Using awk to find a string in a file

Question

So I want to grab a string from a file, file contains data:

----------------------------------------------------
Id   Name      CreationDate         Comment
----------------------------------------------------
1    testing    19.10.11             created by jag

2    develop    19.10.12             created by jag

3    array      19.10.12             created by jaguuuu

4    start123   19.10.12             created by akj

I want to grep both start123 but using only start because following number changes from time to time. So it could be start456, start567. But it will start with start****.

This is what I tried so far:

awk '$0 ~ arr{print NR-1 FS b}{b=$0}' arr="start" /filepath
echo "string found : $arr"

Updating: Also I want to extract only start123 from second column which could be in any row from 1-4 or 1-whatever number. Once I got the string "start123", want to store it in an variable. Sorry for not being clear initially.

So if I try to sort it via comment = created by akj and print out start123 still. I think it will be just an && statement. Will something like this work:

arr=$(awk -v str=start '$2 ~ "^" str "[0-9]*" { print $2 }' /filepath)
 if [ -z "$arr" ]
 then echo "string not found"
 else echo "String found: $arr"
 fi

It is not working for some reason. Any help would be appreciated.

Thanks

Kyle

"grep" is the name of a program that will do what you want. Is there a reason you are using awk? (which will also do what you want, but it's not as easy as using grep) — Dale Wilson
– Dale Wilson, Commented Sep 9, 2014 at 20:58
The example that I referred was using awk, so I have being trying to get this to work. But I don't know what I am doing wrong. I want to know where I am going wrong. — kyle
– kyle, Commented Sep 9, 2014 at 21:00

Barmar · Accepted Answer · 2014-09-09 21:42:45Z

2

You can use the -o option of grep to print just the part of the file that matches the regular expression. In this case, it's start followed by any number of digits.

str=start
if arr=$(grep -o "$str[0-9]*" /filepath)
then echo "string found: $arr"
else echo "string not found"
fi

If you only want to find it in column 2, you can use awk:

arr=$(awk -v str=start '$2 ~ "^" str "[0-9]*" { print $2; exit; }' /filepath)
if [ -z "$arr" ]
then echo "string not found"
else echo "String found: $arr"
fi

edited Sep 9, 2014 at 21:42

answered Sep 9, 2014 at 20:59

Barmar

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10 Comments

kyle Over a year ago

I want to print start123 and store that in a variable.

Barmar Over a year ago

Maybe you should say something like that in the question? There's nothing about printing just the second column or assigning to a variable in there.

kyle Over a year ago

actually my bad again. As I mentioned in the code there could be string such as start123, start456, Can I sort it via comment? I tried as I updated the code. Also once found should stop, It just continues to run. Wonder why???

Barmar Over a year ago

I tried my code and it worked. It matches start followed by any sequence of digits.

kyle Over a year ago

yes, it prints everything that contains start***. I am trying to tweak it where lets say it has start123, start456, start566. Once it hits the first one. for example: start123. it should store it in an variable and stop from going to other one.

|

glenn jackman · Accepted Answer · 2014-09-10 00:17:06Z

1

If you have GNU awk:

gawk 'match($0, /(start[0-9]+)/, m) {print m[1]; exit}'

documentation: http://www.gnu.org/software/gawk/manual/html_node/String-Functions.html#String-Functions

answered Sep 10, 2014 at 0:17

glenn jackman

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2 Comments

kyle Over a year ago

Instead of file path can I have variable?? arr=$(awk -v str=start '$2 ~ "^" str "[0-9]*" { print $2; exit; }' /filepath)

glenn jackman Over a year ago

sure: gawk -v str=start 'match($0, "(" str "[0-9]+", m) {print m[1]; exit}' /filepath

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