Question

我有一个处理JAVA代码的文件，在处理文件时，我想记录所有文件处理细节，比如文件中没有这些行，它有一些＆＃34; Hello＆＃34;正在处理每个文件的单独日志文件中的字符串。

我在这里尝试的代码工作正常，但如果我有2个文件，如file1.txt和file2.txt。记录器在log_file1.log中记录file1.txt详细信息然后记录log_file2.log中的file2.txt它保留log_file1.log内容也在log_file2.log中。

FILE1.TXT

Hello How are you
Hello How are you

log_file1.log

it has 2 lines and 2 Hello Strings

FILE2.TXT

Hello How are you
Hello How are you
Hello How are you

log_file2.log

it has 2 lines and 2 Hello Strings
it has 3 lines and 3 Hello Strings

我试过的代码：

public class DummyLog
{
    private static PatternLayout patternLayout = new PatternLayout("%d{ISO8601}\t%p\t%c\t%m%n");

    public static Logger getLogger(String fileName, Class clazz) throws Exception
        {
            fileName = "D:/New folder/log-properties/" + fileName + ".log";
            Logger logger = Logger.getLogger(clazz);
            FileAppender appender = new DailyRollingFileAppender(patternLayout,
                            fileName, "'.'yyyy-MM-dd");
            logger.addAppender(appender);
            logger.setLevel(Level.DEBUG);
            return logger;
        }
}

主要类别：

public class TestDummyLog
{
    public static void main(String args[]) throws Exception
    {
        for(int i = 2;i > 0;i--)
        {           
                 String loggerName = "Log1_"+new SimpleDateFormat("yyyyddMMHHmmssSSS").format(new Date());
                 Logger logger = DummyLog.getLogger(loggerName, TestDummyLog.class);
                 logger.info("Joseph Micheal TestLogger:" +new SimpleDateFormat("yyyyddMMHHmmssSSS").format(new Date()));
                 new TestLoggerChild().getChileStuff(loggerName);
                 Thread.sleep(6000);
        }

    }
}

儿童班：

public class TestLoggerChild
{
public static Logger logger = null;
public void getChileStuff(String fileName) throws Exception
{
    logger = DummyLog.getLogger(fileName,TestLoggerChild.class);
    logger.info("Child Sample info message"+new SimpleDateFormat("yyyyddMMHHmmssSSS").format(new Date()));
    logger.info("Class :::::: TestLoggerChild.class");
    logger.debug("fileName @@@@@@@@@@@@@:" + fileName);
    logger.info("Sample info message");
    logger.warn("Sample warn message");
    logger.error("Sample error message");
    logger.info("Sample info message ");
    logger.info("Class :::::: TestLoggerChild.class");
}
}

Answer 1

是否需要使用相同的类别（TestLoggerChild.class）？如果你传递自定义参数化字符串，它必须工作：DummyLog.getLogger（fileName，“Category”+ i）;

Log4j创建动态文件

1 个答案: