Re: Shell grep help - how to get only first occurrence of match?

• Subject: Re: Shell grep help - how to get only first occurrence of match?
• From: "Mark J. Reed" <email@hidden>
• Date: Wed, 28 Mar 2007 13:57:24 -0400
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If you have columns of data separated by arbitrary amounts of
whitespace, the command you're looking for is awk, rather than grep.

awk '{print $1}' will print the first column of every line.  $2 for
the second, etc.
$NF (short for "Number of Fields") for the last column, $(NF-1) for
the second-to-last, etc.
$0 is the whole line.

you can stick /pattern/ in front of the {...} block to run it only for
lines matching that pattern, e.g.

awk '/[0-9]/ {print $1}'  will print the first column of all lines
that contain a digit anywhere in the line.  Note that this means it is
never necessary to do grep | awk. :)

I have been trying really hard to find the proper grep syntax for what I
guess is really simple, but I just can't get it right

The text I like to grep is made up of two or more columns numbers separated
by one or more spaces. The following columns may consist of any combination
of numbers and characters .

What I need to do is to get only the first column, which unfortunately may,
or may not, begin with spaces, like this:

12345    647890   kdghjblrttrep 9845698476 lkjlfkdj

So by writing

grep "^[0-9]*" -o

I get all the first columns of those lines that doesn't begin with spaces,
but if I try to include one of more spaces like this

grep  "^ *[0-9]*"  -o

it gives my all columns containing numbers, besides what I really want is
the numbers of the first columns but without the spaces

Can anyone please help ?

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Richard Rönnbäck
Publishing-konsult
email@hidden
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SWEDEN
+46 (0) 8 444 06 08
+46 (0) 70 774 06 08
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Mark J. Reed <email@hidden>

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