awk with regex for delimiter

Question

I want to get the first word in every line from a file. Unfortunately a lot of lines begin with space(s). So I try to get the firs word with the following:

awk -F'[ \t]+' '{print $1}' < MyFile.txt, but it's not working. I try this echo " some string: here" | awk -F'[ \t]+' '{print $1}' and the results is blank line (I thing that it prints empty string). So why this is not working? I want to make it works with the awk command and explicitly passed delimiter (with educational purposes)

Thanks in advance.

Answer 1

awk ignores leading blank spaces when assigning fields and the default command is print. So this should work fine:

awk '{print $1}'

Answer 2

You are specifying a field separator for Awk that is made up of at least a space or a tab character.

You feed the string " some string: here" into Awk and ask it to print the first field.

You get no output because the first field is empty.

The fields in this string are

<1:>   <2:some> <3:string:> <4:here>

$ echo '   some string: here' | gawk 'BEGIN {OFS="|";FS="[ \t]+"}{print $1,$2,$3,$4}'
|some|string:|here

It is more visible with comma-separated input data, like

,some,string:,here

Note also that using a regular expression for FS is an extension to Awk, implemented in GNU Awk and Mawk.

Answer 3

In awk, the default field separator -F " " or BEGIN{FS=" "} follows a special convention

• (1) initial spaces and final spaces are ignored
• (2) splits by [ \t]+

Point (1) -- ignoring initial/final spaces -- just apply when field separator is exactly " ".

This behaviour is in fact what we would expect.

The default awk '{print $1}' works find and so does awk -F' ' '{print $1}'.

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There is another similar convention for the input record separator(RS): when RS="" it stands for paragraph separation:

• (1) separator is one or more empty lines
• (2) initial and final empty lines are ignored.

Point (2) -- ignoring initial/final empty lines -- just apply when input record separator is exactly "".