shell script error

[Sajjad Sahir]

when i am entering mismatching number then it also showing matching.

eg:12 and 13 it showing mathcing
12 and 12 also showing matching.

#!/usr/bin/sh
echo "enter the number"
read number1
echo "enter the number"
read number2
if number1=number2
then
echo "matching"
exit
else number1!=number2
echo "Not matching"
fi


exit

[Yogeeraj_1]

hi,

when working with the variables you should use the $ sign.

e.g.

if $number1 = $number2
then

kind regards
yogeeraj

[Michael Steele_2]

Also capitalize your var.'s for readability.

[Yogeeraj_1]

hi again,

below the modified script:

$ vi script1.sh
#!/usr/bin/sh
echo "enter the number"
read number1
echo "enter the number"
read number2
if [ $number1 = $number2 ]
then
echo "matching"
exit
else [ $number1 != $number2 ]
echo "Not matching"
fi
exit

~
~
~
~
~
~
~
~
~
"script1.sh" 14 lines, 200 characters
$ ./script1.sh
enter the number
12
enter the number
13
Not matching
$


hope this helps!

kind regards
yogeeraj

[Yogeeraj_1 · Accepted solution]

hi again,

below a more readable output:

$ vi script1.sh
#!/usr/bin/sh
echo "enter the number"
read NUMBER1
echo "enter the number"
read NUMBER2
if [ $NUMBER1 = $NUMBER2 ]
then
echo "matching"
exit
else [ $NUMBER1 != $NUMBER2 ]
echo "Not matching"
fi
exit

~
~
~
~
~
~
~
~
~
"script1.sh" 14 lines, 200 characters
$ ./script1.sh
enter the number
12
enter the number
12
matching
$ r
./script1.sh
enter the number
12
enter the number
13
Not matching
$



kind regards
yogeeraj

[Sajjad Sahir]

#!/usr/bin/sh
echo "enter the number"
read NUMBER1
echo "enter the number"
read NUMBER2
if $NUMBER1=$NUMBER2
($NUMBER1 = $NUMBER2 I TESTED THIS ONE ALSO)
then
echo "matching"
else
echo "Not matching"
fi

[Sajjad Sahir]

dear all
still it showing error
i don't why i am still confusing


#!/usr/bin/sh
echo "enter the Number"
read NUMBER1
echo "enter the Number"
read NUMBER2
if [$NUMBER1= $NUMBER2]
then
echo "matching"
exit
else [$NUMBER1 != $NUMBER2]

echo "Not matching"
fi
exit

[Sajjad Sahir]

Yogeeraj what is the problem in my shell script
I wrote in the same way but i am not getting
any parmeter missing or somethign like that
or file permission some thing like that

#!/usr/bin/sh

echo "enter the Number"

read NUMBER1

echo "enter the Number"

read NUMBER2

if [ $NUMBER1= $NUMBER2 ]

then

echo "matching"

exit

else [ $NUMBER1 != $NUMBER2 ]

echo "Not matching"
fi
exit

[Davis Paul]

Hi Sajjad,
I can execute the following script without any error.
#!/usr/bin/sh
echo "enter the Number"
read NUMBER1
echo "enter the Number"
read NUMBER2
if [ $NUMBER1 -eq $NUMBER2 ]
then
echo "matching"
exit
else [ $NUMBER1 != $NUMBER2 ]
echo "Not matching"
fi
exit

For setting the execution permission, use the command:
#chmod u+x filename.
If you want a trial run of the script without setting the execution permition use:
# sh filename
Regards,
Davis Paul.

[Sajjad Sahir]

dear david paul
i gave already that permission
also
./
sh etc...

any how others can excute this one without
erro
but i am getting error
i don't why?
it is amazing me
sajjad

[Yogeeraj_1]

hi again,

it could be a problem with spaces in the script...

am attaching the code in a text file.

Please download and run it.

$ chmod +x script1.sh
$ ./script1.sh


revert

kind regards
yogeeraj

[Davis Paul]

Hi Sajjad,
Please post the error messages which you are getting now.
Regards,
Davis Paul

[Sajjad Sahir]

closed

[Dennis Handly]

>closed

You haven't closed the thread.
Also, if any answers here are useful, you need to assign points before you close it.

[OldSchool]

actually, the test in the else case isn't necessary, as its either matching or it doesn't...since you tested (and failed) match the else should just *happen*

also, the original poster needs to be aware that the tests indicated are for matching strings , so "1" and "01", while numerically equal fail as the strings differ

[Bill Hassell]

It's REALLY important to keep strings and numbers separate. The strings 01 001 and 1 are very different strings. *ALWAYS* compare numbers with the numeric comparison operators:

-eq -ne -gt -ge -lt -le

Note that the strings 1 9 22 777 will compare (sort) in this order:

1
22
777
9

but as numbers, they will sort this way:

1
9
22
777

[Aashique]

Hi,
here is the correct information of the script:

#!/usr/bin/sh
echo "enter the number"
read number1
echo "enter the number"
read number2
if [ $number1 = $number2 ]
then
echo "matching"
exit
else
echo "Not matching"
fi


Thanks & Regards

Aashique

[Yogeeraj_1]

hi Oldschool and Bill,

I thank you for making these clarifications. I was not really aware of these *Intricacies* in shell scripting... :(

it seems that the last poster did not read your post properly, so i will post the new script with corrections made as per your recommendations:


#!/usr/bin/sh
echo "enter the number"
read number1
echo "enter the number"
read number2
if [ $number1 -eq $number2 ]
then
echo "matching"
exit
else [ $number1 -ne $number2 ]
echo "Not matching"
fi

Sorry for inconvenices!

kind regards
yogeeraj

[Sajjad Sahir]

I closed