can not use unix $variable in Fixed search of awk command

Question

I can not use unix $variable in Fiexd search of awk command.

Please see below my commands.

a="NEW_TABLES NEW_INSERT"
b="NEW"
echo $a | awk -v myvar=$b -F'$0~myvar' '{print $2}'

is not returning any output

but if manually enter the $b value there , its working as below

echo $a | awk -v -F'NEW' '{print $2}'

    outputs: 
 TABLES NEW_INSERT

I didn't downvote... Also, it would be good to give a sample of what's in the $a variable. — fedorqui
– fedorqui, Commented Feb 28, 2014 at 16:51
your use of awk is somewhat unorthodox. it would be better if you showed inputs explained what you are trying to achieve. — just somebody
– just somebody, Commented Feb 28, 2014 at 16:57
Also it would be important to indicate what is your expected output. — fedorqui
– fedorqui, Commented Feb 28, 2014 at 16:59

fedorqui · Accepted Answer · 2014-02-28 16:56:02Z

2

This should make it:

$ a="NEW_TABLES NEW_INSERT"

$ echo $a | awk -F"NEW_" '{print $2}'
TABLES 

$ b="NEW_"
$ echo $a | awk -F"$b" '{print $2}'
TABLES

answered Feb 28, 2014 at 16:56

fedorqui

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2 Comments

if i use semi colon ';' in the awk it throws error. awk -F'\;' '{print $1}' Output : awk: warning: escape sequence \;' treated as plain ;'

No need to escape the ;. Use directly awk -F';'

jaypal singh · Accepted Answer · 2014-02-28 17:04:05Z

1

Your quotings are all messed up and you can use your variable to split the line using split function:

a="NEW_TABLES NEW_INSERT"
b="NEW"
echo $a | awk -v myvar="$b" '{split($0,ary,myvar);print ary[2]}'

Outputs:

_TABLES

answered Feb 28, 2014 at 16:57

jaypal singh

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