UNIX for Dummies Questions & Answers

Hi all,

Suppose i wish to have a file which should be something like this:
ROOH20020406080015000003
RLO00220404144142ENG01
RLO00220404123602ENG01
RLO00220404123602ENG02

As u can seeit contains 3 files.

Basically I want the header in the file to be something like this.

here ROOH will remain always(0-3)
20020406080015 (4-17) are YYYYMMDDHHMMSS.
000003(18-23) This a count.

Now my query is basically how do I pad these 0 before 3 .
Morever Tomorrow my count might be 35 or 456 then how can I automate it.

Please help if there is a way to do that.

I am able to get upto 17 and then I get 3.
I have also tried it like this.
TOTAL_FILES=`ls $DESTINATION |wc -l`
COUNT="000000"
FILE_COUNT=0
FILE_COUNT=`expr $FILE_COUNT + $TOTAL_FILES`

But still I get the file_count as 3.

Your help & suggestion will be highly appreciated.

Thanx
ROOH

In ksh, typeset -Z6 COUNT will give you 6-digit output.

Hi Jimbo,

Thans for ur suggestion I was successful in doing it.
It worked Thanks a billion.

Thanx.
Rooh