Shell script help - almost got it...

[sc0rpi0n]

Hello everyone!

Funny thing happened to me... I had a script I worked on and one day my HD crashed and I lost it... (yes I know I should of had a backup lol). So now I'm trying to recreate this script but I can't seem to remember how I managed to get a certain function (command) to work... Maybe someone can help me out... Here are the details:

I have the following block of script...

#!/bin/bash

sn=5
var1=`echo $1 |wc -L`
res=$(($sn-$var1))

echo -n $1 > tmp1
COUNTER=1

while [ $COUNTER -lt $res ]
do
printf " " >> tmp1
let COUNTER+=1
done

Basically what this will do is if $1 passed to the script from is for example: "John" it will calculate and add 1 white space after "John" into the tmp1 file.

Now my question is, what if $1 was more than 5 characters? Say for example: "Amanda" That would be 6 characters. What my script, originally did, was just take the first 5 characters if the word was longer than (in this case the sn= variable) 5. So the result in the tmp1 file would be "Amand".

I vaguely remember using "sed" to accomplish this, but I could be wrong...

Can someone shed some light on how I can possibly do this? Man I can't believe I lost the damn script lol

Thanks in advance!!

Scorp

[Cabhan]

You may have used sed to do this, but there is a simpler way, built into Bash directly.

It's called a substring. We can specify that we only want the first five characters:

#!/bin/bash

sn=5
var=${1:0:$sn}

linelen=`echo "$var" |wc -L`
res=$(($sn-$linelen))

echo -n "$var" > tmp1
COUNTER=1

while [ $COUNTER -lt $res ]
do
    printf " " >> tmp1
    let COUNTER+=1
done

The bolded lines are lines that I changed. Basically, ${var:start:len} takes a substring of length len from starting index start (indices are 0-based). So here, I say that I want the $sn characters beginning at position 0. And I get only them.

Does that make sense?

[sc0rpi0n]

Hey

Thanks for the reply! Much appreciate it... I'm a little confused...

see my comments below:

#!/bin/bash

sn=5
var=${1:0:$sn} linelen=`echo "$var" |wc -L`
^^^ I'm assuming this is not a one liner...
res=$(($sn-$linelen))

echo -n "$var" > tmp1
COUNTER=1

while [ $COUNTER -lt $res ]
do
    printf " " >> tmp1
    let COUNTER+=1
done

In all of that where is my "name"? When I ran the script all I got was 4 white spcaces the tmp1 file... I'm not quite sure how this is suppose to work.

Thanks!

Scorp

[Kieren]

Using what Cabhan has said here are some examples:

ksearle@plyw9x214:~> var=kieren
ksearle@plyw9x214:~> echo ${var:0:3}
kie
ksearle@plyw9x214:~> echo ${var:0:9}
kieren

Hope this helps

[Franklin52]

With printf you can accomplish this with 1 line:

#!/bin/bash

printf "%-5.5s\n" $1 >> tmp1

Regards