Shell Programming and Scripting

I need help in removing a leading zero in a particular position.
For eg.: XYZ*04567472*0099*020091231*0123*0.12

In the above line, I want to replace "*0123" with "123" and "0.12" with ".12". I want to remove the leading zero only in position number 4 and 5 (the bolded segments)

I was able to replace 0.12 with .12 using --- sed 's/\*00*\./\*\./g'

But, I am unable to replace "0123" with "123". I tried using
sed '/^XYZ\*/s/\*00*/\*/4'. But it is not working.
Using, sed '/^XYZ\*/s/\*00*/\*/g' removes all leading zeros (which I dont want).

Any help with this is much appreciated.

Thanks,
Ananth.

Code:

sed 's/^\([^*][^*]*[*][^*][^*]*[*][^*][^*]*[*][^*][^*]*\)[*]00*\([^*][^*]*\)[*]00*\(\..*\)/\1*\2*\3/'

I think that should work with most any sed variant.

Try this:

Code:

xx='XYZ*04567472*0099*020091231*0123*0.12'
echo $xx | sed 's/\(\([^\*]*\*\)\{4\}\)0*\([^\*]*\*\)0*\(.*\)/\1\3\4/'

cjcox, thanks, but sorry it is not working. :-(. I tried replacing \1\2\3 with other combinations, but still does not work.
edidataguy, yes it does work. I have a couple of questions:
1. (\([^\*]*\*\)\{4\}\) --> This means that we are searching for 4 occurrences of "*" from the beginning. After this we search for "0". But what is the last "*" for?
([^\*]*\*\ <---
2. Why do you use \1\3\4, why not \1\2\3\4?

---------- Post updated at 07:56 PM ---------- Previous update was at 07:41 PM ----------

Unfortunately, when I was testing it, I ran into some issues.

XYZ*004567472*0099*0020091231*00001*00012.50 became
XYZ*004567472*00*0020091231*1*12.50 (lost the 99 in the 2nd segment)

XYZ*03988651*099*020091231*01*012.50 became
XYZ*03988651*09*020091231*1*12.50

XYZ*03676802*00099*20091231*001*0012.50 became
XYZ*03676802*00*20091231*1*12.50 (seems to be retaining only 2 digits in the 2nd segment)

try awk.. anyone can make it shorter?

Code:

-bash-3.2$ cat test
XYZ*04567472*0099*020091231*0123*0.12
-bash-3.2$ awk -F'*' '{printf $1"*"$2"*"$3"*"$4"*"substr($5,2)"*"substr($6,2)}' test
XYZ*04567472*0099*020091231*123*.12
-bash-3.2$

This will need a lot of expl. to make. Google the net.
I will try my level best to explain.

1. (\([^\*]*\*\)\{4\}\) --> This means that we are searching for 4 occurrences of "*" from the beginning.
Yes. (Actually, I missed the "From the bginning" part. For your case, it does not matter. Still I have added it.)

Code:

echo $xx | sed 's/^\(\([^\*]*\*\)\{4\}\)0*\([^\*]*\*\)0*\(.*\)/\1\3\4/'

1b. After this we search for "0".
Yes.

1c. But what is the last "*" for? ([^\*]*\*\ <---
This is the part that says 'Search for chars that are not "*"' -->[^\*]
Any number of the above --> *
Followed by a literal * --> \* (\* go togather)

2. Why do you use \1\3\4, why not \1\2\3\4?
Each "(" in the FIRST part is represented by a number in the SECOND part. (s/FIRST/SECOND/)
In "s/^\(\([^\*]*\*\)\{4\}\)" I am skipping the second \2 because it is with in the first "\(\(".
You go ahead and add the "\2" also, as "\1\3\2\4", you will understand.

3. Unfortunately, when I was testing it, I ran into some issues.
I have no issues, it works fine for me.
You sure you have not missed something?
Try again with the new code. (I doubt it.)

To remove first leading 0

Code:

awk -F'*' -v OFS="*" '{sub( /^0/, "", $5); sub( /^0/, "", $6); print }'

Or to remove all leading 0's

Code:

awk -F'*' -v OFS="*" '{sub( /^0*/, "", $5); sub( /^0*/, "", $6); print }'