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it is doing nothing :o

aaaah here I go again :(


VAR_1="123456"
VAR_2="XXXXXX"
/usr/xpg4/bin/awk '/$VAR_1/ && !flag++{print " $VAR_2";next}1' file >output
put me result:

/usr/xpg4/bin/awk: /$VAR_1/: unknown...

THAAAAAAAAAAAANKS A LOT :)

awk way solve my problem :)

thanks again a lot :) you saved lot of my neurons :D

best regards :)

hm under solaris it is still not working... still the same mistake :(

I tryied it under debian linux and there it is running but, script will just put in stdout XXXXXX not in the file...

any...

Thanks, but RESULT: awk: syntax error near line 1

maybe I should mentioned that I`m working under Solaris (SunOS 5.9)... :)

Hi all,

I will be glad if you can help me.

I have this kind of document:

$cat mistake_file
546488
123456
546887
132548
123456
546887
324645
123456
123456

and I want to replace...