awk command to print only selected rows in a particular column specified by column name

Shell Programming and Scripting
1

Dear All,

I have a data file input.csv like below. (Only five column shown here for example.)

Data1,StepNo,Data2,Data3,Data4
2,1,3,4,5
3,1,5,6,7
3,2,4,5,6
5,3,5,5,6

From this I want the below output

Data1,StepNo,Data2,Data3,Data4
2,1,3,4,5
3,1,5,6,7

where the second column StepNo contents are '1'.
I used the below simple script to get this output.

awk -F, '$2==1' input.csv

But many times the second column is not always StepNo as there are many versions of input files with varying column positions and total number of columns.
So I need a script(with awk) to get my required output by specifying column name instead of referring from column number .

Thanks in advance.
Sidda

2

Try this:

col="Something"
awk -v col="$col" ' NR==1 {for(i=1; i<=NF ; i++){ if($1==col) {break} } ; next}
                          {print $i} '  infile.csv
3 
awk -F, 'NR==1{for(i=1;i<=NF;i++){if($i=="StepNo")x=i;}} NR>1{if($x=="1")print;}' input_file
1 Like
4

Hi Jim,
Something missing in this code to get my required output.
But the one liner code given by msabhi is perfect for my requirement with little change (removing NR>1 in the print section to print my headers also in the output).

Thank you very much any way for your quick reply.
Sidda

5

Edited Jim's solution.....

col="StepNo"
awk -F , -v col="$col" ' NR==1 {for(i=1; i<=NF ; i++){ if($i==col){ n=i ; print; break; } }} {if ( $n == "1" ) { print } }'  file