我想以编程方式从Java程序启动Maven发行版。该网页显示了一般的操作方法。这就是我所做的：

    final URL url = new URL("http://jenkins/job/MyProject/m2release/submit");

    final HttpURLConnection urlConnection = (HttpURLConnection) url.openConnection();
    urlConnection.setRequestMethod("POST"); 
    urlConnection.setDoOutput(true);

    final String userpass = "User:Password";
    final String authentication = "Basic " + DatatypeConverter.printBase64Binary(userpass.getBytes());
    urlConnection.setRequestProperty("Authorization", authentication);

    try (final OutputStream os = urlConnection.getOutputStream();
            final BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(os, "UTF-8"));) {
        writer.write("releaseVersion=1.2.3&");
        writer.write("developmentVersion=1.2.4-SNAPSHOT&");
        // writer.write("isDryRun=on&"); // uncomment for dry run
        writer.write("scmUsername=User&");
        writer.write("scmPassword=Password&");
        writer.write("scmCommentPrefix=[test]&");
        writer.write("json={\"releaseVersion\":\"1.2.3\",\"developmentVersion\":\"1.2.4-SNAPSHOT\",\"isDryRun\":false}&");
        writer.write("Submit=Schedule Maven Release Build");

        writer.flush();
    }

    urlConnection.connect();

    try (BufferedReader reader = new BufferedReader(
            new InputStreamReader(urlConnection.getInputStream(), "UTF-8"))) {
        for (String line; (line = reader.readLine()) != null;) {
            System.out.println(line);
        }
    }

该论坛建议“在发布之前，请先查看一下表格，您就应该能够提出cURL请求”，这就是我所做的。发行至少开始。

我现在不知道如何逃避一切。浏览器将空格显示为“ +”，但是如果我以这种方式发送数据，它将无法正常工作。实际上，“”，“ +”或“％20”都不用作空格。

我仍然可以Unable to commit files进入构建，因此我可以肯定用户名/密码/注释前缀有问题。即使已发送登录，Jenkins本身也会返回登录页面（“需要身份验证”）。

在Jenkins上触发Maven版本的正确方法是什么？