linux使用Java8发送curl命令

Question

下面是我的代码，我试图使用java将curl命令发送到指定的url，但当我运行代码时，它不会执行并发送curl命令。请帮助修复它或告诉我哪里出错：

String command="curl 'http://ipaddress:port/smshttpquery/qs?REQUESTTYPE=SMSSubmitReq&USERNAME=myapp&PASSWORD=app&PHON=XXXXXXXXX&MESSAGE=TEST&ADDRESS=MYAPP&TYPE=4'";
    ProcessBuilder processBuilder=new ProcessBuilder(command.split(" "));
    try {
        processBuilder.start();
    } catch (IOException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }
}

我想要运行的linux命令是

curl 'http://ipaddress:port/smshttpquery/qs?REQUESTTYPE=SMSSubmitReq&USERNAME=myapp&PASSWORD=app&PHON=XXXXXXXXX&MESSAGE=TEST&ADDRESS=MYAPP&TYPE=4'

Answer 1

假设您向我们展示了所有代码，您将看不到任何结果，这仅仅是因为您的代码没有读取流程的输出流（process.getInputStream（））

然而，在通往成功的道路上还有其他陷阱

Answer 2

public static String HttpCall(String link) throws IOException{
    String response = "";
    String beginPoint = link;
    URL url = new URL(beginPoint);

    try (BufferedReader reader = new BufferedReader(new InputStreamReader(url.openStream(), "UTF-8"))) {
        String line = "";
        while ((line = reader.readLine()) != null) {
            response += line;
        }
        reader.close();
        response = response.replace("\n", "").replace("\r", "");
    }
    return response;

}