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Print selected lines from file in order

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# 1  
Old 08-04-2011
Print selected lines from file in order
I need to extract selected lines from a log file, I can use grep to pull one line matching 'x' or matching 'y', how can I run through the log printing both matching lines in order top to bottom.

i.e

Code:
line 1 xyz - not needed
line 2 User01 - needed
line 3 123 - not needed
line 4 Info - needed
line 5 abc - not needed
line 6 .... - not needed
line 7 .... - not needed
line 8 xyz - not needed
line 9 User02 - needed
line 10 123 - not needed
line 11 Info - needed
line 12 abc - not needed
line 13.... - not needed
line 14 .... - not needed
cont .....

so as described above I only want to print lines containing 'User' and 'Info' in consecutive order for hundreds of users from one log

Thanks
Last edited by radoulov; 08-04-2011 at 08:44 PM.. Reason: Code tags.
# 2  
Old 08-04-2011
Code:
grep -E 'x|y'

# 3  
Old 08-04-2011
MySQL
Quote:
Originally Posted by DGPickett
Code:
grep -E 'x|y'

Thanks for reply, I tried your suggestion

Code:
grep -E 'x|y' xyz.log

I am still only getting 'x' printed, no 'y'

---------- Post updated at 07:57 AM ---------- Previous update was at 07:41 AM ----------

Quote:
Originally Posted by DGPickett
Code:
grep -E 'x|y'

Ignore the previous, it worked fine. Case was causing the problem.

Thanks very much for response.Smilie
Last edited by radoulov; 08-04-2011 at 08:44 PM.. Reason: Code tags.
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