Getting stacktrace in logger

74,316

Solution 1

You have to use the two argument form

log.error("my logging message", exception)

See http://www.devdaily.com/blog/post/java/how-print-exception-stack-trace-using-log4j-commons for more details.

Solution 2

Change your logging statement to:

log.error("Exception is: ", e);

Solution 3

It is actualy log4j that prevents the printing of the fulltime stacktrace. You should however set the exception as a second parameter for the error method.

Solution 4

If you use the below than e.toString() will be called which calls e.getMessage()

log.error("Exception is:::" + e);

However, to print full stack trace you can use the below:

log.error("Exception is:::" + ExceptionUtils.getStackTrace(e));

where ExceptionUtils is imported as org.apache.commons.lang3.exception.ExceptionUtils

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Geek
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Geek

Updated on July 09, 2022

Comments

  • Geek
    Geek almost 2 years

    I am using log4j to log my exceptions. I want to log whatever I get in e.printStackTrace();
    My code looks like this:

    try {
    
    } catch(Exception e) {
        log.error("Exception is:::" + e);
    }
    

    But the content I get logged looks like this:

    2012-02-02 12:47:03,227 ERROR [com.api.bg.sample] - Exception in unTech:::[Ljava.lang.StackTraceElement;@6ed322
    2012-02-02 12:47:03,309 ERROR [com.api.bg.sample] - Exception is :::java.lang.IndexOutOfBoundsException: Index: 0, Size: 0
    

    But the content I expect is:

    java.io.IOException: Not in GZIP format
    at java.util.zip.GZIPInputStream.readHeader(Unknown Source)
    at java.util.zip.GZIPInputStream.<init>(Unknown Source)
    at java.util.zip.GZIPInputStream.<init>(Unknown Source)
    at com.api.bg.sample.unGZIP(sample.java:191)
    at com.api.bg.sample.main(sample.java:69)
    

    I tried e.getMessage(), e.getStackTrace(); however I don't get the full stacktrace. Any suggestions?