cut column

Shell Programming and Scripting
1

I have a file as below,

$vi myfile

aaa;20071217
bbb;20070404
ccc;20070254
"

if I want to cut the column 9-12 of the first line , the output should be 1217 , can advise how to write a script to get the result ? thx

p.s. can a script that have only ONE line could do that ?

2 
echo 'aaa;20071217' | sed 's/.*\(....\)/\1/'
3

Simply:

cut -c 9-12

or:

awk '{print substr($0, 9, 4)}'

Regards

4

<prompt>$ cut -b 9- myfile > cutfile
<prompt>$ cat cutfile
or
<prompt>$ cat myfile | cut -b 9-

5

thx reply ,

I tried it , the result is as below,

1217
0404
0254

if I only want the get the date in FIRST line , the result is 1217 , can advise how to modify the script ? thx

6 
awk 'NR==1{print substr($0, 9, 4);exit}'

Regards