Java ProcessBuilder 传递参数

Question

我有一个带有一个参数的 shell 脚本，如下所示：

测试文件

#!/bin/bash
echo "Shell Demo";
echo "Hello $0";

现在我想使用 ProgressBuilder 执行这个脚本并传递参数。 java代码如下：

 public void testShell() throws Exception {
        String shPath = "./test.sh";
        // want to pass a value "Jack" to shell script
        ProcessBuilder builder = new ProcessBuilder(shPath, "Jack");
        Process result = builder.start();
        result.waitFor();
        BufferedReader stdInput = new BufferedReader(new InputStreamReader(result.getInputStream()));
        String output;
        while ((output = stdInput.readLine()) != null) {
            System.out.println(output);
        }
    }

输出：

Shell Demo
Hello ./test.sh

我想要的输出是：

Shell Demo
Hello Jack

Answer 1

您将要删除该result.waitFor并指定要使用的执行程序，即）bash（您也可以使用 sh），除了您在正确的路径上。

public String[] createExecutionString(String process, String...params) {
    final List<String> executor = new ArrayList<>();
    executor.add("bash"); /* cmd on windows */
    executor.add("-c"); /* /c on windows */
    executor.add(process);
    for (String param : params) {
        executor.add(param);
    }
    return executor.toArray(new String[0]);
}

public void testShell() throws Exception {
    String shPath = "./test.sh";
    // want to pass a value "Jack" to shell script
    ProcessBuilder builder = new ProcessBuilder(createExecutionString(shPath, "Jack"));
    Process result = builder.start();
    BufferedReader stdInput = new BufferedReader(new InputStreamReader(result.getInputStream()));
    String output;
    while ((output = stdInput.readLine()) != null) {
        System.out.println(output);
    }
}