how to select lines from file?

[Henk Geurts]

Hi guys..
I have a file with lines like :

aabbddcc1234xxxx
aa1234cc2345yyyy
aa1234xx3456zzzz

How can i select the lines that have certain string in caracters 10-14 ?

[Rita C Workman]

man grep

grep aa1234cc /path/filename

Just a quick thought,
Rita

[Henk Geurts]

oke .. I have to make it more clear..

aabbddcc1234xxxx
aa1234cc2345yyyy
aa1234xx3456zzzz

when i to select the line with caracters 8-11 to be 1234 , grep will also show other lines.. as grep does not use the place of the caracters in the line..

Can awk do this ?
What will be the command.. ?
regards

[sreekanthtm]

Sorry qust: is not clear for me...

[Rita C Workman]

How about:

grep '.......1234*' /path/filename

That's seven dots or periods then 1234 and anything else after that...

Rgrds,
Rita

[john korterman]

Hi Henk,

another quick thought:

grep -E "........1234*" ./infile


regards,
John K.

[James R. Ferguson · Accepted solution]

Hi Henk:

Using your data and using 'awk' [because that's one way] you could do:

# awk '{if (substr($0,9,4)==1234) {print}}' file
aabbddcc1234xxxx

Regards!

...JRF...

[Henk Geurts]

Thanks James..
This was what I was looking for!!!

regards,
Henk

[Dennis Handly]

>when I select the line with characters 8-11 to be 1234

(Typically line positions start with 1, your example seems to be 0.)
An old thread about it:
http://forums.itrc.hp.com/service/forums/questionanswer.do?threadId=76665

Or for your case for columns 9-12.
grep '^.\{8\}1234' file