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mierdatuti

03-20-2014 03:57 AM

Redirect ouput to the same script file with date

Hi,
I'm trying to redirect output of a script file to the same file with date and log extension but I cant.
I try with :

Code:

/arq/uti/xxxxxx.sh > /arq/log/"$(basename "$1")"_`date +%Y%m%d_%H%M%S`.log

Shows me:

Code:

_20140320_095532.log

And I would like:
xxxxxx__20140320_095532.log
Could you help me please?
I'm using ksh.
Thanks

colucix

03-20-2014 04:06 AM

The name of the script is $0, not $1. Even better

Code:

$(readlink -f "$0")

but take in mind that this works inside the script itself. If you use $0 directly in the command line, it will be substituted by the name of the shell (ksh) not the script.

A way to tell the script to redirect all the output to a file could be

Code:

#!/bin/ksh

#

exec > /arq/log/"$(basename "$(readlink -f "${0/.sh/}")")"_$(date +%Y%m%d_%H%M%S).log

mierdatuti

03-20-2014 04:20 AM

Quote:

Originally Posted by colucix (Post 5137924)

The name of the script is $0, not $1. Even better

Code:

$(readlink -f "$0")

Code:

#!/bin/ksh

#

exec > /arq/log/"$(basename "$(readlink -f "$0")")"_$(date +%Y%m%d_%H%M%S).log

Thanks!
if I execute:

Code:

/arq/uti/xxxxx.sh > /arq/log/"$(basename "$(readlink -f "$0")"_$(date +%Y%m%d_%H%M%S).log

dont ends and it shows me an >

colucix

03-20-2014 04:47 AM

Well, you missed a closing parenthesis and a closing double quotes, therefore it shows the secondary prompt waiting for further input. Anyway, to be more clear that command line doesn't work as you expect, since $0 will be substituted by the name of the running shell, which is ksh and not from the name of the script.

To be substituted by the name of the script the value $0 must be inside the script itself. This the reason why I suggested the exec command as the first statement of your code. It will redirect all the output from subsequent statements to the file specified. In other word your ksh script should begin with:

Code:

#!/bin/ksh

#

exec > /arq/log/"$(basename "$(readlink -f "${0/.sh/}")")"_$(date +%Y%m%d_%H%M%S).log

and then you can run it simply as

Code:

/arq/uti/xxxxx.sh

mierdatuti

03-20-2014 04:59 AM

Quote:

Originally Posted by colucix (Post 5137940)

Code:

#!/bin/ksh

#

exec > /arq/log/"$(basename "$(readlink -f "${0/.sh/}")")"_$(date +%Y%m%d_%H%M%S).log

and then you can run it simply as

Code:

/arq/uti/xxxxx.sh

Thanks but I have too much scripts and I won't edit them.
There is no alternative?
Thanks, sorry for my english!

descendant_command

03-20-2014 05:42 AM

What about (for lots)

Code:

for i in *.sh; do $i > /arq/log/"$(basename -s .sh $i)"_$(date +%Y%m%d_%H%M%S).log; done

?

or singly

Code:

i=script.sh; $i > /arq/log/"$(basename -s .sh $i)"_$(date +%Y%m%d_%H%M%S).log

colucix

03-20-2014 06:01 AM

Yes. You have two alternatives:

1) Run the script as

Code:

set -- /arq/uti/xxxxx.sh && /arq/uti/xxxxx.sh > /arq/log/"$(basename ${1/.sh/})_$(date +%Y%m%d_%H%M%S).log

this will set the first argument of the current shell as the name of the script you're going to run.

2) Edit your scripts using sed and the -i option. For example, suppose you have all your scripts under /arq and they are suffixed with .sh

Code:

find /arq -name \*.sh -exec sed -i.bck '1iexec > /arq/log/"$(basename "$(readlink -f "${0/.sh/}")")"_$(date +%Y%m%d_%H%M%S).log' {} \;

This will add the exec statement as the very first line of each .sh script under /arq. If your script have a sha-bang (that is #!/bin/ksh) as the first line, you may want to insert the text at the second line and in that case change 1 with 2 in the sed command.

Please note the option

Code:

-i.bck

it is a way to force sed to save a backup copy of the original file, in case something goes wrong. You will find those .sh.bck files that you can easily remove with a find command.

mierdatuti

03-20-2014 06:16 AM

Quote:

Originally Posted by descendant_command (Post 5137962)

What about (for lots)

Code:

for i in *.sh; do $i > /arq/log/"$(basename -s .sh $i)"_$(date +%Y%m%d_%H%M%S).log; done

?

or singly

Code:

i=script.sh; $i > /arq/log/"$(basename -s .sh $i)"_$(date +%Y%m%d_%H%M%S).log

When I do:

Code:

i="/arq/uti/xxxxx.sh"; $i > /arq/log/"$(basename -s .sh $i)"_$(date +%Y%m%d_%H%M%S).log

basename: invalid option -- s
Try `basename --help' for more information.

---------- Post added 03-20-14 at 06:17 AM ----------

Quote:

Originally Posted by colucix (Post 5137973)

Yes. You have two alternatives:

1) Run the script as

Code:

set -- /arq/uti/xxxxx.sh && /arq/uti/xxxxx.sh > /arq/log/"$(basename ${1/.sh/})_$(date +%Y%m%d_%H%M%S).log

this will set the first argument of the current shell as the name of the script you're going to run.

Secondary prompt with these command :-(

descendant_command

03-20-2014 06:23 AM

Quote:

basename: invalid option -- s
Try `basename --help' for more information.

Works here

Code:

basename --version

basename (GNU coreutils) 8.21

colucix

03-20-2014 06:30 AM

Quote:

Originally Posted by mierdatuti (Post 5137978)

Secondary prompt with these command :-(

My fault, this time. Corrected version:

Code:

set -- /arq/uti/xxxxx.sh && /arq/uti/xxxxx.sh > /arq/log/"$(basename "${1/.sh/}")"_$(date +%Y%m%d_%H%M%S).log

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