Java-使用Runtime.getRuntimeexec无法启动临时文件启动的进程

问题描述

通过传递一组环境变量来解决。

解决方法

启动的进程Runtime.getRuntime().exec(cmdLine,envp,workingDirectory);无法创建临时文件。

从mvn启动报价：

 Caused by: java.io.IOException: �ܾ���ʡ�
    at java.io.WinNTFileSystem.createFileExclusively(Native Method)
    at java.io.File.createTempFile(File.java:1879)

完整日志

演示代码段遇到相同的错误。

        String mavenPath = "D:\\Progs\\springsource\\apache-maven-3.0.4\\bin\\mvn.bat";
        String mavenOptions  = "-X compile exec:java -Dexec.mainClass=runclass.Runme";

        String[] cmdLine = new String[2];
        cmdLine[0] = mavenPath;  //cmdLine.add(mavenPath);
        cmdLine[1] = mavenOptions;      //cmdLine.add(mavenOptions+" compile exec:java -Dexec.mainClass="+packageClass);

        String[] envp = new String[2];
        //Map<String,String> envm = new HashMap<String,String>();
        envp[0] = "JAVA_HOME=" + System.getProperty("java.home"); //System.getenv("JAVA_HOME");
        envp[1] = "M2_HOME=" + System.getenv("MAVEN_HOME");

        File workingDirectory = null;
        String currentDir = new File(".").getAbsolutePath();
        log(currentDir);
        String userDir = System.getProperty("user.dir"); //User working directory ; "user.home"     User home directory
        workingDirectory = new File(userDir);       
        log(workingDirectory.toString());

        //
        Runtime rt = Runtime.getRuntime();
        Process proc = rt.exec(cmdLine,workingDirectory);
        InputStream stdout = proc.getInputStream();
        InputStream stderr = proc.getErrorStream();
        InputStreamReader isr = new InputStreamReader(stdout);
        InputStreamReader isr2 = new InputStreamReader(stderr);
        BufferedReader br = new BufferedReader(isr);
        BufferedReader br2 = new BufferedReader(isr2);

更新：

传递TMP和TEMP环境变量无济于事。
通过 null而不是envp也无济于事。

如果envp为null，则子进程继承当前进程的环境设置。

java java