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cut vs. sed vs. awk ?

 
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# 1  
Old 09-28-2004
Question cut vs. sed vs. awk ?
hi again...need new help guysSmilie

the file contains following infos...

users/abc/bla1.exe
newusers/defgh/ik/albg2.exe
users2/opww/ertz/qqwertzu/rwerwew.exe

how to get the file content into...

users/abc/
newusers/defgh/ik/
users2/opww/ertz/qqwertzu/

with...

[sed -e ´s/.exe/ /g´<infile>outfile]

you can erase the point and the extension

by the way, when you make

[sed -e ´s/.....exe/ /g´<infile>outfile]

you can also delete chars, but then it would be like..

users/abc/
newusers/defgh/ik/al
users2/opww/ertz/qqwertzu/rwe

anybody knows a trick? thx...

secondary I just don't know if there's another chance by using awk or cut.
# 2  
Old 09-28-2004
What about a little script like...

Code:
#!/bin/bash

while read line
do    
   path=${line%/*}  
   echo $path
done < infile > outfile

exit 0

Cheers
ZB
# 3  
Old 09-28-2004
MySQL nice !
works fine, thx....

some problems like..

script interpreter "/bin/bash" not found
script interpreter link resolves to "/usr/bin/bash"

..but with #!/bin/sh it works perfectly on my systemSmilie

by the way I'm litte confused by the different shells...
# 4  
Old 09-28-2004
Yep, that's just because my "bash" is in a different place to yours. You can change the shebang to #!/usr/bin/bash and it should work okay.

Cheers
ZB
# 5  
Old 09-28-2004
Lightbulb
could you pleaz interpret what the line

path=${line%/*}

does exactly?

the loop is nearly traceable and per "$" it gets the info from the file and so on..I know.

but what about the syntax {line%/*} ? it's not really clear. why it cuts exactly at the right position ? thx..
# 6  
Old 09-28-2004
From man bash
Quote:
${parameter%word}
${parameter%%word}
The word is expanded to produce a pattern just as in pathname
expansion. If the pattern matches a trailing portion of the
expanded value of parameter, then the result of the expansion is
the expanded value of parameter with the shortest matching pat-
tern (the ``%'' case) or the longest matching pattern (the
``%%'' case) deleted. If parameter is @ or *, the pattern
removal operation is applied to each positional parameter in
turn, and the expansion is the resultant list. If parameter is
an array variable subscripted with @ or *, the pattern removal
operation is applied to each member of the array in turn, and
the expansion is the resultant list.
So basically, discard the shortest pattern matching "/*" (i.e. a forward slash followed by "anything") from the right hand side of the variables value, and return the rest.

Cheers
ZB
 
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