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Missing Assigned Variable within logic operator

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# 1  
Old 03-12-2010
Missing Assigned Variable within logic operator
Hey ,

I'm trying to perform the following command, however it cannot read the variable assigned earlier. I'm not sure why this happen. Please help thanks

Code:
while :
do
echo "what's ur name? (if none just press [ENTER])"
read name
changeName = echo $name | sed "s/on/ey/"
echo $changeName  #this variable still can be read
if [ -z $name ]; then
 echo "you don't have a name goodbye"
 exit 0
else
 echo "your name will change from: $name to $changeName" #now changeName cannot be read why?
fi
done

# 2  
Old 03-12-2010
MySQL
To assign a command output to the variable you need to use the backtick(`).and also you should not give space while assign the output to a variable.

Code:
while :
do
echo "what's ur name? (if none just press [ENTER])"
read name
changeName=`echo $name | sed "s/on/ey/"`
echo $changeName  #this variable still can be read
if [ -z $name ]; then
 echo "you don't have a name goodbye"
 exit 0
else
 echo "your name will change from: $name to $changeName" #now changeName cannot be read why?
fi
done

# 3  
Old 03-13-2010
Hey,

Thanks for replying but I've tried using the back tick and i didn't give a space when assigning variable but instead i'm getting this error

./testVar: line 7: jey: command not found

the changeName=`echo $name | sed "s/on/ey/"` command cannot be read if i used the back tick

please helpp
# 4  
Old 03-13-2010
Code:
changeName=`echo "$name" | sed "s/on/ey/"`

Code:
if [ -z "$name" ]; then

Use double quotes for the variable $name. Because your input may have space characters.
Then only it will preserve the white spaces
Last edited by thillai_selvan; 03-13-2010 at 01:14 AM..
# 5  
Old 03-13-2010
I thing you have gave space characters in your input , to avoid this use the double quotes.

Code:
while :
do
echo "what's ur name? (if none just press [ENTER])"
read name
changeName=`echo "$name" | sed "s/on/ey/"`
echo $changeName  #this variable still can be read
if [ -z "$name" ]; then
 echo "you don't have a name goodbye"
 exit 0
else
 echo "your name will change from: $name to $changeName" #now changeName cannot be read why?
fi
done

# 6  
Old 03-13-2010
Hey Guys,

thanks for the fast replies, even though i used double quotes it still doesn't wor and my input doesn't have any white spaces

here is the code suggested
Code:
#!/bin/sh

while :
do
echo "what's ur name? (if none just press [ENTER])"
read name
changeName= `echo "$name" | sed "s/on/ey/"`
echo $changeName  #this variable still can be read
if [ -z "$name" ]; then
 echo "you don't have a name goodbye"
 exit 0
else
 echo "your name will change from: $name to "$changeName" ? [y/n]"
 read ans
fi
done


This is the ouput I got:

what's ur name? (if none just press [ENTER])
jon
./testVar: line 7: jey: command not found

your name will change from: jon to ? [y/n]



I'm confused never had this problem before
# 7  
Old 03-13-2010
Quote:
Originally Posted by sexyTrojan
Hey Guys,

thanks for the fast replies, even though i used double quotes it still doesn't wor and my input doesn't have any white spaces

here is the code suggested
Code:
#!/bin/sh

while :
do
echo "what's ur name? (if none just press [ENTER])"
read name
changeName= `echo "$name" | sed "s/on/ey/"`
echo $changeName  #this variable still can be read
if [ -z "$name" ]; then
 echo "you don't have a name goodbye"
 exit 0
else
 echo "your name will change from: $name to "$changeName" ? [y/n]"
 read ans
fi
done

This is the ouput I got:

what's ur name? (if none just press [ENTER])
jon
./testVar: line 7: jey: command not found

your name will change from: jon to ? [y/n]



I'm confused never had this problem before
Code:
#!/bin/sh

while :
do
echo "what's ur name? (if none just press [ENTER])"
read name
changeName=`echo "$name" | sed "s/on/ey/"`
echo $changeName  #this variable still can be read
if [ -z "$name" ]; then
 echo "you don't have a name goodbye"
 exit 0
else
 echo "your name will change from: $name to "$changeName" ? [y/n]"
 read ans
fi
done

I just removed the space after the assignment
operator as in the below line and it is working for me...

Code:
changeName=`echo "$name" | sed "s/on/ey/"`

And it will keep on looping till you don't put a name...
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