Substituting variable value in AWK /start/,/stop/


 
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Old 04-16-2009
Quote:
Originally Posted by whomi
Hi vgersh99
I have tried with, as u have suggestd bt its giving me below error..........

Apr 16
Apr 15
oracheck.sh: Apr 15: not found
oracheck.sh: Apr 16: not found
awk: syntax error near line 1
awk: bailing out near line 1



awk -v var="$YESTERDAY" -v var1="$TODAY" '$0~var,$0~var1' $HOME/alert_out.log > $HOME/alert_work.log
awk -v var=`"$YESTERDAY"` -v var1=`"$TODAY"` '/$0~var/,/$0~var1/' $HOME/alert_out.log > $HOME/alert_work.log

I have tried out with both this optionss.....
but in both the cases its giving me same error
the 'oracheck.sh' errors have nothing to do with the question you're asking originally - check the $PATH.
As far as 'awk' errors are concerned - as always, if you're on Solaris, use either /usr/bin/nawk or /usr/xpg4/bin/awk instead of plain old '/usr/bin/awk'.
Good luck.
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