[Q] simple shell command question ..

[Tony, Lim]

I would like to split some characters that have a 9 digit, and same character - 1 "m", 8 numeric characters such as
$ cat test.txt
m12345678
m76543210
m09876543
m98765432
...

I would like to split this 9 digit to 8 digit.
(want to remove last character such as
m1234567
m7654321
...

Would you recommand any command to do this ?
like awk, sed etc ...

I am not sure what option is available to this at awk.

Thanks in advance.

[Pete Randall]

What about the cut command? Man cut for details.


Pete

[James R. Ferguson]

Hi Tony:

TMTOWTDI (There's More Than One Way To Do It). Here's one using shell commands and 'cut':

while read LINE
do
echo ${LINE}|cut -c1-8
done < filein > fileout

Regards!

...JRF...

[Luk Vandenbussche]

cut -c1-8 test.txt > output.txt

[James R. Ferguson]

Hi (again) Tony:

For fun, and brevity, you could do:

# perl -pli.old -e 'chop' file

This reads and chops the last character of every line (which is what you wanted); makes a backup copy of your input file as "file.old" and replaces "file" with the modified copy.

Regards!

...JRF...

[Peter Nikitka]

Hi,

don't know what your statement about 'm' means - I guess only strings starting with 'm' and length 9 should be modified.

awk '$1 ~ /^m/ && length($1)==9 {print substr($1,1,8);next}
{print}' test.txt

mfG Peter

[Sandman!]

How about giving sed a try...

# sed '/^m/s/\(.*\).$/\1/g' test.txt

OR with awk as...

# awk '{sub(/.$/,"");print $0}' test.txt

cheers!

[Sandman!]

Update to my last post as I forgot to include the conditional evaluation that the string should start with an "m" and should have exactly nine characters.

sed way...

# sed -n '/^m.\{8\}$/ s/\(.*\).$/\1/p' test.txt

awk way...

# awk '$0~/^m/ && length($0)==9 {sub(/.$/,"");print $0}' test.txt

cheers!

[Sandman!]

Yet another oops! as the string should have exactly nine characters, starting with "m" and followed by eight numerics...so here's the final version:

sed version...

# sed -n '/^m[0-9]\{8\}$/ s/\(.*\).$/\1/p' test.txt

awk version...

# awk '$0~/^m[0-9]{8}$/ {sub(/.$/,"");print $0}' test.txt

hope it helps!!!

[curt larson_1]

just do it in the shell:

while read var restOfLine
do
print "${var%?}"
done < yourFile