Shell Programming and Scripting

Hi all,
I am newbie and hope that you can help me to rename a file
If I have a file name Perform.01222006.12345.Log now I would like to backup another file with another name like perform-20060112.dat

This is a flat file, and I want to collect some field, then put it in a new file from Perform.mmddyyyy.ID.Log to perform-yyyymmdd.dat how do I do that ? can you show me how you write the code.

Thanks,

P/S: I will use ksh shell , and I am sorry if I post in a wrong forum .

It looks like you are asking about five different things.

Sentence #1: Use the mv command to rename files.
Sentence #2: I think you mean copy a file use cp

Code:

cp Perform.01222006.12345.Log perform-20060112.dat

Do you want #1 and #2 as part of the "collect some field". You need to give us exact requirements. Sample of the old file, sample of new file, and file names.

Hi Jim,
Thanks for replying, my case is not sample as mv or cp . I think .
Okay here is the problem:
In a directory DATA have alot of file name Perform.mmddyyyy.ID.Log where mm is month, dd is day and yyyy is year like Perform.02072006.1234.Log, Perform.02062006.4321.Log, Perform.02082006.4321.Log ...
Each file is a flat file, separate by a pipe like
ID|Location|Date|Hostname|Age|Sex

Then how I write a ksh shell to find all files in DATA directory which are new in the last 2 weeks, and take only 4 fields
ID|Date|Hostname|Age
then write it in a NEWDATA directory, and the file will be named perform-yyyymmdd.dat

Example:
Perform.02062006.4321.Log have
ID|Location|Date|Hostname|Age|Sex
1|SFO|02/06/2006|hawkeye|35|M
2|LAX|02/06/2006|sf49ers|30|M
3|OAK|02/06/2006|goraiders|27|F
4|PIT|02/06/2006|steeler|35|M

and write to a file name perform-20060206.dat
1|02/06/2006|hawkeye|35
2|02/06/2006|sf49ers|30
3|02/06/2006|goraiders|27
4|02/06/2006|steeler|35

How can you do that ? Thanks for sharing your knowlege .

Hi,

I have solution for this.

Suppose you have a file with "Perform.01222006.12345.Log" and want to rename
as "Perform-01222006.dat".

Solution:
Go to the directory where u have the src-files
under the directory type this command
ls -1 Perform*|awk -F"\." '{print $_ " " $1"-"$2".dat"}'|xargs mv

this will change all the files as per your requirment.

-regards
SubbuMalepati

The OP wants to extract some part of the data and redirect it to another file in another directory. Maybe this is the way to go about it:

Code:

cd /full/path/to/DATA
for file in Perform*Log; do
awk -F\| '{if(NR!=1) print $1"|"$3"|"$4"|"$5}' $file > /full/path/to/NEWDATA/$(echo $i|awk -F. '{print
 tolower($1)"-"$2".dat"}');
done

Thanks for replying and showed me, but
if we use the code
ls -1 Perform*|awk -F"\." '{print $_ " " $1"-"$2".dat"}'|xargs mv

It will give me Perform-mmddyyyy.dat

and if I use:

awk -F\| '{if(NR!=1) print $1"|"$3"|"$4"|"$5}' $file > /full/path/to/NEWDATA/$(echo $i|awk -F. '{print
tolower($1)"-"$2".dat"}');

Then I also get perform-mmddyyyy.dat .

Am I right ?? I need the file name as perform-yyyymmdd.dat