How to split a string in bash delimited by tab

Question

I'm trying to split a tab delimitted field in bash.

I am aware of this answer: how to split a string in shell and get the last field

But that does not answer for a tab character.

I want to do get the part of a string before the tab character, so I'm doing this:

x=`head -1 my-file.txt` echo ${x%\t*} 

But the \t is matching on the letter 't' and not on a tab. What is the best way to do this?

Thanks

Answer 1

If your file look something like this (with tab as separator):

1st-field   2nd-field 

you can use cut to extract the first field (operates on tab by default):

$ cut -f1 input 1st-field 

If you're using awk, there is no need to use tail to get the last line, changing the input to:

1:1st-field     2nd-field 2:1st-field     2nd-field 3:1st-field     2nd-field 4:1st-field     2nd-field 5:1st-field     2nd-field 6:1st-field     2nd-field 7:1st-field     2nd-field 8:1st-field     2nd-field 9:1st-field     2nd-field 10:1st-field    2nd-field 

Solution using awk:

$ awk 'END {print $1}' input 10:1st-field 

Pure bash-solution:

#!/bin/bash  while read a b;do last=$a; done < input echo $last 

outputs:

$ ./tab.sh  10:1st-field 

Lastly, a solution using sed

$ sed '$s/\(^[^\t]*\).*$/\1/' input 10:1st-field 

here, $ is the range operator; i.e. operate on the last line only.

For your original question, use a literal tab, i.e.

x="1st-field    2nd-field" echo ${x%   *} 

outputs:

1st-field 

Answer 2

Use $'ANSI-C' strings in the parameter expansion:

$ x=$'abc\tdef\tghi' $ echo "$s" abc     def     ghi $ echo ">>${x%%$'\t'*}<<" >>abc<<