Shell Programming and Scripting

Hi
Im trying to do the following:

Code:

grep -H  Date: out/* | sed  's/':'/ /' | awk '$4 ~ /^[Aa-Zz]/ {print $1}' | while read VARIABLE; do
awk '{print $1,$3,$2}' $VARIABLE | sed (take stdin and replace a string in $VARIABLE)
done

What this is basically doing is finding all files with Date: in them where the 4th column starts with a letter, then passes out the file name to awk which re-arranges certain columns within the file.

What I need to do then is pipe the output of awk to sed where it replaces a string with the output of awk.

I cant find out if sed allows you to define the replacement sting as stdin.

Thanks

you could use a read command to take input from stdin and use that variable in your sed command.

If you need to read stdin for every execution of while, then put the read command insider loop. as the awk and sed is also a part of loop, this should work.

I feel something is wrong with the logic?
Output of awk inside the while will give you more than one value(of course it depends on the file) and which value will you chose to replace using sed and with what?

regards,
Ahamed

Ok yeh i just realised that this wont work, I was testing with files that only had Date: blah blah in them but normally they will have other data as well which will stil get caught up. Thanks for the heads up.

Any idea of a better way to do this?
The whole goal of this is that I have a line in multiple files with:

Date: Thu, Aug 23 2001 21:27:04 -0400

As an example. What im trying to do is determine if a file has a date format of month day year and then swap them around so it would be instead:

Date: Thu, 23 Aug 2001 21:27:04 -0400

So just simply swapping the two columns around for files that have the wrong date format.

Try this

Code:

sed 's/\(Date:.*, \)\([Aa-Zz]..\) \([0-9].\)\(.*\)/\1\3 \2\4/g' inputfile

regards,
Ahamed

thanks, that is much simpler and works perfectly. thank you

Ive ben looking around for something else to deal with the same line of data.

Sometimes the data will be formatted like so:

Date: Thu, Aug 23 2001 09:27 PM -0400

Is it possible to calculate values in sed, so say is column 7 Starts with P then add 12 to column 6 to make it 24 hr format so 21:27:00

Try this for your second scenario

Code:

awk '/PM/{split($6,_1,":");$6=_1[1]+12":"_1[2]}/AM|PM/{$6=$6":00";$7=$8;NF=7}1' infile

regards,
Ahamed