e)
{
e.printStackTrace();
}
回答by Rohit Mandiwal
Here is the LIGHT WEIGHTED solution which worked for me with no external HTTPCore and such libs. I was facing issue of 64K methods so have no option left to avoid HTTPCore libraries
这是轻量级解决方案，它对我有用，没有外部 HTTPCore 和此类库。我遇到了 64K 方法的问题，所以别无选择来避免 HTTPCore 库
import java.util.List;
import java.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.net.HttpURLConnection;
import java.net.URL;
import java.net.URLConnection;
import java.util.ArrayList;
import java.util.List;
/**
* This utility class provides an abstraction layer for sending multipart HTTP
* POST requests to a web server.
*
* @author www.codejava.net
*/
public class MultipartUtility {
private final String boundary;
private static final String LINE_FEED = "\r\n";
private HttpURLConnection httpConn;
private String charset;
private OutputStream outputStream;
private PrintWriter writer;
/**
* This constructor initializes a new HTTP POST request with content type
* is set to multipart/form-data
*
* @param requestURL
* @param charset
* @throws IOException
*/
public MultipartUtility(String requestURL, String charset)
throws IOException {
this.charset = charset;
// creates a unique boundary based on time stamp
boundary = "===" + System.currentTimeMillis() + "===";
URL url = new URL(requestURL);
httpConn = (HttpURLConnection) url.openConnection();
httpConn.setUseCaches(false);
httpConn.setDoOutput(true); // indicates POST method
httpConn.setDoInput(true);
httpConn.setRequestProperty("Content-Type",
"multipart/form-data; boundary=" + boundary);
httpConn.setRequestProperty("User-Agent", "CodeJava Agent");
httpConn.setRequestProperty("Test", "Bonjour");
outputStream = httpConn.getOutputStream();
writer = new PrintWriter(new OutputStreamWriter(outputStream, charset),
true);
}
/**
* Adds a form field to the request
*
* @param name field name
* @param value field value
*/
public void addFormField(String name, String value) {
writer.append("--" + boundary).append(LINE_FEED);
writer.append("Content-Disposition: form-data; name=\"" + name + "\"")
.append(LINE_FEED);
writer.append("Content-Type: text/plain; charset=" + charset).append(
LINE_FEED);
writer.append(LINE_FEED);
writer.append(value).append(LINE_FEED);
writer.flush();
}
/**
* Adds a upload file section to the request
*
* @param fieldName name attribute in <input type="file" name="..." />
* @param uploadFile a File to be uploaded
* @throws IOException
*/
public void addFilePart(String fieldName, File uploadFile)
throws IOException {
String fileName = uploadFile.getName();
writer.append("--" + boundary).append(LINE_FEED);
writer.append(
"Content-Disposition: form-data; name=\"" + fieldName
+ "\"; filename=\"" + fileName + "\"")
.append(LINE_FEED);
writer.append(
"Content-Type: "
+ URLConnection.guessContentTypeFromName(fileName))
.append(LINE_FEED);
writer.append("Content-Transfer-Encoding: binary").append(LINE_FEED);
writer.append(LINE_FEED);
writer.flush();
FileInputStream inputStream = new FileInputStream(uploadFile);
byte[] buffer = new byte[4096];
int bytesRead = -1;
while ((bytesRead = inputStream.read(buffer)) != -1) {
outputStream.write(buffer, 0, bytesRead);
}
outputStream.flush();
inputStream.close();
writer.append(LINE_FEED);
writer.flush();
}
/**
* Adds a header field to the request.
*
* @param name - name of the header field
* @param value - value of the header field
*/
public void addHeaderField(String name, String value) {
writer.append(name + ": " + value).append(LINE_FEED);
writer.flush();
}
/**
* Completes the request and receives response from the server.
*
* @return a list of Strings as response in case the server returned
* status OK, otherwise an exception is thrown.
