Question

我必须从休息电话下载一个Zip文件并提取其内容（一些PDF文件和一个PNG文件）。

我正在使用Java Spring。

怎么办？

Answer 1

您可以使用Spring的RestTemplate下载文件

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

使用标准java或第三方库提取内容。

示例实用程序，改编自此处：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();
    }
}

然后你就像这样使用这个实用程序：

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

Answer 2

请求主体应该是这样的：

@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 {

在控制器中，你可以这样做：

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();

提取使用HTTP GET请求

2 个答案: