Shell Programming and Scripting

Hello All,
I have a csv file with 3 columns. The file which looks like this

Code:

47850000,100,233
23560000,10000,456
78650000,560000,54
34000000,3456,3

The first column has 4 trailing zeros. I have to remove 4 trailing zeroes from 1st field. The output file should appear as follows.

Code:

4785,100,233
2356,10000,456
7865,560000,54
3400,3456,3

How can I achieve this using sed?

Maybe something like this ?

Code:

$ 
$ 
$ cat -n f9
     1 47850000,100,233
     2 23560000,10000,456
     3 78650000,560000,54
     4 34000000,3456,3
$ 
$ sed 's/^\([0-9][0-9][0-9][0-9]\)[^,]*\(,.*\)$/\1\2/' f9
4785,100,233
2356,10000,456
7865,560000,54
3400,3456,3
$ 
$

tyler_durden

Code:

sed 's/^\(.[^,]*\)0000,\(.*\)/\1,\2/' file

Code:

awk -F "," '{ if ($1~/0000$/) {$1=substr($1,1,length($1)-4)}}1' OFS="," urfile

Code:

sed 's/0000,/,/' file

i think it would remove 0's from each field

it shd be

Code:

sed 's/0000,/,/1' file

and that would still be wrong if the first field doesnot have any zero then it would remove from the subsequent fields

The 1 subsitution flag is redundant. By default, SED always subs only the first instance, unless you give it a number or g. There is no difference at all between "sed 's/0000,/,/1'" and "sed 's/0000,/,/'".

The original poster stated that the first column contains 4 trailing zeroes. My sed will remove them. If only some of the rows in the first column contain the trailing zeroes, then the mistake is in the original poster's problem statement for being imprecise. I assume that the problem is stated correctly and choose not to overcomplicate my proposed solution.

Regards,
Alister

---------- Post updated at 11:30 PM ---------- Previous update was at 11:25 PM ----------

However, if it did need to handle a first field that may not contain the four trailing zeroes, the correct sed would be:

Code:

sed 's/^\([^,]*\)0000,/\1,/' file