How to compare two strings using if


 
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# 1  
Old 02-23-2010
How to compare two strings using if

Hi,

Here is my script

Code:
#!/bin/ksh
echo $pick_typ

if [[ "$pick_typ" = "CUS" || "$pic_typ" = "ORD" || "$pic_typ" = TRR" ]];then
  echo "inside if"
else
  echo "outside if"
fi

when ever i pass CUS as parameter to this script am getting the correct value CUS, however if i pass ORD as parameter it is not coming inside if it is echoing else "Outside if" statement .

I have tried several options but none worked

Please help me

Thanks

Bhargav

Last edited by Franklin52; 02-23-2010 at 09:57 AM.. Reason: Please indent your code and use code tags!
# 2  
Old 02-23-2010
Maybe you mean:

Code:
#!/bin/ksh
pick_typ="$1"
echo "$pick_typ"

if [ "$pick_typ" = "CUS" -o "$pic_typ" = "ORD" -o "$pic_typ" = "TRR" ]
then
         echo "inside if"
else
         echo "outside if"
fi

# 3  
Old 02-23-2010
I have tried this option as well still am getting the else echo "Outside if "
# 4  
Old 02-23-2010
Please post the current version of your script.

I'm expecting that you run the script as:
Code:
./scriptname CUS
./scriptname ORD
./scriptname TRR

Is this right?
# 5  
Old 02-23-2010
Yes I run the script like that by passing the either of the one as Parameter

./test CUS or ./test ORD

Thanks
# 6  
Old 02-23-2010
[[ $a == z* ]] # True if $a starts with an "z" (pattern matching)
[[ $a == "z*" ]] # True if $a is equal to z* (literal matching).
[ $a == z* ] # File globbing and word splitting take place.
[ "$a" == "z*" ] # True if $a is equal to z* (literal matching).

try the below:-

Code:
#!/bin/ksh
pick_typ="$1"
echo "$pick_typ"

if [ "$pick_typ" == "CUS" -o "$pic_typ" == "ORD" -o "$pic_typ" == "TRR" ]
then
         echo "inside if"
else
         echo "outside if"
fi

# 7  
Old 02-23-2010
k is missing in the variable name
Code:
#!/bin/ksh
pick_typ="$1"
echo "$pick_typ"

if [ "$pick_typ" = "CUS" -o "$pick_typ" = "ORD" -o "$pick_typ" = "TRR" ]
then
         echo "inside if"
else
         echo "outside if"
fi

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