Easy One - First 8 Characters Only Please 3

[jmarkus]

How do I get only the first 8 characters of a string to print?

I'm trying:
echo "12345678.msytring.ext" | sed '/[A-Z,a-a,0-9][A-Z,a-a,0-9][A-Z,a-a,0-9][A-Z,a-a,0-9][A-Z,a-a,0-9][A-Z,a-a,0-9][A-Z,a-a,0-9][A-Z,a-a,0-9]/';

But I'm obviously not doing it right.

Thanks,
Jeff

 

[PHV]

echo "$string" | cut -c1-8
echo "$string" | sed 's!^\(.\)\{8\}.*!\1!'
echo "$string" | awk '{print substr($0,1,8)}'
In ksh:
typeset -L8 newstring="$string"

Hope This Help, PH.
Want to get great answers to your Tek-Tips questions? Have a look at FAQ219-2884

 

[Ghodmode]

echo "12345678abcd" | awk '{ print substr($0, 0, 8) }'

--
-- GhodMode

 

[bigoldbulldog]

expr "$string" : "\(.\{8\}\)"

Cheers,
ND

bigoldbulldog@hotmail.com

 

[aigles]

With bash : ${string:0:8}

Jean Pierre.

 

[Ygor]

Or...

expr substr $string 1 8

 

[uptodate]

ksh93 also supports:

#var=1234567890
#print ${var:0:3}
#123