如何在Java中使用cURL

问题描述

寻找一种简单的方法来在Java中复制以下Linux cUrl命令:

Looking for an easy way to replicate the following Linux cUrl command in java:

我需要通过HTTP/Curl将文件"/home/myNewFile.txt"上传到Http服务器(在这种情况下为人工制品或)

I need to upload the file "/home/myNewFile.txt" via HTTP / Curl to a Http server (which in this case is artifact or)

curl -u myUser:myP455w0rd! -X PUT "http://localhost:8081/artifactory/my-repository/my/new/artifact/directory/file.txt" -T /home/myNewFile.txt

curl -u myUser:myP455w0rd! -X PUT "http://localhost:8081/artifactory/my-repository/my/new/artifact/directory/file.txt" -T /home/myNewFile.txt

提前谢谢！

推荐答案

首先，将URLConnection强制转换为HttpURLConnection.

First, cast your URLConnection to an HttpURLConnection.

• 对于curl的-X选项，请使用 setRequestMethod .
• 对于curl的-T选项，请使用 setDoOutput(true)， getOutputStream()和 Files.copy .
• 对于curl的-u选项，设置Authorization 请求标头到"Basic "(包括空格)，后跟 user + ":" + password的base 64编码形式.这是 RFC 2616:HTTP 1.1 规范和 RFC 2617:HTTP身份验证.

• For curl’s -X option, use setRequestMethod.
• For curl’s -T option, use setDoOutput(true), getOutputStream(), and Files.copy.
• For curl’s -u option, set the Authorization request header to "Basic " (including the space) followed by the base 64 encoded form of user + ":" + password. This is the Basic Authentication Scheme described in the RFC 2616: HTTP 1.1 specification and RFC 2617: HTTP Authentication.

总而言之，代码如下所示:

In summary, the code would look like this:

URL url = new URL("http://localhost:8081/artifactory/my-repository/my/new/artifact/directory/file.txt");

HttpURLConnection conn = (HttpURLConnection) url.openConnection();

String auth = user + ":" + password;
conn.setRequestProperty("Authorization", "Basic " +
    Base64.getEncoder().encodeToString(
        auth.getBytes(StandardCharsets.UTF_8)));

conn.setRequestMethod("PUT");
conn.setDoOutput(true);
try (OutputStream out = conn.getOutputStream()) {
    Files.copy(Paths.get("/home/myNewFile.txt"), out));
}

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