Getting the file name passed to a shell script


 
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# 1  
Old 10-28-2010
Getting the file name passed to a shell script

Hi all

I want to use the filename passed into a shell script within the shell it self. The file will be passed as shown below

Code:
./myScript < myFile.file

i tried $1 but i think since myFile is not passed as a command line arg its not saving the file name in $1

any help is appreciated.
# 2  
Old 10-28-2010
You are correct

Can you call it like this?:
Code:
./myScript myFile.file < myFile.file

At least you will have it in $1
# 3  
Old 10-29-2010
thankx for the reply

Well i can but its against the policy of my workplace. I would prefer if there was a simpler way of getting the filename in the way that i am calling the script
# 4  
Old 10-29-2010
The filename is never "passed into the script". By the time your script is running, all the redirection has already happened, it never gets told a thing about it. You're trying to go backwards from a file descriptor to a filename, difficult to impossible in a shell.

What's the filename needed for? Maybe there's another way to accomplish what you want without it.
# 5  
Old 10-29-2010
well basically im passing it to
Code:
awk

to generate a report. But before doing that i want to enter into a Log file the name of the file being processed

eg: Processing file <FILENAME>
...
...

i also need to use it within awk to print a few lines eg

Code:
{"head -n "FNR" <FILENAME> | tail -n 1"|getline temp } print ""temp"">> "Log"


Last edited by sridanu; 10-29-2010 at 02:33 AM..
# 6  
Old 10-29-2010
If file name and location is fixed, then store it in variable and pass the variable to script.
Use variable in main script whenever u want using $variable.
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