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Print a number up to last significant digit after decimal point

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Top Forums Shell Programming and Scripting Print a number up to last significant digit after decimal point
# 8  
Old 08-04-2012
Code:
 echo "$x" | awk '{sub("0*$","",$0);sub("\\.$",".0",$0)}1'

# 9  
Old 08-04-2012
Thanks burtas11. That serves the purpose.
# 10  
Old 08-04-2012
burtas11 has shown you how to do it with awk.

For sed try:
Code:
echo "$x" | sed '/[1-9]0*$/s/0*$//
/[.]0*$/s/0*$/0/'

And, at least for ksh, this should work in a shell script:
Code:
y="${x%%*([0])}"
if [ "$y" == "${y%.}." ]
then	y="${y}0"
fi
echo $y

# 11  
Old 08-04-2012
In bash, according to what you like better in case of round integers (most important parts in bold):

Code:
r=0.9000
s=10.0000
t=91.0900
shopt -s extglob

for i in $r $s $t; do echo ${i%%?(.)*(0)}; done
0.9
10
91.09

for i in $r $s $t; do a=${i%%*(0)}; echo ${a/%./.0}; done
0.9
10.0
91.09

--
Bye
# 12  
Old 08-05-2012
Quote:
Originally Posted by bartus11
Code:
 echo "$x" | awk '{sub("0*$","",$0);sub("\\.$",".0",$0)}1'

This works fine. But when I use
Code:
x=`echo "$x" | awk '{sub("0*$","",$0);sub("\\.$",".0",$0)}1'`

I get the following error message.

awk: warning: escape sequence `\.' treated as plain `.'

and also I don't get the expected value when I do echo $x
Last edited by Franklin52; 08-06-2012 at 03:50 AM.. Reason: Please use code tags for data and code samples
# 13  
Old 08-05-2012
When you're using command substitution with commands that contain single quotes, you need to know that the quoting rules change. $(...) is almost always easier to get right without having to think about it too much as opposed to `...`.
The command:
Code:
x=$(echo "$x" | awk '{sub("0*$","",$0);sub("\\.$",".0",$0)}1')

will do what you wanted.
# 14  
Old 08-05-2012
Quote:
Originally Posted by Don Cragun
When you're using command substitution with commands that contain single quotes, you need to know that the quoting rules change. $(...) is almost always easier to get right without having to think about it too much as opposed to `...`.
The command:
Code:
x=$(echo "$x" | awk '{sub("0*$","",$0);sub("\\.$",".0",$0)}1')

will do what you wanted.
Yes, it works now. Thanks. Smilie
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