在java中执行ProcessBuilder后获取。class文件名和路径

Question

{"code": "class Demo{  \r\n" + 
            "    public static void main(String args[]){  \r\n" + 
            "     System.out.println(\"Hello Java\");  \r\n" + 
            "    }  \r\n" + 
            "}",
"language": "java"}

ProcessBuilder processBuilder = new ProcessBuilder(new String[] { "javac", fullPath });
//fullPath is C:\temp\BNkHZk.java
Process process = processBuilder.start();

processBuilder = new ProcessBuilder(new String[] { "javac", fullPath });
Process process = processBuilder.start();
            if (process.getErrorStream().read() != -1) {
                System.out.println("Compilation Errors" + process.getErrorStream());
            }
            BufferedReader reader = new BufferedReader(new InputStreamReader(process.getInputStream()));
            String line = "";

            while ((line = reader.readLine()) != null) {
                System.out.Println(line);
            }

    private void ExecuteCode(String language, String code){
    String tempFileName = RandomStringUtils.randomAlphanumeric(6);
    // Code to creating the file and save under "C:\\temp\\"
    String fullPath = "C:\\temp\\" + tempFileName+ "." + language;
    try {
            fileWriter = new FileWriter(new File(fullPath));
            fileWriter.write(code);
        } catch (Exception e) {
            System.out.println("IO Exception while creating new file");
            e.printStackTrace();
        } finally {
            if (fileWriter != null) {
                fileWriter.close();
            }
        }
    
    ProcessBuilder processBuilder = new ProcessBuilder(new String[] { "javac", fullPath });
                ////fullPath is C:\temp\BNkHZk.java
            Process process = processBuilder.start();

            if (process.getErrorStream().read() != -1) {
                BufferedReader reader = new BufferedReader(new InputStreamReader(process.getErrorStream()));
                String line = "";

                while ((line = reader.readLine()) != null) {
                    System.out.println(line);
                }
            }
 //the above execution will create a new **Demo.class** file.
            int exitcode = -1;
            try {
                exitcode = process.waitFor();
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
            if (exitcode == 0) {
                (
                    processBuilder  = new ProcessBuilder(new String[] { "java","tempFileName"});//BNkHZk
                        Process process1 = processBuilder.start();
                        BufferedReader reader = new BufferedReader(new InputStreamReader(process.getInputStream()));
                        String line = "";

                        while ((line = reader.readLine()) != null) {
                            System.out.println(line + "\n");
                        }
            }
            
}

注意：由于我将以字符串的形式获取文件内容，所以不能保存与.class名称相同的文件名。是否有任何变通方法将文件名保存为。类名？？任何变通建议都是赞赏和欢迎的。提前谢了。

Answer 1

有两个不同的问题。第一个是类文件的命名。类的名称必须与文件的名称匹配。类demo->Demo.java，这就是为什么不能使用随机文件名。编译运行良好，但不是运行。

第二个是ProccessBuilder的处理。我建议采用以下模式：

ProcessBuilder pb = new ProcessBuilder("javac.exe", "Demo.java");
pb.directory(Paths.get("path_to_java_file").toFile());
pb.redirectErrorStream(true);
Process process = pb.start();
try (var infoStream = process.getInputStream()) {
    infoStream.transferTo(System.out);
}
status = process.waitFor();

现在您可以看到，如果编译运行正确。