Shell Programming and Scripting

Hello,

I have a list of words.. ranging from 4 to any characters long.. to not more than 20 though.

How can I select only first seven letters of the list of words?


example:-

wwwwwwwwww
eeeee
wererreetf
sdsarddrereewtewt
sdsdsds
sdsd
ereetetttt
ewtwertwrttrttrtrtwtrww


I want that for characters longer than 7 letters.. the results should only should first seven letters.. and for those less than 7 letters it should show as it is.



Please advise.

Code:

echo '123456789' | sed 's/^\(.\{7\}\).*/\1/'

Code:

awk '{print substr( $0,0,7 )}' in_file

Cheers
ZB

Thanks a lot... both works fine...

1) awk '{print substr( $0,0,7 )}' in_file ........ shows first 6 letters..probably 7 should be 8

2) echo '123456789' | sed 's/^\(.\{7\}\).*/\1/' ... works as expected.

awk ' { printf "%-.7s\n",$0 } ' in_file

Prints the first 7 for me...

Code:

$ cat > tmp 
wwwwwwwwww
eeeee
wererreetf
sdsarddrereewtewt
sdsdsds
sdsd
ereetetttt
ewtwertwrttrttrtrtwtrww
$ awk '{print substr( $0,0,7 )}' tmp
wwwwwww
eeeee
wererre
sdsardd
sdsdsds
sdsd
ereetet
ewtwert

Cheers,
ZB

Ok thanks