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Extracting filenames

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# 1  
Old 05-07-2010
Extracting filenames
Hi

I need to pull out the name of the file from the path.

See, here is my loop that gets the files:

Code:
dsxdir="/var/local/dsx/import"
 
for dsxfile in $dsxdir/*.dsx;
do

dsxlog $reverb --info --module="$module" "$dsxfile"
$dsximp $norule $oprange --dsn=$dsn --dbname=$dbname --user=datasafe --password=datasafe --dsxfile=$dsxfile

done

When I try to process only the filename, I get the entire path:

Code:
 
Now processing /var/local/dsx/import/gen_cell_orig_001.dsx

I only want "gen_cell_orig_001.dsx"

Can you please help.
# 2  
Old 05-07-2010
Have you tried /usr/bin/basename ?

Code:
# which basename
/usr/bin/basename
# basename /var/adm/messages
messages

HTH
# 3  
Old 05-07-2010
Not necessary to use an external command:

Code:
file="/var/local/dsx/import/gen_cell_orig_001.dsx"

Now processing ${file##*/}

# 4  
Old 05-07-2010
Thanks Franklin52 .... it works like a charm .....
Can you please explain to me exactly what happens in that call ... sorry but I'm not too good with scripting
# 5  
Old 05-07-2010
Quote:
Originally Posted by ladyAnne
Thanks Franklin52 .... it works like a charm .....
Can you please explain to me exactly what happens in that call ... sorry but I'm not too good with scripting
Have a read of:

Learning the Korn Shell, 2nd Edition: Chapter 4: Basic Shell Programming

or:

Manipulating Strings
This User Gave Thanks to Franklin52 For This Post:
ladyAnne
# 6  
Old 05-07-2010
$ echo "/var/local/dsx/import/gen_cell_orig_001.dsx"| cut -d '/' -f6
gen_cell_orig_001.dsx
This User Gave Thanks to mr_harish80 For This Post:
ladyAnne
# 7  
Old 05-07-2010
Another well known approach

Code:
$> file="/var/local/dsx/import/gen_cell_orig_001.dsx"; echo $(basename "$file")
gen_cell_orig_001.dsx

This User Gave Thanks to kchinnam For This Post:
ladyAnne
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