java - 如何下载 REST API 响应文件？

Question

我有一个包含以下详细信息的 API:

GET /REST/sql_snapshot/2003-03-01.sql.gz
HTTP/1.1 Host: api.application.cap.cams.net
Authorization: Basic asdwqfasft

The response from API shown below omits the message body, which contains binary compressed SQL data.

HTTP/1.1 200 OK
Date: Wed, 05 Mar 2003 10:19:46 GMT
Server: Apache/1.3.22 (Unix) (Red-Hat/Linux)
Content-Type: application/octet-stream

我最初有这段代码

URL location = new URL("https://api.application.cap.cams.net/REST/sql_snapshot/2003-03-01.sql.gz");
HttpsURLConnection connection = (HttpsURLConnection) location.openConnection();
connection.setHostnameVerifier(HostnameVerifierFactory getHostnameVerifier());
connection.setSSLSocketFactory(SSLConfigurerFactory.getConfigurer().getSSLSocketFactory());
connection.connect();
// after this I will retrieve the input stream of the connection. Then write the file (zip file).

我是否做错了什么，因为我无法获取输入流，因为连接的响应代码是-1。我知道这个响应代码，但我不完全确定我是如何得到这个的。这是从 REST API 调用检索和下载文件的正确方法吗？

Answer 1

就您而言，您只想下载文件，而不必担心进行休息调用。您可以执行以下操作(不需要外部库):

import java.io.InputStream; 
import java.net.URI; 
import java.nio.file.Files; 
import java.nio.file.Paths;
...
public void downloadFile(String url) {
    try (InputStream inputStream = URI.create(url).toURL().openStream()) {
        Files.copy(inputStream, Paths.get(url.substring(url.lastIndexOf('/') + 1)));
    }catch(Exception ex) {
        ex.printStackTrace();
    }
}

用法:

downloadFile("https://api.application.cap.cams.net/REST/sql_snapshot/2003-03-01.sql.gz")

保存:

2003-03-01.sql.gz

这会将文件保存在与您的项目相同的目录中。如果您想将其放置在特定位置，则必须修改 Paths.get(...) 并添加您的输出目录。

What happens here?

Get filename from URL: url.substring(url.lastIndexOf('/') + 1).

In order to download the file, you need to read it first. InputStream.

Once you've read it by URI.create(url).toURL().openStream().

We save what we've read in our stream to disk using Files.copy(...)