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Input:

201103 1 /mnt/hdd/PUB/SOMETHING
201102 7 /mnt/hdd/PUB/SOMETH ING
201103 11 /mnt/hdd/PUB/SO METHING
201104 3 /mnt/hdd/PUB/SOMET HING
201106 1 /mnt/hdd/PUB/SOMETHI NG

Desired output:

201103 01 /mnt/hdd/PUB/SOMETHING
201102 07 /mnt/hdd/PUB/SOMETH ING
201103 11 /mnt/hdd/PUB/SO METHING
201104 03 /mnt/hdd/PUB/SOMET HING
201106 01 /mnt/hdd/PUB/SOMETHI NG

How can I add a 0 if there is only a single digit, e.g. 1 in the "day" part? I need this date format: YYYYMM DD.

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5 Answers 5

Reset to default
18

Another solution: awk '{$2 = sprintf("%02d", $2); print}'

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15 
$ sed 's/\<[0-9]\>/0&/' ./infile
201103 01 /mnt/hdd/PUB/SOMETHING
201102 07 /mnt/hdd/PUB/SOMETH ING
201103 11 /mnt/hdd/PUB/SO METHING
201104 03 /mnt/hdd/PUB/SOMET HING
201106 01 /mnt/hdd/PUB/SOMETHI NG
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7
  • Can you explain how this works? This is the first time I'm looking at the \<[0-9]\> construct which I think is the one responsible for matching the single digits but not sure what this construct is called. Thanks.
    – sasuke
    Mar 12, 2011 at 13:16
  • 2
    \< means: start of a 'word' ... [0-9] means a single digit from 0 to 9 ... \> means: end of a 'word' ... word: a token which is whitespace delimited (or begins/ends at start/end of the line, for \< and \> respectively)... PS. I just tried punctuation marks.. they are also delimiters.
    – Peter.O
    Mar 12, 2011 at 13:46
  • 1
    You can also do this without capturing parentheses: & in the replacement string will use the matched LHS -- sed 's/\<[0-9]\>/0&/'
    – glenn jackman
    Mar 12, 2011 at 14:33
  • Oh, wasn't aware that <> is a word boundary in shell regex syntax. Come to think of it, even `sed 's/\b[0-9]\b/0&/' also works. Thank you both. :)
    – sasuke
    Mar 12, 2011 at 15:18
  • @sasuke: <> is a feature of extended regex (not of the shell, as such)... depending on which version and which options you use, sed and the shell can both use either extended or standard regex ... standard regex uses \<\>
    – Peter.O
    Mar 12, 2011 at 17:19
2

Here is a (non-sed) way to use bash with extended regex..
This method, allows scope to do more complex processing of individual lines. (ie. more than just regex substitutions) 

while IFS= read -r line ; do
    if [[ "$line" =~ ^(.+\ )([0-9]\ .+)$ ]]  
    then echo "${BASH_REMATCH[1]}0${BASH_REMATCH[2]}" 
    else echo "$line"
    fi
done <<EOF
201103 1 /mnt/hdd/PUB/SOMETHING
201102 7 /mnt/hdd/PUB/SOMETH ING
201103 11 /mnt/hdd/PUB/SO METHING
201104 3 /mnt/hdd/PUB/SOMET HING
201106 1 /mnt/hdd/PUB/SOMETHI NG
EOF

output:

201103 01 /mnt/hdd/PUB/SOMETHING
201102 07 /mnt/hdd/PUB/SOMETH ING
201103 11 /mnt/hdd/PUB/SO METHING
201104 03 /mnt/hdd/PUB/SOMET HING
201106 01 /mnt/hdd/PUB/SOMETHI NG
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1

I would do something like this:

sed -E 's/ ([0-9]) / 0\1 /' ./input

This grabs lonely numbers, strips them of whitespace with a group ' ([0-9]) ', then places them back in with 0 and whitespace padding ' 0\1 '.

The -E option allows for modern RegEx expressions on OSX (so you don't have to use "\" so often), -r does the same thing on the linux systems I've tested.

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-1 
while read a b c
do 
new_format=$(printf "%02d" $b)
echo "$a $new_format $c"
done </tmp/input
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