"sed -n expression " fails. Why? - Shell Programming and Scripting

grahamb · #1 (**permalink**) 12-14-2005

Hi Everyone!

I am trying to extract records between date makers in a log file. My selection string doesn't work, though. Please help.

This test works, if I manually enter the two date-strings (output=28 lines):

Code:

# sed -n '/\[20051213\]/,/\[20051214\]/p' ./access_log

But if the date string is a variable, it will not work:
eg1 (output=ZERO lines):

Code:

# DF="20051213"; DT="20051214"; export DF DT
# sed -n '/\[`echo ${DF}`\]/,/\[`echo ${DT}`\]/p' ./access_log

eg2, using --file= (output= sed error message):

Code:

# DF="20051213"; DT="20051214"; export DF DT
# echo "'/\[${DF}\]/,/\[${DT}\]/'" > /tmp/sed_file
# cat /tmp/sed_file
# '/\[20051213\]/,/\[20051214\]/'  ## note: contents of /tmp/sed_file
# sed -n --file=/tmp/sed_file ./tangofreaks_access_log
# sed: file /tmp/sed_file line 1: Unknown command: `''

What is going wrong???

???
Please help! Thanks

GrahamB

vgersh99 · #2 (**permalink**) 12-14-2005

and this? - no need for 'echo':

Code:

DF="20051213"; DT="20051214"; sed -n "/\[${DF}\]/,/\[${DT}\]/p" ./access_log

grahamb · #3 (**permalink**) 12-14-2005

Thanks vgersh99 !
I would never have dreamt to use the sed string with out the single quotes. Well it worked;

thank you

GrahamB

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