Reputation: 377
I have a file that looks like:
>ref_frame=1
TPGIRYQYNVLPQGWKGSPAIFQSSMTKILEPFRKQNPDIVIYQYMDDLYVGSD
>ref_frame=2
HQGLDISTMCFHRDGKDHQQYSKVA*QKS*SLLENKIQT*LSINTWMICM*DLT
>ref_frame=3
TRD*ISVQCASTGMERITSNIPK*HDKNLRAF*KTKSRHSYLSIHG*FVCRI*
>test_3_2960_3_frame=1
TPGIRYQYNVLPQGWKGSPAIFQSSMTKILEPSRKQNPDIVIYQYMDDLYVGSD
I want to assign a bash variable so that echo $variable
gives test_3_2960
The line/row that I want to assign the variable to will always be line 7. How can I accomplish this using bash?
so far I have:
variable=`cat file.txt | awk 'NR==7'`
echo $variable
= >test_3_2960_3_frame=1
Upvotes: 1
Views: 91
Reputation: 15246
No pipes needed here...
$: variable=$(awk -F'[>_]' 'NR==7{ OFS="_"; print $2, $3, $4; exit; }' file)
$: echo $variable
test_3_2960
-F
is using either >
or _
as field separators, so your data starts in field 2.
OFS="_"
sets the Output Field Separator, but you could also just use "_" instead of commas.
exit
keeps it from wasting time bothering to read beyond line 7.
Upvotes: 2
Reputation: 218
If you wish to continue with awk
$ variable=$(awk 'NR==7' file.txt | awk -F "[>_]" '{print $2"_"$3"_"$4}')
$ echo $variable
test_3_2960
Upvotes: 1
Reputation: 11207
Using sed
$ variable=$(sed -En '7s/>(([^_]*_){2}[0-9]+).*/\1/p' input_file)
$ echo "$variable"
test_3_2960
Upvotes: 3