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: Andrei Tchijov <email@hidden>

Date: Wed, 28 Mar 2007 13:19:33 -0400

Delivered-to: email@hidden

There are other useful commands which can come to the rescue (echo command used just to generate test data):

1) To remove all spaces at the beginning of the string

echo -e " aaa aaa\nbbb bbb\n ccc ccc" | sed -e "1,\$s/^[ ]*//"

2) To "normalize" spaces (many to one )

echo -e " aaa aaa\nbbb bbb\n ccc ccc" | sed -e "1,\$s/^[ ]*//" | sed -e "1,\$s/ [ ]*/ /g"

3) To select 1st colum

echo -e " aaa aaa\nbbb bbb\n ccc ccc" | sed -e "1,\$s/^[ ]*//" | sed -e "1,\$s/ [ ]*/ /g" | cut "-d\ " -f1

4) Optionally, if you need to remove duplicates

... | sort -u

NOTE: If your want to compare data as numbers, use "sort -n -u".

On Mar 28, 2007, at 11:53 AM, Richard Rönnbäck wrote:

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
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References:
	>Shell grep help - how to get only first occurrence of match? (From: Richard Rönnbäck <email@hidden>)