script substitute with regular expression

[Franky Leeuwerck_1]

Hi,

I need some help on the following script problem.

How can I change a fraction at the beginning of a line in this way :

'Dec 3 09' should be changed into 'Dec 3 09' .
'Dec 12 09' should stay that way.

The first 3 characters ( month) can be different, also the last two digits (day of the month).

Thanks in advance.
Franky

[Franky Leeuwerck_1]

It is not clearly visible in the web page, but the string to be replaced contains two spaces between the month and the day of the of the month number.

Franky

[Mark Grant]

if I understand you correctly, you want to remove an extra space which is between one of those three "words"

If so, then this will return the fields without the extra space

echo "Dec 02 09"| awk '{printf "%s %s %s", $1,$2,$3}'

Please clarify if you need something else.

[Elmar P. Kolkman]

In that case I think you need the following:
sed 's| *| |g'

This will replace all multiple spaces to a single space.

Or, if the month is at the start of the line, it can be done in a better way:
sed 's|^\([A-Za-z]*\) |\1 |g'
(Mind the double space after the closing bracket!)

[Franky Leeuwerck_1]

Hello Mark,

Thanks for your quick reply.

No, that's not want I need. Sorry, for the unclear explanation.

This is what I need. I will mark the space with 'SPACE' to make the example clear.

'DecSPACE3SPACE9' should be replaced with
'DecSPACESPACE3SPACE9'

'DecSPACE10SPACE9' should stay that way.

Franky

[Franky Leeuwerck_1]

Hi Elmar,

See my reply to Mark.

[Mark Grant · Accepted solution]

Ahh, got you :)

In that case

echo "Dec 2 09" | awk '{printf "%s %2s %s", $1,$2,$3}'

Regards

[Franky Leeuwerck_1]

Thanks Mark !
That's what I needed.

Regards,
Franky

[john korterman]

Hi,
try this, using your infile as $1:


#!/usr/bin/sh
typeset -R2 day
while read month day number
do
echo "$month" "$day" "$number"
done <$1

regards,
John K.