Question

I have a text file like list:

-----> 2012-02-13 19:29:27.325 <http-0.0.0.0-9090-1> at org.jboss.resource.adapter.jdbc.WrappedPreparedStatement.executeQuery(WrappedPreparedStatement.java:236) 
  at org.hibernate.jdbc.AbstractBatcher.getResultSet(AbstractBatcher.java:187) 
  at org.hibernate.loader.Loader.getResultSet(Loader.java:1787)
  .... (many lines starts with at)

5. select * from mytable

-----> 2012-02-13 19:31:27.325 <http-0.0.0.0-9090-1> at 
... (many lines start with at, just like above)

I want to print the block of lines starting with -----> 2012-02-13 if the line starting with 5. select contains the keyword mytable.

How to do that with awk?

Answer 1

Try this:

awk 'BEGIN {RS="----->"} /5\. select \* from mytable/ { printf("%s %s",RS,$0)}' INPUT_FILE

If I had understood correctly what you need (print the block, if 5. select * from mytable is in there). If you want a reversed output, try:

awk 'BEGIN {RS="----->"} ! /5\. select \* from mytable/ { printf("%s %s",RS,$0)}' INPUT_FILE

See it in action here.