[SOLVED] bash extracting with sed or grep

doni · 03-20-2015, 09:34 AM

Hi Guys
One more question:

awk '/ORA/{print RS,$1;for(i=1;i<=NF;i++)if($i ~ /ORA/)print $i}' FS='\n' RS='SQL>' logfile

Does anyone know how can I pass variable to awk - instead of ORA I would like to have variable from bash:

grep ORA logfile.txt | while read err; do

awk '/$err/{print RS,$1;for(i=1;i<=NF;i++)if($i ~ /$err/)print $i}' FS='\n' RS='SQL>' logfile

done

Tried with awk -v but didn't work

grail · 03-20-2015, 09:41 AM

Well you could do all of that in an awk script instead, but to pass variables you can look at the -v option

danielbmartin · 03-20-2015, 10:06 AM

Quote:

Originally Posted by doni

... Does anyone know how can I pass variable to awk ...
Tried with awk -v but didn't work

awk -v does work. Maybe you made a keying error. Once again, you are asking a question without giving enough background. Show us what you tried. Tell us what you mean by "didn't work." That vague wording could mean ...
- you got a syntax error
- you got no output
- you got incorrect output
- your computer "hung"
- you smelled smoke and the monitor went blank

Daniel B. Martin

ntubski · 03-20-2015, 10:11 AM

You can't use awk variables in regexp literals, you have to use Computed Regexps instead.

Code:

awk -vvar=foo '/var/'     # this searches for the string "var"
awk -vvar=foo '$0 ~ var'  # this searches for the string "foo"
# if you're just searching for a simple string, you don't need a regexp as such:
awk -vvar=foo '$0 == var' # this searches for the string "foo"