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Top Forums Shell Programming and Scripting Missing Assigned Variable within logic operator Post 302403526 by karthigayan on Friday 12th of March 2010 10:07:57 PM
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

 

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ECHO(1) 						    BSD General Commands Manual 						   ECHO(1)

NAME
echo -- write arguments to the standard output SYNOPSIS
echo [-n] [string ...] DESCRIPTION
The echo utility writes any specified operands, separated by single blank (' ') characters and followed by a newline (' ') character, to the standard output. The following option is available: -n Do not print the trailing newline character. This may also be achieved by appending 'c' to the end of the string, as is done by iBCS2 compatible systems. Note that this option as well as the effect of 'c' are implementation-defined in IEEE Std 1003.1-2001 (``POSIX.1'') as amended by Cor. 1-2002. Applications aiming for maximum portability are strongly encouraged to use printf(1) to sup- press the newline character. Some shells may provide a builtin echo command which is similar or identical to this utility. Most notably, the builtin echo in sh(1) does not accept the -n option. Consult the builtin(1) manual page. EXIT STATUS
The echo utility exits 0 on success, and >0 if an error occurs. SEE ALSO
builtin(1), csh(1), printf(1), sh(1) STANDARDS
The echo utility conforms to IEEE Std 1003.1-2001 (``POSIX.1'') as amended by Cor. 1-2002. BSD
April 12, 2003 BSD
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