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Remove trailing 0 from the field

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# 1  
Old 09-18-2012
Question Remove trailing 0 from the field
Hi Freinds,

I have file1.txt as below

file1.txt

Code:
1521894~~-0.400~201207
1521794~~-0.486~201207
152494~~-0.490~201207
152154894~~-0.490~201207
1521894354~~-0.489~201207

expected output :
Code:
1521894~~-0.4~201207
1521794~~-0.486~201207
152494~~-0.49~201207
152154894~~-0.49~201207
1521894354~~-0.489~201207

Please help. I am very new to unix.
Last edited by Scott; 09-18-2012 at 08:54 AM.. Reason: Code tags
# 2  
Old 09-18-2012
Try:
Code:
awk -F"~" -vOFS="~" '{sub("0*$","",$3)}1' file

This User Gave Thanks to bartus11 For This Post:
i150371485
# 3  
Old 09-18-2012
With sed:
Code:
sed 's/0*\(~[0-9]*\)$/\1/' <file

--
Bye
This User Gave Thanks to Lem For This Post:
i150371485
# 4  
Old 09-18-2012
@Lem ,@bartus11 thanks for reply. it worked good . but if i have a value 9.000 the output which i am getting is 9. but expcted is 9
# 5  
Old 09-18-2012
Building on @Lem's and @bartus11's solutions, try
Code:
awk -F"~" -vOFS="~" '{sub("[.0]*$","",$3)}1' file
or
sed 's/[\.0]*\(~[0-9]*\)$/\1/' <file

btw, Bartus11's suggestion will work if field4 contains e.g. letters while Lem's will fail on that.
This User Gave Thanks to RudiC For This Post:
i150371485
# 6  
Old 09-18-2012
An awk solution:

Code:
 
awk -f b.awk infile

where b.awk:

Code:
 
{
 FS=l="~";
 for (i=1; i<=NF; i++) {
    if ($(i) ~ /[.]/) sub("[.0]*$", "", $(i));
    if (i==NF) l="\n";
    printf $i l;
 }
}

# 7  
Old 09-18-2012
Quote:
Originally Posted by i150371485
@Lem ,@bartus11 thanks for reply. it worked good . but if i have a value 9.000 the output which i am getting is 9. but expcted is 9
This should work better:
Code:
sed 's/\.*0*\(~[^~]*\)$/\1/' <file

--
Bye
This User Gave Thanks to Lem For This Post:
i150371485
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