* @throws IOException
*/
public List<String> finish() throws IOException {
List<String> response = new ArrayList<String>();
writer.append(LINE_FEED).flush();
writer.append("--" + boundary + "--").append(LINE_FEED);
writer.close();
// checks server's status code first
int status = httpConn.getResponseCode();
if (status == HttpURLConnection.HTTP_OK) {
BufferedReader reader = new BufferedReader(new InputStreamReader(
httpConn.getInputStream()));
String line = null;
while ((line = reader.readLine()) != null) {
response.add(line);
}
reader.close();
httpConn.disconnect();
} else {
throw new IOException("Server returned non-OK status: " + status);
}
return response;
}
}
USAGE
用法
private void uploadMedia() {
try {
String charset = "UTF-8";
File uploadFile1 = new File("/sdcard/myvideo.mp4");
String requestURL = Data.BASE_URL+Data.URL_UPLOAD_REACTION_TEST;
MultipartUtility multipart = new MultipartUtility(requestURL, charset);
// multipart.addHeaderField("User-Agent", "CodeJava");
// multipart.addHeaderField("Test-Header", "Header-Value");
multipart.addFormField("friend_id", "Cool Pictures");
multipart.addFormField("userid", "Java,upload,Spring");
multipart.addFilePart("uploadedfile", uploadFile1);
List<String> response = multipart.finish();
Log.v("rht", "SERVER REPLIED:");
for (String line : response) {
Log.v("rht", "Line : "+line);
}
} catch (Exception e) {
e.printStackTrace();
}
}
PHP Code to accept upload
PHP代码接受上传
<?php
$friend_id = $_REQUEST['friend_id'];
$userid = $_REQUEST['userid'];
echo 'friend_id : '.$friend_id. ' userid '.$userid;
move_uploaded_file($_FILES['uploadedfile']['tmp_name'], "./uploads/".$_FILES["uploadedfile"]["name"]);
?>
回答by Hpsaturn
More easy, light (32k), and many more performance:
更简单、更轻便 (32k) 和更多性能：
Android Asynchronous Http Client library: http://loopj.com/android-async-http/
Android 异步 Http 客户端库：http: //loopj.com/android-async-http/
Implementation:
执行：
How to send a “multipart/form-data” POST in Android with Volley
回答by Jóni
Try this:
尝试这个：
public void SendMultipartFile() {
Log.d(TAG, "UPLOAD: SendMultipartFile");
DefaultHttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost( <url> );
File file = new File("/sdcard/spider.jpg");
Log.d(TAG, "UPLOAD: setting up multipart entity");
MultipartEntity mpEntity = new MultipartEntity(HttpMultipartMode.BROWSER_COMPATIBLE);
Log.d(TAG, "UPLOAD: file length = " + file.length());
Log.d(TAG, "UPLOAD: file exist = " + file.exists());
try {
mpEntity.addPart("datafile", new FileBody(file, "application/octet"));
mpEntity.addPart("id", new StringBody("1"));
} catch (UnsupportedEncodingException e1) {
Log.d(TAG, "UPLOAD: UnsupportedEncodingException");
e1.printStackTrace();
}
httppost.setEntity(mpEntity);
Log.d(TAG, "UPLOAD: executing request: " + httppost.getRequestLine());
Log.d(TAG, "UPLOAD: request: " + httppost.getEntity().getContentType().toString());
HttpResponse response;
try {
Log.d(TAG, "UPLOAD: about to execute");
response = httpclient.execute(httppost);
Log.d(TAG, "UPLOAD: executed");
HttpEntity resEntity = response.getEntity();
Log.d(TAG, "UPLOAD: respose code: " + response.getStatusLine().toString());
if (resEntity != null) {
Log.d(TAG, "UPLOAD: " + EntityUtils.toString(resEntity));
}
if (resEntity != null) {
resEntity.consumeContent();
}
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
}
回答by Chris
I highly recommend Loopj.
我强烈推荐Loopj。
I have successfully used it to upload multiple files at once, including different mime types. Simply do this:
我已经成功地使用它一次上传多个文件，包括不同的 mime 类型。只需这样做：
File myVideo = new File("/path/to/myvideo.mp4");
File myPic = new File("/path/to/mypic.jpg");
RequestParams params = new RequestParams();
try {
params.put("profile_picture", myPic);
params.put("my_video", myVideo);
} catch(FileNotFoundException e) {}
For large or many files you might have to increase the timeout amount else the default timeout is used which might be too short:
对于大文件或许多文件，您可能必须增加超时量，否则使用可能太短的默认超时：
client.setTimeout(500000) //make this the appropriate timeout in milliseconds
Please see this links for a full description of loopj and how to use it, by far the easiest async http library I have come across:
有关 loopj 的完整说明以及如何使用它，请参阅此链接，这是迄今为止我遇到的最简单的异步 http 库：
http://loopj.com/android-async-http/http://loopj.com/android-async-http/doc/com/loopj/android/http/AsyncHttpClient.html
http://loopj.com/android-async-http/ http://loopj.com/android-async-http/doc/com/loopj/android/http/AsyncHttpClient.html
回答by karthik vishnu kumar
Remove all your httpclient, httpmime dependency and add this dependency compile 'commons-httpclient:commons-httpclient:3.1'. This dependency has built in MultipartRequestEntity so that you can easily upload one or more files to the server
删除所有 httpclient、httpmime 依赖项并添加此依赖项compile 'commons-httpclient:commons-httpclient:3.1'。此依赖项已内置于 MultipartRequestEntity 中，因此您可以轻松地将一个或多个文件上传到服务器
public class FileUploadUrlConnection extends AsyncTask<String, String, String> {
private Context context;
private String url;
private List<File> files;
public FileUploadUrlConnection(Context context, String url, List<File> files) {
this.context = context;
this.url = url;
this.files = files;
}
@Override
protected String doInBackground(String... params) {
HttpClient client = new HttpClient();
PostMethod post = new PostMethod(url);
HttpClientParams connectionParams = new HttpClientParams();
post.setRequestHeader(// Your header goes here );
try {
Part[] parts = new Part[files.size()];
for (int i=0; i<files.size(); i++) {
Part part = new FilePart(files.get(i).getName(), files.get(i));
parts[i] = part;
}
MultipartRequestEntity entity = new MultipartRequestEntity(parts, connectionParams);
post.setRequestEntity(entity);
int statusCode = client.executeMethod(post);
String response = post.getResponseBodyAsString();
Log.v("Multipart "," "+response);
if(statusCode == 200) {
return response;
}
} catch (IOException e) {
e.printStackTrace();
}
return null;
}
You can also add the request and response timeout
您还可以添加请求和响应超时
client.getParams().setParameter(CoreConnectionPNames.CONNECTION_TIMEOUT, 10000);
client.getParams().setParameter(CoreConnectionPNames.SO_TIMEOUT, 10000);
回答by VV W
public class Multipart{
private final Map<String, String> headrs;
private String url;
private HttpURLConnection con;
private OutputStream os;
private String delimiter = "--";
private String boundary = "TRR" + Long.toString(System.currentTimeMillis()) + "TRR";
public Multipart (String url, Map<String, String> headers) {
this.url = url;
this.headrs = headers;
}
public void connectForMultipart() throws Exception {
con = (HttpURLConnection) (new URL(url)).openConnection();
con.setRequestMethod("POST");
con.setDoInput(true);
con.setDoOutput(true);
con.setRequestProperty("Connection", "Keep-Alive");
for (Map.Entry<String, String> entry : headrs.entrySet()) {
con.setRequestProperty(entry.getKey(), entry.getValue());
}
con.setRequestProperty("Content-Type", "multipart/form-data; boundary=" + boundary);
con.connect();
os = con.getOutputStream();
}
public void addFormPart(String paramName, String value) throws Exception {
writeParamData(paramName, value);
}
public void addFilePart(String paramName, String fileName, byte[] data) throws Exception {
os.write((delimiter + boundary + "\r\n").getBytes());
os.write(("Content-Disposition: form-data; name=\"" + paramName + "\"; filename=\"" + fileName + "\"\r\n").getBytes());
os.write(("Content-Type: application/octet-stream\r\n").getBytes());
os.write(("Content-Transfer-Encoding: binary\r\n").getBytes());
os.write("\r\n".getBytes());
os.write(data);
os.write("\r\n".getBytes());
}
public void finishMultipart() throws Exception {
os.write((delimiter + boundary + delimiter + "\r\n").getBytes());
}
public String getResponse() throws Exception {
InputStream is = con.getInputStream();
byte[] b1 = new byte[1024];
StringBuffer buffer = new StringBuffer();
while (is.read(b1) != -1)
buffer.append(new String(b1));
con.disconnect();
return buffer.toString();
}
private void writeParamData(String paramName, String value) throws Exception {
os.write((delimiter + boundary + "\r\n").getBytes());
os.write("Content-Type: text/plain\r\n".getBytes());//;charset=utf-8
os.write(("Content-Disposition: form-data; name=\"" + paramName + "\"\r\n").getBytes());
;
os.write(("\r\n" + value + "\r\n").getBytes());
}
}
Then call below
然后在下面调用
Multipart multipart = new Multipart(url__, map);
multipart .connectForMultipart();
multipart .addFormPart(entry.getKey(), entry.getValue());
multipart .addFilePart(KeyName, "FileName", imagedata);
multipart .finishMultipart();
回答by Igor
You can you use GentleRequest, which is lightweight library for making http requests(DISCLAIMER: I am the author):
您可以使用 GentleRequest，它是用于发出 http 请求的轻量级库（免责声明：我是作者）：
Connections connections = new HttpConnections();
Binary binary = new PacketsBinary(new
BufferedInputStream(new FileInputStream(file)),
file.length());
//Content-Type is set to multipart/form-data; boundary=
//{generated by multipart object}
MultipartForm multipart = new HttpMultipartForm(
new HttpFormPart("user", "aplication/json",
new JSONObject().toString().getBytes()),
new HttpFormPart("java", "java.png", "image/png",
binary.content()));
Response response = connections.response(new
PostRequest(url, multipart));
if (response.hasSuccessCode()) {
byte[] raw = response.body().value();
String string = response.body().stringValue();
JSONOBject json = response.body().jsonValue();
} else {
}
Feel free to check it out: https://github.com/Iprogrammerr/Gentle-Request
回答by RecencyEffect
For posterity, I didn't see okhttp mentioned. Related post.
对于后代，我没有看到提到 okhttp。 相关帖子。
Basically you build up the body using a MultipartBody.Builder, and then post this in a request.
基本上，您使用 MultipartBody.Builder 构建主体，然后将其发布到请求中。
Example in kotlin:
kotlin 中的示例：
val body = MultipartBody.Builder()
.setType(MultipartBody.FORM)
.addFormDataPart(
"file",
file.getName(),
RequestBody.create(MediaType.parse("image/png"), file)
)
.addFormDataPart("timestamp", Date().time.toString())
.build()
val request = Request.Builder()
.url(url)
.post(body)
.build()
httpClient.newCall(request).enqueue(object : okhttp3.Callback {
override fun onFailure(call: Call?, e: IOException?) {
...
}
override fun onResponse(call: Call?, response: Response?) {
...
}
})
