java - ProcessBuilder 中的逃逸空间

Question

我无法在 Eclipse 中执行以下代码:

public static void main(String[] arg){
    String path="C:\\Users\\my dir\\SendMailPS.ps1";
    ProcessBuilder processBuilderObject
                = new ProcessBuilder("powershell",path);
    try {
        processBuilderObject.start();
    } catch (IOException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }
}

但是如果路径为C:\\Users\\SendMailPS.ps1，我就能够执行它。那么问题出在空格上，我该如何解决这个问题？

编辑:我也尝试过这样

public static void main(String[] arg){
    String path="C:\\Users\\my dir\\SendMailPS.ps1";
    try {
        Runtime.getRuntime().exec("powershell "+path);
    } catch (IOException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }
}

但是没有用。然后我直接从命令提示符尝试

>powershell
> C:\Users\SendMailPS.ps1

这给了我输出。但下面的行给了我错误

>powershell
> C:\Users\my dir\SendMailPS.ps1

错误:

C:\Users\my : The term 'C:\Users\my' is not recognized as the name of a cmdlet, function, script file, or operable program. Check the spelling of the name, or if a path was included, verify that the path is correct and try again.

Answer 1

String path="C:\\Users\\my dir\\SendMailPS.ps1";
ProcessBuilder processBuilderObject
            = new ProcessBuilder("powershell",path);

您实际上在这里做的是运行一个调用 SendMailPS 脚本的单行 powershell 脚本。一行脚本受到 powershell 脚本解析的影响，这导致了您的问题。

尝试以这种方式运行脚本:

String path="C:\\Users\\my dir\\SendMailPS.ps1";
ProcessBuilder processBuilderObject
            = new ProcessBuilder("powershell", "-File", path);

这明确告诉 Powershell 将指定的文件作为脚本运行。

这里不要使用字符串连接:

// Don't do this
ProcessBuilder processBuilderObject
            = new ProcessBuilder("powershell -File " + path); // Don't do this
// Don't do this

尝试以这种方式运行它会给您带来更多麻烦。