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# 1  
Old 06-13-2009
Help in for loop.
Hi,

I have a file lets say FILE1 which contains list of files and the contents of this file are not going to change over time.

FILE1
---------

feedset1_${DATA_DATE1}.dat
feedset2_${DATA_DATE1}.dat
feedset3_${DATA_DATE1}.dat
accounts_${DATA_DATE2}.dat
customers_${DATA_DATE2}.dat
clients_${DATA_DATE2}.dat

Here DATA_DATE1= sysdate in YYYYMMDD
DATA_DATE2= (sysdate-1) in YYYYMMDD
I require a output file(FILE_OUT) on a daily basis whose content will be like below for sysdate='14Jun2009'.The content of this file will change everyday according to the date conditions.

FILE_OUT
----------
feedset1_20090614.dat
feedset2_20090614.dat
feedset3_20090614.dat
accounts_20090613.dat
customers_20090613.dat
clients_20090613.dat

Please let me know how can I achieve this in for loop...Smilie

Thanks.
# 2  
Old 06-14-2009
You may want to check if your shell supports basic date arithmetic. In GNU Bash, for example, you can do this (assuming sysdate = 14-Jun-2009):

Code:
$
$ date "+%Y%m%d"
20090614
$
$ date -d "-1 day" "+%Y%m%d"
20090613
$

You can then use grep to look for these dates in FILE1:

Code:
$
$ cat file1
feedset1_${20090614}.dat
feedset2_${20090614}.dat
feedset3_${20090614}.dat
accounts_${20090613}.dat
customers_${20090613}.dat
clients_${20090613}.dat
xyz_${20090612}.dat
$
$ grep -E "`date \"+%Y%m%d\"`|`date -d \"-1 day\" \"+%Y%m%d\"`" file1 | sed 's/[{}$]//g'
feedset1_20090614.dat
feedset2_20090614.dat
feedset3_20090614.dat
accounts_20090613.dat
customers_20090613.dat
clients_20090613.dat
$
$

Or maybe a bit cleaner:

Code:
grep -E "$(date '+%Y%m%d')|$(date -d '-1 day' '+%Y%m%d')" file1 | sed 's/[{}$]//g'

tyler_durden
# 3  
Old 06-14-2009
Quote:
Originally Posted by durden_tyler
You may want to check if your shell supports basic date arithmetic.

I don't know of any shell that supports date arithmetic (unless you write scripts to do it).
Quote:
In GNU Bash, for example, you can do this (assuming sysdate = 14-Jun-2009):
Code:
$
$ date "+%Y%m%d"
20090614
$
$ date -d "-1 day" "+%Y%m%d"
20090613


Code:
$ ps $$
  PID  TT  STAT      TIME COMMAND
84615  p0  Ss     0:00.86 -bash (bash3.0.16)
$ date -d "-1 day" "+%Y%m%d"
usage: date [-jnu] [-d dst] [-r seconds] [-t west] [-v[+|-]val[ymwdHMS]] ... 
            [-f fmt date | [[[[[cc]yy]mm]dd]HH]MM[.ss]] [+format]

It's the date command that does the arithmetic, not the shell. Sysems whose date command do not use that syntax may have GNU date installed as gdate. *BSD systems' date command uses different syntax for date arithmetic. Most versions have none.
-----Post Update-----

Quote:
Originally Posted by 46019
Hi,

I have a file lets say FILE1 which contains list of files and the contents of this file are not going to change over time.

FILE1
---------

feedset1_${DATA_DATE1}.dat
feedset2_${DATA_DATE1}.dat
feedset3_${DATA_DATE1}.dat
accounts_${DATA_DATE2}.dat
customers_${DATA_DATE2}.dat
clients_${DATA_DATE2}.dat

Here DATA_DATE1= sysdate in YYYYMMDD
DATA_DATE2= (sysdate-1) in YYYYMMDD
I require a output file(FILE_OUT) on a daily basis whose content will be like below for sysdate='14Jun2009'.The content of this file will change everyday according to the date conditions.

FILE_OUT
----------
feedset1_20090614.dat
feedset2_20090614.dat
feedset3_20090614.dat
accounts_20090613.dat
customers_20090613.dat
clients_20090613.dat

Please let me know how can I achieve this in for loop...Smilie

Why do you want a for loop? That will slow the script down.

Code:
file=FILE1

DATA_DATE1=$(date +%Y%m%d)
DATA_DATE1=$(date -d -1day +%Y%m%d) ## Use whatever works for you

eval "printf '%s\n' \"$( cat "$file" )\""

Last edited by cfajohnson; 06-14-2009 at 04:27 PM..
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