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How do I pass a wild card as an argument


 
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Top Forums UNIX for Dummies Questions & Answers How do I pass a wild card as an argument
# 1  
Old 03-28-2007
How do I pass a wild card as a variable

Hi,

I would like to pass a wild card as part of an argument. But when I do it the script views the wild card as text.
Example:
sFile=MG1A*
sort $sFile > $sFile.sorted

What I get is MG1A*.sorted

The problem is I am passed a series of files where the first few characters like "MG1A" are constant. The additional part of the file name is a date time stamp. I will never know the exact date time stamp. I need to do a search for MG1A* which will find me one file. How can I write my script so the variable $sFile will know that the "*" in MG1A* is a wild card not text?

I know I can write code to run "ls" on a directory and get all file names then write a loop or do statement to do my sort based upon that list regardless of the file name. But I have more files in the dir than I want to grab. I also know I can run the "ls" command output the file names in the dir to a file and grep for the part fo the file names I am looking for. I also think this can be done if I pput the whole comamnd iBut current design restricts me form going that path. It would require a whole re-write and I am trying to avoid that.

Thank you.

Last edited by eja; 03-28-2007 at 11:37 AM..
# 2  
Old 03-28-2007
Quote:
Originally Posted by eja
Hi,

I would like to pass a wild card as part of an argument. But when I do it the script views the wild card as text.
Example:
sFile=MG1A*
sort $sFile > $sFile.sorted

What I get is MG1A*.sorted

The problem is I am passed a series of files where the first few characters like "MG1A" are constant. The additional part of the file name is a date time stamp. I will never know the exact date time stamp. I need to do a search for MG1A* which will find me one file. How can I write my script so the variable $sFile will know that the "*" in MG1A* is a wild card not text?

I know I can write code to run "ls" on a directory and get all file names then write a loop or do statement to do my sort based upon that list regardless of the file name. But I have more files in the dir than I want to grab. I also know I can run the "ls" command output the file names in the dir to a file and grep for the part fo the file names I am looking for. I also think this can be done if I pput the whole comamnd iBut current design restricts me form going that path. It would require a whole re-write and I am trying to avoid that.

Thank you.
try changing (ksh)
sFile='MG1A*'
sort $sFile > $sFile.sorted

Be sure and use single quotes - it will not expand out the wildcard at the assignment line.
# 3  
Old 03-28-2007
works great. thanks. I tried everthing bu. feel dumb now. I tried "" and `` did not think of single qoutes
 
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