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Passing argument not retrieving.

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# 1  
Old 10-23-2015
Passing argument not retrieving.
Hi
I have a script which am trying to pass an argument which am trying to call using $1 but its not taking the value inside the if loop as it showing the error as if: Empty if..
Any help on this will be helpful.



Code:
#!/usr/bin/csh
echo $1
if ('$1' == "pp")
then
echo "Printing $1"
endif

rogerben
# 2  
Old 10-23-2015
man csh:
Quote:
Variable substitution is suppressed inside of single quotes
---------- Post updated at 17:13 ---------- Previous update was at 17:12 ----------

But that wouldn't explain the "Empty..." message. Does it echo $1 correctly?
# 3  
Old 10-23-2015
Quote:
Originally Posted by RudiC
man csh:

---------- Post updated at 17:13 ---------- Previous update was at 17:12 ----------

But that wouldn't explain the "Empty..." message. Does it echo $1 correctly?
yes echo $1 working fine.
rogerben
# 4  
Old 10-23-2015
Quote:
Originally Posted by rogerben
Hi
I have a script which am trying to pass an argument which am trying to call using $1 but its not taking the value inside the if loop as it showing the error as if: Empty if..
Any help on this will be helpful.



Code:
#!/usr/bin/csh
echo $1
if ('$1' == "pp")
then
echo "Printing $1"
endif

The csh syntax is a bit different. Try:


Code:
#!/usr/bin/csh
echo $1
if ($1 == "pp") then
    echo "Printing $1"
endif

# 5  
Old 10-26-2015
Code:
#!/usr/bin/csh
echo $1    ---> Value printing
if ($1 == "pp")
then
echo "Printing $1"
endif

I tried the same code but still am getting the error like "if: Empty if"
rogerben
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