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Case structure combined with standard input

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# 1  
Old 03-06-2013
Case structure combined with standard input
Hi folks,

I am new to bash scripting so please excuse my question.

Is there any chance to combine a case structure with the read command?

Like
Code:
case (read -p "" variable) in
x)


Thx!
# 2  
Old 03-06-2013
Removed: wrong code.
Last edited by radoulov; 03-06-2013 at 10:44 AM..
# 3  
Old 03-06-2013
Thx for ur reply!

Unfortunately it is not working like I want it to...
Code:
#case $(read -p "" datacheck) in
#1) echo "yes" ;;
#2) echo "no" ;;
#*) echo "wrong";;
#esac

the answer is alway "wrong"

if I write it like this

Code:
read -p "" datacheck
#case $datacheck in
#1) echo "yes" ;;
#2) echo "no" ;;
#*) echo "wrong";;
#esac

the outcome depends on the standard input like it should be

Smilie
Last edited by radoulov; 03-06-2013 at 10:39 AM..
# 4  
Old 03-06-2013
Yes, sorry. And why don't you use the second version (the working one)?
# 5  
Old 03-06-2013
What, exactly, does read -p "" datacheck print to the screen?

What, exactly, would $( ) give for a command that prints nothing to the screen?

This is why your case statement doesn't work.
# 6  
Old 03-07-2013
Quote:
Originally Posted by Corona688
What, exactly, does read -p "" datacheck print to the screen?


What, exactly, would $( ) give for a command that prints nothing to the screen?

This is why your case statement doesn't work.
Well, I misunderstood the -p "". So it is just "read variable".

Code:
read variable

would just use the input as the new variable

But I still don't get why
Code:
case $(read variable)

does not work.

Would you please be so kind and explain it to me?
# 7  
Old 03-07-2013
See Corona688's statement in post#5.
Command substitution makes stdout of program/command available for e.g. variable assignments or shell interpretation. read var doesn't print anything to stdout, so nothing can be substituted. Make it case $(read var; echo $var) in ...
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