将多个 byte[] 放入单个响应主体中。

huangapple
go评论138阅读模式
英文:

put multiple byte[] into a single reponse body

问题

我试图在响应主体中一次性发送多个文件。我的问题是我无法将多个数组列表连接成一个数组列表，然后稍后能够将其重新分离为多个文件。

这是我的代码（不起作用）：

  1. List<PDDocument> documents = splitter.split(PDDocument.load(documentData));
  2. ArrayList<byte[]> newDocuments = new ArrayList<>();
  3. for (PDDocument doc : documents) {
  4.     ByteArrayOutputStream os = new ByteArrayOutputStream();
  5.     doc.save(os);
  6.     newDocuments.add(os.toByteArray());
  7.     os.close();
  8. }
  10. t.sendResponseHeaders(200, newDocuments.toArray().length);
  11. OutputStream responseBody = t.getResponseBody();
  12. responseBody.write(newDocuments.toArray());
  13. responseBody.close();

所以我的问题是：

如何使用Java 11 HTTP服务器将多个文件发送回单个HTTP响应？

谢谢！

更新：

在借助Joni修复了我的代码后，我面临另一个问题：
生成的Zip文件损坏：

这是代码：

  1. Splitter splitter = new Splitter();
  2. List<PDDocument> documents = splitter.split(PDDocument.load(documentData));
  4. t.sendResponseHeaders(200, 0);
  5. t.getResponseHeaders().set("Content-Type", "application/zip");
  7. OutputStream responseBody = t.getResponseBody();
  8. ByteArrayOutputStream baos = new ByteArrayOutputStream();
  9. ZipOutputStream zos = new ZipOutputStream(baos);
  11. int counter = 1;
  12. for (PDDocument doc : documents) {
  13.     ZipEntry zipEntry = new ZipEntry("document" + counter);
  14.     zos.putNextEntry(zipEntry);
  15.     ByteArrayOutputStream docOs = new ByteArrayOutputStream();
  16.     doc.save(docOs);
  17.     docOs.close();
  18.     zos.write(docOs.toByteArray());
  19.     zos.closeEntry();
  20.     zos.finish();
  21.     zos.flush();
  23.     counter++;
  24. }
  25. zos.close();
  26. baos.close();
  28. responseBody.write(baos.toByteArray());
  29. responseBody.flush();
  30. responseBody.close();
英文:

I'm trying to send multiple files at onces in my Response body.
My issue is that i was not able to concat multiples Array List into one that i'm later able to re seperate into multiple files.

This is my code (that is not working) :

  1.        List&lt;PDDocument&gt; documents =  splitter.split(PDDocument.load(documentData));
  2.             ArrayList&lt;byte[]&gt; newDocuments = new ArrayList&lt;&gt;();
  3.             for (PDDocument doc : documents)
  4.             {
  5.                 ByteArrayOutputStream os = new ByteArrayOutputStream();
  6.                 doc.save(os);
  7.                 newDocuments.add(os.toByteArray());
  8.                 os.close();
  11.             }
  12.             t.sendResponseHeaders(200,newDocuments.toArray().length);;
  13.             OutputStream responseBody = t.getResponseBody();
  14.             responseBody.write(newDocuments.toArray());
  15.             responseBody.close();

So my question is :

How to send back multiple files into a single http reponse using java 11 http server ?

Thank you !

UPDATE :

After fixing my code with the help of Joni i'm facing another issue :
The Zip that is generated is corrupted :

This is the code :

  1.    Splitter splitter = new Splitter();
  2.             List&lt;PDDocument&gt; documents =  splitter.split(PDDocument.load(documentData));
  5.             t.sendResponseHeaders(200, 0);
  6.             t.getResponseHeaders().set(&quot;Content-Type&quot;, &quot;application/zip&quot;);
  8.             OutputStream responseBody = t.getResponseBody();
  9.             ByteArrayOutputStream baos = new ByteArrayOutputStream();
  10.             ZipOutputStream zos = new ZipOutputStream(baos);
  13.             int counter = 1;
  14.             for (PDDocument doc : documents)
  15.             {
  16.                 ZipEntry zipEntry = new ZipEntry(&quot;document&quot; + counter);
  17.                 zos.putNextEntry(zipEntry);
  18.                 ByteArrayOutputStream docOs = new ByteArrayOutputStream();
  19.                 doc.save(docOs);
  20.                 docOs.close();
  21.                 zos.write(docOs.toByteArray());
  22.                 zos.closeEntry();
  23.                 zos.finish();
  24.                 zos.flush();
  26.                 counter++;
  27.             }
  28.             zos.close();
  29.             baos.close();
  32.             responseBody.write(baos.toByteArray());
  33.             responseBody.flush();
  36.             responseBody.close();

答案1

得分: 2

你不能在HTTP响应中发送多个文件。

您可以将多个文件放入一个“压缩存档”文件中，例如ZIP文件，然后发送该文件。例如：

  1. t.getResponseHeaders().set("Content-Type", "application/zip");
  2. t.sendResponseHeaders(200, 0);
  3. OutputStream responseBody = t.getResponseBody();
  4. ZipOutputStream zos = new ZipOutputStream(responseBody);
  5. int counter = 1;
  6. for (PDDocument doc : documents)
  7. {
  8.     ZipEntry zipEntry = new ZipEntry("document" + counter);
  9.     zos.putNextEntry(zipEntry);
  10.     doc.save(zos);
  11.     zos.closeEntry();
  12.     counter++;
  13. }
  14. zos.close();
英文:

You cannot send multiple files in a HTTP response.

What you can do is put multiple files in one "compressed archive" file such as a ZIP file, and send that instead. For example:

  1.         t.getResponseHeaders().set(&quot;Content-Type&quot;, &quot;application/zip&quot;);
  2.         t.sendResponseHeaders(200, 0);
  3.         OutputStream responseBody = t.getResponseBody();
  4.         ZipOutputStream zos = new ZipOutputStream(responseBody);
  5.         int counter = 1;
  6.         for (PDDocument doc : documents)
  7.         {
  8.             ZipEntry zipEntry = new ZipEntry(&quot;document&quot;+counter);
  9.             zos.putNextEntry(zipEntry);
  10.             doc.save(zos);
  11.             zos.closeEntry();
  12.             counter++;
  13.         }
  14.         zos.close();

huangapple
  • 本文由 发表于 2020年8月24日 20:11:42
  • 转载请务必保留本文链接：https://go.coder-hub.com/63560805.html
  • com.sun.net.httpserver
  • file
  • http
  • java
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定