2016-05-02 81 views

回答

0

你可以使用Spring的RestTemplate下載文件

RestTemplate templ = new RestTemplate(); 
byte[] downloadedBytes = templ.getForObject(url, byte[].class); 

用標準java或第三方庫提取內容。

實施例實用,適於從這裏:http://www.codejava.net/java-se/file-io/programmatically-extract-a-zip-file-using-java

package com.test; 

import java.io.BufferedOutputStream; 
import java.io.ByteArrayInputStream; 
import java.io.File; 
import java.io.FileOutputStream; 
import java.io.IOException; 
import java.util.zip.ZipEntry; 
import java.util.zip.ZipInputStream; 

public class ZipHelper { 

    private static final int BUFFER_SIZE = 4096; 

    public static void unzip(byte[] data, String dirName) throws IOException { 
     File destDir = new File(dirName); 
     if (!destDir.exists()) { 
      destDir.mkdir(); 
     } 
     ZipInputStream zipIn = new ZipInputStream(new ByteArrayInputStream(data)); 
     ZipEntry entry = zipIn.getNextEntry(); 

     while (entry != null) { 
      String filePath = dirName + File.separator + entry.getName(); 
      if (!entry.isDirectory()) { 
       // if the entry is a file, extracts it 
       extractFile(zipIn, filePath); 
      } else { 
       // if the entry is a directory, make the directory 
       File dir = new File(filePath); 
       dir.mkdir(); 
      } 
      zipIn.closeEntry(); 
      entry = zipIn.getNextEntry(); 
     } 
     zipIn.close(); 
    } 

    private static void extractFile(ZipInputStream zipIn, String filePath) throws IOException { 
     BufferedOutputStream bos = new BufferedOutputStream(new FileOutputStream(filePath)); 
     byte[] bytesIn = new byte[BUFFER_SIZE]; 
     int read = 0; 
     while ((read = zipIn.read(bytesIn)) != -1) { 
      bos.write(bytesIn, 0, read); 
     } 
     bos.close(); 
    } 
} 

然後使用此工具像這樣:

ZipHelper.unzip(downloadedBytes, "/path/to/directory"); 
+0

謝謝,我需要解碼響應? r認爲圖書館使用了? – KPN24

+0

對不起,你是什麼意思「r認爲」? –

+0

對不起,'or',類似那樣?:Inflater decompressor = new Inflater(); \t \t \t decompressor.setInput(file); \t \t \t ByteArrayOutputStream bos = new ByteArrayOutputStream(file.length); \t \t \t byte [] buf = new byte [1024]; \t \t \t而{ \t \t \t詮釋計數= decompressor.inflate(BUF)(decompressor.finished()!); \t \t \t bos.write(buf,0,count); \t \t \t} \t \t \t bos.close(); \t \t byte [] decompressedData = bos.toByteArray(); – KPN24

0

的requesdt體應該是這樣的:

@RequestMapping(value = "/push/{id}", method = RequestMethod.POST, consumes = MediaType.APPLICATION_OCTET_STREAM_VALUE, produces=MediaType.APPLICATION_JSON_VALUE) 
@ResponseStatus(HttpStatus.OK) 
public String pushTransactions(@PathVariable("id") String id, @RequestBody byte[] str) throws MessagingException, JsonParseException, JsonMappingException, IOException { 

,並在控制器,你可以這樣做:

GZIPInputStream gis = new GZIPInputStream(new ByteArrayInputStream(str)); 
     ByteArrayOutputStream out = new ByteArrayOutputStream(); 
     int len; 
     while ((len = gis.read(buffer)) > 0) { 
      out.write(buffer, 0, len); 
     } 

     gis.close(); 
     out.close(); 
+0

@RequestBody字節[] SRT拋出異常:該註釋是不允許的此位置。 – KPN24

+0

@FrancescoPerfetti再次檢查我的示例。我已編輯 – Neron

+0

無論如何感謝您的關注。 – KPN24