如何使用JAX-RS从Java服务器端返回Zip文件？ [英] How can I return a Zip file from my Java server-side using JAX-RS?

问题描述

我想使用JAX-RS从我的服务器端java返回一个压缩文件到客户端。

I want to return a zipped file from my server-side java using JAX-RS to the client.

我尝试了以下代码，

@GET
public Response get() throws Exception {

    final String filePath = "C:/MyFolder/My_File.zip";

    final File file = new File(filePath);
    final ZipOutputStream zop = new ZipOutputStream(new FileOutputStream(file);

    ResponseBuilder response = Response.ok(zop);
    response.header("Content-Type", "application/zip");
    response.header("Content-Disposition", "inline; filename=" + file.getName());
    return response.build();
}

但我得到的例外如下，

SEVERE: A message body writer for Java class java.util.zip.ZipOutputStream, and Java type class java.util.zip.ZipOutputStream, and MIME media type application/zip was not found
SEVERE: The registered message body writers compatible with the MIME media type are:
*/* ->
  com.sun.jersey.core.impl.provider.entity.FormProvider

有什么问题，如何解决这个问题？

What is wrong and how can I fix this?

推荐答案

您正在泽西岛委派如何序列化ZipOutputStream的知识。所以，与您的合作您需要为ZipOutputStream实现自定义MessageBodyWriter。相反，最合理的选择可能是将字节数组作为实体返回。

You are delegating in Jersey the knowledge of how to serialize the ZipOutputStream. So, with your code you need to implement a custom MessageBodyWriter for ZipOutputStream. Instead, the most reasonable option might be to return the byte array as the entity.

您的代码如下所示：

@GET
public Response get() throws Exception {
    final File file = new File(filePath);

    return Response
            .ok(FileUtils.readFileToByteArray(file))
            .type("application/zip")
            .header("Content-Disposition", "attachment; filename=\"filename.zip\"")
            .build();
}

在本例中，我使用来自 Apache Commons IO 将File转换为byte []，但您可以使用其他实现。

In this example I use FileUtils from Apache Commons IO to convert File to byte[], but you can use another implementation.

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