How do you parse a filename in bash?

Question

I have a filename in a format like:

system-source-yyyymmdd.dat

I'd like to be able to parse out the different bits of the filename using the "-" as a delimiter.

Barn

Answer 1

You can use the cut command to get at each of the 3 'fields', e.g.:

$ echo "system-source-yyyymmdd.dat" | cut -d'-' -f2
source

"-d" specifies the delimiter, "-f" specifies the number of the field you require

Thermolabile

I'm curious why you added the # prompt. Normally, that prompt indicates the root or superuser. In generally, I'd think stuff like trying out the cut command would be better done as a regular user. I'd have used the $ prompt. – Julianejuliann

Oh, yeah - good point. I must admit, I was logged in as root at the time and simply went for it - a bad habit, I know. Having said that, I think echo and cut are two of the least harmful commands :) But, for the sake of completeness, I'll certainly update the example right away. Cheers. – Thermolabile

how can i assign the output to a variable? if i do var = "system-source-yyyymmdd.dat" | cut -d'-' -f2 does not work. – Paradies

Fantastic - an answer in 3 minutes - that's quicker than phoning a friend! - originally added as an answer but that was deleted 6 years after I asked the question :) – Barn

Answer 2

A nice and elegant (in my mind :-) using only built-ins is to put it into an array

var='system-source-yyyymmdd.dat'
parts=(${var//-/ })

Then, you can find the parts in the array...

echo ${parts[0]}  ==> system
echo ${parts[1]}  ==> source
echo ${parts[2]}  ==> yyyymmdd.dat

Caveat: this will not work if the filename contains "strange" characters such as space, or, heaven forbids, quotes, backquotes...

Preconize

Answer 3

Depending on your needs, awk is more flexible than cut. A first teaser:

# echo "system-source-yyyymmdd.dat" \
    |awk -F- '{printf "System: %s\nSource: %s\nYear: %s\nMonth: %s\nDay: %s\n",
              $1,$2,substr($3,1,4),substr($3,5,2),substr($3,7,2)}'
System: system
Source: source
Year: yyyy
Month: mm
Day: dd

Problem is that describing awk as 'more flexible' is certainly like calling the iPhone an enhanced cell phone ;-)

Pursuant

Answer 4

Use the cut command.

e.g.

echo "system-source-yyyymmdd.dat" | cut -f1 -d'-'

will extract the first bit.

Change the value of the -f parameter to get the appropriate parts.

Here's a guide on the Cut command.

Infraction

Answer 5

Another method is to use the shell's internal parsing tools, which avoids the cost of creating child processes:

oIFS=$IFS
IFS=-
file="system-source-yyyymmdd.dat"
set $file
IFS=$oIFS
echo "Source is $2"

Stumble

Answer 6

The simplest (and IMO best way) to do this is simply to use read:

$ IFS=-. read system source date ext << EOF
> foo-bar-yyyymmdd.dat
> EOF
$ echo $system
foo
$ echo $source $date $ext
bar yyyymmdd dat

There are many variations on that theme, many of which are shell dependent:

bash$ IFS=-. read system source date ext <<< foo-bar-yyyymmdd.dat

echo "$name" | { IFS=-. read system source date ext
   echo In all shells, the variables are set here...; }
echo but only in some shells do they retain their value here

Steffy