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Shell Programming and Scripting

Test decimal number

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# 1  
Old 07-31-2009
Test decimal number
Hi,

I would like test if a number is a decimal number or not
# 2  
Old 07-31-2009
one way
Code:
isdec()
{
    echo "$1" | tr -dc '[:digit:]'  | read  value
    if [[ $(( ${#1} - ${value} )) -eq 1 ]] ; then   # all numbers except one character
          [[ echo "$1" | grep -q '\.' ]] && echo 'ok' || echo 'not ok'  # one char is .
    fi
}
var=zz3
isdec $var
var=1.234
isdec $var

# 3  
Old 07-31-2009
Try...
Code:
echo $number | awk '$0==$0+0 && match($0,0)!=1 {print "Decimal number"}'

$0==$0+0 checks that its a number(but not Hexadecimal) and match($0,0)!=1 checks its not octal.
# 4  
Old 07-31-2009
Hi, i try something like this :
Code:
nombre=13.5

if (test $nombre = {*.*}); then
	echo "ici"
else
	echo "la bas"
fi
exit 0;

But it doesn't works..... Whatever value I put in nombre, it returns "la bas". I think I don't have the correct syntaxe to say " a string composed of blablabla.blablabla" (where blabla are digits)
I tried
Quote:
[*.*]
,
Quote:
{*.*}
and many other things, but it still don't works.

so is it possible to do what I want to do? If yes, could someone tell me the right way to write it?

regards
# 5  
Old 07-31-2009
Sorry, i would like know , after a division , if result is a number integer or not,

( 300 and 301 are value)

echo 300 / 2 | bc -l
150.00000000000000000000
--> it 's ok

echo 301 / 2 |bc -l
150.50000000000000000000
--> it's ko
# 6  
Old 07-31-2009
Use pattern matching
if [ `echo $num / 2 | bc -l | grep -c "[0-9]*\.00000000000000000000" ` = 1 ]
then
echo "OK"
else
echo "KO"
fi
# 7  
Old 07-31-2009
Something like this?

Code:
echo <number> | awk '$0-int($0){print $0 " = KO";next}{print $0 " = OK"}'
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