17

我正在尝试为 ProcessBuilder 对象添加环境变量，但是当我在 ProcessBuilder 中调用该新变量时出现错误。这就是我构建流程的方式

public class OTU
{
    public static void main(String[] args) throws Exception
    {
        ProcessBuilder pb = new ProcessBuilder();
        Map<String, String> env = pb.environment();
        //set environment variable u
        env.put("u", "util/");

        pb.command("echo $u");
        Process p = pb.start();
        String output = loadStream(p.getInputStream());
        String error  = loadStream(p.getErrorStream());
        int rc = p.waitFor();
        System.out.println("Process ended with rc=" + rc);
        System.out.println("\nStandard Output:\n");
        System.out.println(output);
        System.out.println("\nStandard Error:\n");
        System.out.println(error);
    }

    private static String loadStream(InputStream s) throws Exception
    {
        BufferedReader br = new BufferedReader(new InputStreamReader(s));
        StringBuilder sb = new StringBuilder();
        String line;
        while((line=br.readLine()) != null)
            sb.append(line).append("\n");
        return sb.toString();
    }
}

我得到错误

Exception in thread "main" java.io.IOException: Cannot run program "$u": java.io.IOException: error=2, No such file or directory
    at java.lang.ProcessBuilder.start(ProcessBuilder.java:475)
    at ca.utoronto.siseq_1_2.OTU.main(OTU.java:22)
Caused by: java.io.IOException: java.io.IOException: error=2, No such file or directory
    at java.lang.UNIXProcess.<init>(UNIXProcess.java:164)
    at java.lang.ProcessImpl.start(ProcessImpl.java:81)
    at java.lang.ProcessBuilder.start(ProcessBuilder.java:468)
    ... 1 more

如果我只是为此进程设置变量，我不明白为什么会出现错误。请帮助我如何设置环境变量，以便我可以在 ProcessBuilder 中使用它。

4

2 回答 2

23

Alfredo O 的示例为您提供了正确的想法。您需要告诉 ProcessBuilder 使用什么程序来执行您的命令。在这种情况下，bash 带有“-c”开关，它告诉 bash 将接下来的内容（即“echo $u”）解释为命令。

import java.io.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Map;

public class OTU {
    public static void main(String[] args) throws Exception {
        ProcessBuilder pb = new ProcessBuilder("/bin/bash", "-c", "echo $u");
        Map<String, String> env = pb.environment();
        // set environment variable u
        env.put("u", "util/");

        Process p = pb.start();
        String output = loadStream(p.getInputStream());
        String error = loadStream(p.getErrorStream());
        int rc = p.waitFor();
        System.out.println("Process ended with rc=" + rc);
        System.out.println("\nStandard Output:\n");
        System.out.println(output);
        System.out.println("\nStandard Error:\n");
        System.out.println(error);
    }

    private static String loadStream(InputStream s) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(s));
        StringBuilder sb = new StringBuilder();
        String line;
        while ((line = br.readLine()) != null)
            sb.append(line).append("\n");
        return sb.toString();
    }
}

这会产生以下输出：

Process ended with rc=0

Standard Output:

util/


Standard Error:
于 2012-06-26T02:17:39.990 回答
5

这在 Windows 中对我有用：

@Test
public void testProcessBuilder() throws IOException {
    ProcessBuilder processBuilder = new ProcessBuilder("cmd.exe", "/C", "echo Hello %name%");
    Map<String, String> environment = processBuilder.environment();
    environment.put("name", "Alfredo Osorio");
    Process p = processBuilder.start();
    String line;
    BufferedReader r = new BufferedReader(new InputStreamReader(p.getInputStream()));
    while ((line = r.readLine()) != null) {
        System.out.println(line);
    }
    r.close();
}

输出：

你好阿尔弗雷多·奥索里奥

正如您在 Windows 中看到的，您使用 %environmentVariable% 而不是 $environementVariable

于 2012-06-25T23:28:45.440 回答