shell script for extracting line of file using awk

Question

I want, the selected lines of file to be print in output file side by side separated by space. Here what I have did so far,

for file in SAC*

do

awk 'FNR==2 {print $4}' $file >>exp 

awk 'FNR==3 {print $4}' $file >>exp

awk 'FNR==4 {print $4}' $file >>exp

awk 'FNR==5 {print $4}' $file >>exp

awk 'FNR==7 {print $4}' $file >>exp

awk 'FNR==8 {print $4}' $file >>exp

awk 'FNR==24 {print $0}' $file >>exp 

done

My output is:

XV

AMPY

BHZ

2012-08-15T08:00:00

2013-12-31T23:59:59

I want output should be

XV AMPY BHZ 2012-08-15T08:00:00  2013-12-31T23:59:59

Modify all print $4 to printf "%s ",$4 in your awk command — CWLiu
– CWLiu, Commented Jul 10, 2017 at 9:20

James Brown · Accepted Answer · 2017-07-10 12:17:55Z

1

First the test data (only 9 rows, tho):

Then the awk. No need for that for loop in shell, awk can handle multiple files:

$ awk '
BEGIN { 
    ORS=" "
    a[2];a[3];a[4];a[5];a[7];a[8]  # list of records for which $4 should be outputed
}
FNR in a { print $4 }              # output the $4s
FNR==9 { printf "%s\n",$0 }        # replace 9 with 24
' file file # ...                  # the files you want to process (SAC*)
24 34 44 54 74 84 1 2 3 94
24 34 44 54 74 84 1 2 3 94

answered Jul 10, 2017 at 12:17

James Brown

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