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Shell Programming and Scripting

Count number of digits in a word

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# 8  
Old 01-06-2009
Code:
#!/bin/ksh

WORD=abcd1234

[[ $(expr $WORD : '.*\([0-9]\+\)')  -eq 5 ]] && echo OK || echo KO

# 9  
Old 01-06-2009
Code:
$ WORD=abcd1234
$ count=${#WORD}
$ echo $count
8

# 10  
Old 01-06-2009
awesome!
AWESOME Replies!!!

This worked in bash
Quote:
Originally Posted by vgersh99
Code:
#!/bin/ksh

WORD=abcd1234

[[ $(echo "${WORD}" | awk '{print gsub("[0-9]", "")}')  -eq 5 ]] && echo OK || echo KO

Also Good Idea...

Quote:
Originally Posted by cfajohnson
Code:
$ WORD=abcd1234
$ count=${#WORD}
$ echo $count
8

Thank you...
# 11  
Old 01-06-2009
Code:
#! /usr/bin/perl

$str="asf123afsdfjlk1243ljk4356jkl57";
@arr=$str=~m/[0-9]+/g;
print join "|",@arr;

# 12  
Old 01-07-2009
Quote:
Originally Posted by ./hari.sh
AWESOME Replies!!!

This worked in bash
Code:
[[ $(echo "${WORD}" | awk '{print gsub("[0-9]", "")}')  -eq 5 ]] && echo OK || echo KO


In bash you don't need an external command for that:

Code:
WORD=abcd1234
temp=${WORD//[!0-9]/}
echo ${#temp}

# 13  
Old 01-07-2009
Quote:
Originally Posted by cfajohnson

In bash you don't need an external command for that:

Code:
WORD=abcd1234
temp=${WORD//[!0-9]/}
echo ${#temp}

Great!! Smilie
Thank you!
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