How to grep a portion of line

Shell Programming and Scripting
1

Mysql log has something like:

I want to grep only the portion "ernie-1328697839.1233158" from each line. How to do this?

2

Suppose you have the log ina file called "file", here's the code you want to execute:

awk -F "'" '{ print $2 }' file | awk -F "-%" '{ print $1 }'
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3

Now i want a script that will scan the mysql.log file and if it finds 5 consecutive instances of the same line then it would execute certain commands.

4 
if [ `sed "s/.*\(ernie-1328697839.1233158\).*/\1/" mysql.log | wc -l` -ge 5 ]; then echo yes; fi
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5

To be more precise:

1) The script will scan the mysql.log file in real time(something like tail -F mysql.log)
2) If it encounters 5 consecutive identical lines then it would invoke some commands.

Any help will be highly appreciated.

6
  1. Your requirements keep changing over the posts.
  2. One way to achieve requirement mentioned in post #5 is to write a script that will keep scanning the file until a certain condition is met.
7

1) Extremely sorry for that.
2) How? Can you help with the basic logic.