Open command and linux path problem 1

[veteq]

I am kinda confused on this one...(I am new to perl)

in linux I use a variable to hold the location of the log files

$WD_LOGDIR = /var/log/WS

when I try to incorporate this variable to the open command, it creates a file in the "current location" that included the path name.

open (FILEOUT, "\$WD_LOGDIR\\progfile.log") or ....

what am I missing?

please help

 

[mlibeson]

$WD_LOGDIR = '/var/log/WS';

open (FILEOUT, "< $WD_LOGDIR/progfile.log") || die "ERR\n";

The character "<" will tell the open to read instead of write to the file.

The character ">" will tell the open to write a new file instead of read from the file.

The characters ">>" will tell the open to append instead of read the file or write a new file.



Michael Libeson

 

[veteq]

Thank you Michael, I will try that....


Veteq

 

[veteq]

Sorry, I didn't post the correct line of code....


open (FILEOUT, ">> $WD_LOGDIR/progfile.log") or die "not foun...";

the $WD_LOGDIR is a shell specified location and varies from machine to machine...I need to get the linux value into perl and use that value in the open file path.....

 

[mlibeson]

try

$ENV{'WD_LOGDIR'}


Michael Libeson

 

[veteq]

that worked, thank you Micheal for all your help....appreciated


Veteq

 

[mlibeson]

your welcome

Michael Libeson