[SOLVED] bash string manipulation

Garrett85 · 01-07-2013, 08:25 AM

echo ${NUM:0:2}

The above is just an example of the kind of code I'M using. I need a way to represent the end of the string. Is there a way to start from the right side of a line of text instead of the left? What I'M trying to do here is grab the last three digits on end, then the next three and so on. NUM in this case will be a user input so there is no way to hard code the end position. Thanks.

Guttorm · 01-07-2013, 08:44 AM

Code:

NUM=123456789
DIGITS=3
echo ${NUM: -$DIGITS}

Garrett85 · 01-07-2013, 11:13 AM

Thanks. Now I need help on the following.

OLDSTRING=$(( ${NUM}-${NEWSTRING} )) # I need to set OLSTRING to NUM - NEWSTRING

My syntax is all wrong here if someone could give me a hand. Thanks.

Quote:

Originally Posted by Guttorm

Code:

NUM=123456789
DIGITS=3
echo ${NUM: -$DIGITS}

colucix · 01-07-2013, 11:15 AM

Quote:

Originally Posted by Garrett85

OLDSTRING=$(( ${NUM}-${NEWSTRING} )) # I need to set OLSTRING to NUM - NEWSTRING

My syntax is all wrong here if someone could give me a hand. Thanks.

Why do you think the syntax is wrong? It looks correct to me. In the arithmetic operator you can omit the $ sign to evaluate variables, but it's not mandatory. What the problem is exactly?

Guttorm · 01-07-2013, 11:20 AM

Not sure if I understand, but

Code:

echo ${NUM: 0: -$DIGITS}

will give you the string without the last 3 characters.

Garrett85 · 01-07-2013, 12:51 PM

Nevermind I got it.

NEWSTRING=${NUM: -$DIGITS}
OLDSTRING=${NUM:0:${NEWSTRING}}

Okay, now I see that it is working but with unintended results. I was hoping to set OLDSTRING minues the digits in new string but instead I'M getting OLDSTING - 1. If I were to input 1000 I would want NEWSTRING to hold 000 and old string to hold everything that is left over to the left, in this case, 1. Any ideas?

Code:

 for n in ${PROCESSGROUPS};
        do
            DIGITS=3
            echo ${NUM: -$DIGITS}
            NEWSTRING=${NUM: -$DIGITS}
            OLDSTRING=$(( ${NUM}-${NEWSTRING} )) # I need to set OLSTRING to NUM - NEWSTRING
            echo ${OLDSTRING}
        done

Quote:

Originally Posted by colucix

Why do you think the syntax is wrong? It looks correct to me. In the arithmetic operator you can omit the $ sign to evaluate variables, but it's not mandatory. What the problem is exactly?

danielbmartin · 01-07-2013, 02:33 PM

[QUOTE=Garrett85;4864801]

Quote:

If I were to input 1000 I would want NEWSTRING to hold 000 and old string to hold everything that is left over to the left, in this case, 1.

This may help...
Code:

Code:

num=1234
echo; echo "num=" $num
let remainder=($num/1000)
echo; echo "remainder=" $remainder
let modulus=($num%1000)
echo; echo "modulus=" $modulus

Test results:

Code:

num= 1234

remainder= 1

modulus= 234

Daniel B. Martin

konsolebox · 01-07-2013, 09:34 PM

Quote:

Originally Posted by Garrett85

If I were to input 1000 I would want NEWSTRING to hold 000 and old string to hold everything that is left over to the left, in this case, 1. Any ideas?

I think Guttorm already gave the idea:

Code:

NEWSTRING=${NUM:(-3)}

OLDSTRING=${NUM:0:(-3)}
# or
OLDSTRING=${NUM:0:(-${#NEWSTRING})}