UNIX for Dummies Questions & Answers

Hello all!

I'm trying to put together a small script that will take in a file name and attach a datestamp to the end of it (but before the file type extension).

To illustrate...

Before:
filename.txt
anotherfilename.txt

After:
filename_20090724.txt
anotherfilename_20090724.txt

Searching around the forums I've found some pieces that I think are parts of the solution:

Code:

 
date +%Y%m%d ---->gets me my datestamp format
expr index string . ------>gets me the character position of '.'

So now my question is how to substring the file name from the beginning to the period and also substring from the period to the end...giving me my two halves of the full name so that I can concatenate my datestamp in between. Any help is very appreciated.

Thanks!
Glen

Try this:

Code:

echo 'filename.txt' | 
awk -F"." -v d=$(date "+%Y%m%d") '{print $1 "_" d FS $2}'

That works perfectly! Thanks for the response, I figured there was a pretty simple one line answer to this (I usually make things more complicated then they should be). Now for added complexity, I'd like to store this in a variable so that I can use it in an 'mv' statement (i.e. mv filename.txt $variable):

Code:

 
export temp = `echo 'filename.txt' | awk -F"." -v d=$(date "+%Y%m%d") '{print $1 "_" d FS $2}'`
echo $temp

This almost works, I get the following error messages:

Code:

test.shl: line 4: export: `=': not a valid identifier
test.shl: line 4: export: `filename_20090724.txt': not a valid identifier

I seem to be doing something wrong with the variable assignment. Any suggestions?

Thanks in advance!
Glen

Remove the spaces around the equal sign =.
Why do you export the variables? It's not necessary if you use the variables in the current shell.

Regards

I was working off an example script which is why the export is there. Now that I know it's not necessary I've taken it off I'm pretty new at this so I'm kind of learning as I go.

Removing the spaces did the trick, thanks! One more question, I'm trying to implement this whole apparatus inside of a loop statement so that I can read all these filenames from a list file. From an example I found online I put together this:

Code:

 
#!/bin/ksh
 
listfile=myfiles.lst
 
for i in ${listfile}
 
do
 
  temp=`echo $i | awk -F"." -v d=$(date "+%Y%m%d") '{print $1 "_" d FS $2}'`
  mv $i $temp
 
done

The problem is that it works but in the wrong way, instead of applying the datestamp to the items in the list, it takes the actual list file and applies the datestamp to it: myfiles_20090724.lst. I'm guessing I'm doing something wrong in the first line of the for statement where I'm trying to isolate a single line item in the variable $i. Can you shed a little light on the subject?

Thanks again,
Glen

You're using the filename instead of the content of the file and it's more efficient to call the date command once:

Code:

#!/bin/ksh

dd=$(date "+%Y%m%d")

while read i
do
  temp=`echo $i | awk -F"." -v d=$dd '{print $1 "_" d FS $2}'`
  mv "$i" "$temp"
done < myfiles.lst

Actualy this command is sufficient to do the job:

Code:

awk -F"." -v d=$(date "+%Y%m%d") '{system("mv " $0 " " $1 "_" d FS $2)}' myfiles.txt

That's very interesting, so the list file name is actually fed into the loop (or command) at the very end of the script as opposed to the front. Is there a specific reason for this? Or is it just the way the syntax works? This is all very informative for me, I appreciate it!

Also, I'd like to put in some error handling. Specifically to watch for file names that would not be present. Would I be looking for the program code variable "$?"? I'm seeing from other examples that checking $? for a non-zero value indicates that the command failed.