Change new filename with date ??

Shell Programming and Scripting
1

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 .

2

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

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.

3

Hi Jim,
Thanks for replying, my case is not sample as mv or cp . I think :slight_smile: .
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 .

4

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

5

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:

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
6

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

7

I think the following Script will accomplish the required task. I tried to execute the blowtorch's scripts but it doesn't work so I modified the code to be as follows:

cd  /full/path/to/DATA
for file in Perform*Log; do
newFile=`ls -1 $file|awk -F. '{ \
	y = substr($2,5,4); \
	m = substr($2,1,2); \
	d = substr($2,3,2); \
	printf "%s-%s%s%s.dat",$1,y,m,d \
	}'`
awk -F\| '{if(NR!=1) print $1"|"$3"|"$4"|"$5}' $file >  /full/path/to/NEWDATA/$newFile
done
8

Thank you very much , I have learned it from you :slight_smile:
Here is the code if someone want to learn

cd /full/path/to/DATA
for file in Perform*Log; do
newFile=`ls -1 $file|awk -F. '{ \
y = substr($2,5,4); \
m = substr($2,1,2); \
d = substr($2,3,2); \
printf "%s-%s%s%s.dat",tolower($1),y,m,d \
}'`
awk -F\| '{if(NR!=1) print $1"|"$3"|"$4"|"$5}' $file > /full/path/to/NEWDATA/$newFile
done

Thanks again

9

Just wonder if we look at the flie

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

The field $3 is date 02/06/2006 and we write the output is file name perform-20060206.dat, its contain
1|02/06/2006|hawkeye|35
2|02/06/2006|sf49ers|30
3|02/06/2006|goraiders|27
4|02/06/2006|steeler|35

Any idea ? I just want to learn with diffrerent way .
Thanks,

10 
echo '02/06/2006' | nawk '{n=split($1, da, "/"); print da[3] da[1] da[2]}'