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Skipping every other parameter

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PistolPete1999

Technical User
Jan 14, 2003
3
US
How do i skip every other parameter? like for example

REVERSEM K B Z D E K "G," M O

it should display O G, E Z and skips the first two parameters

*REVERSEM is the name of the script

i got the first two parametrs removed working but struggled with skipping every other parameter


A=$*
B=""

shift 2

for i in $A
do
B="$i $B"
done
echo $B

Any help would be appreciated
 
first purge ", then print it:
echo "K B Z D E K \"G,\" M O" | sed -e 's/"//;s/\([A-Z]\) \([A-Z]\) \([A-Z]\) \([A-Z]\) \([A-Z]\) \([A-Z]\) \([A-Z],\) \([A-Z]\) \([A-Z]\)/\9 \7 \5 \3/'



-----------
Ohne Intelligenz bist Du ein Fiasko,
ohne Technik ein Amateur und
ohne Herz eine Maschine.

[ Wladimir Horowitz ]
 
Hi,
you can do:
#!/bin/ksh
i=$#
while [ $i -gt 2 ]
do
B=$B" "$(($i))
((i=i-2))
done
echo $B
Regards Boris
 
Hey Boris,

thanks for replying, its a good script but when i changed the $# to $*, i got a msg saying too many arguements and i noticed that ure using KORN, im using bash, unless im wrong, they re different right?

Any Solutions?

Thanks
 
#!/bin/ksh

#str='K B Z D E K "G," M O'

typeset -i iter=${#}

while [ ${iter} -gt 2 ]
do
eval "printf \"\${$iter} \" "
iter=$((iter-2))
done
echo
vlad
+----------------------------+
| #include<disclaimer.h> |
+----------------------------+
 
Hey Fellas,

finally figured the right code for this...

#!/bin/sh
clear
shift 2

last=$#

while [ $last -gt 0 ]
do
eval 'v=\$$last&quot;
echo -n &quot;$v&quot;
last=$(($last-2))
done
echo




thanks to those people who tried to help out :)
 
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