Question

通过command = touch aaa传递时，下面的代码可能是什么问题

它抛出/bin/sh: 1: touch aaa: not found

这是代码

boolean isWindows = System.getProperty("os.name")
        .toLowerCase().startsWith("windows");
String [] cmd ={"-c", command};
CommandLine cmdLine = new CommandLine("/bin/sh");
cmdLine.addArguments( cmd,false );
if (isWindows) {
    DefaultExecutor exec = new DefaultExecutor();
    exec.setExitValue(0);
    exec.setWorkingDirectory(new File(System.getProperty("user.home")));
    int exitCode = exec.execute(cmdLine);
    return new Response(exitCode, "");

} else {
    DefaultExecutor exec = new DefaultExecutor();
    exec.setExitValue(0);
    exec.setWorkingDirectory(new File(System.getenv("HOME")));
    int exitCode = exec.execute(cmdLine);
    return new Response(exitCode, "");
}

Answer 1

您确定touch和aaa之间是否有适当的间隔（0x20）？

当您尝试从sh运行带有参数的不存在的命令时，来自shell的“未找到”消息不包含参数：

% sh -c 'touchx aaa'        
sh: 1: touchx: not found
% dash -c 'touchx aaa'      
dash: 1: touchx: not found
% bash -c 'touchx aaa'
bash: touchx: command not found
% busybox sh -c 'touchx aaa'
sh: touchx: not found

尝试从命令行运行没有Java代码的命令，然后查看它是否可以从那里运行。

/ bin / sh：1：找不到触摸

1 个答案: