command line parameters


 
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# 1  
Old 11-18-2011
command line parameters

hi again

question on how to change code so that script will prompt to enter parameter if they are missing. . .

code I have so far :

Code:
#!/bin/bash

two="200"
three=500

if [ $one=100 ]; then
echo " first line parameter is $one "
else
echo -n " first parameter is missing , please write parameter "
read -e one
echo "first parameter is $one "
fi
echo " second line parameter is $two "
echo " third line parameter is $three "

exit 0

it just prints parameters $two and $three without asking user to enter $one
what is wrong here?
thanks
# 2  
Old 11-18-2011
Code:
[ -z "$VAR1" ] && read -p "Enter VAR1 value:" VAR1
[ -z "$VAR2" ] && read -p "Enter VAR2 value:" VAR2

# 3  
Old 11-19-2011
I was trying with [ ] and if statements but your way is simple and works, great stuff!

---------- Post updated at 07:20 AM ---------- Previous update was at 07:10 AM ----------

I was trying with [ ] and if statements but your way is simple and works, great stuff!

actually it works great when there is no parameter VAR# predefined but when parameter is predefined script don show it . . . .
PHP Code:
#!/bin/bash
# write a script that takes command line variables and prints out the
# number of parameters passed in and lists them out on separated lines

VAR1=
VAR2="200"
VAR3=500




[ -"$VAR1] && read -"Enter VAR1 value:" VAR1
[ -"$VAR2] && read -"Enter VAR1 value:" VAR2         
[ -"$VAR3] && read -"Enter VAR1 value:" VAR3

exit 
like here it prompts to enter VAR1
but VAR2 and VAR3 are predefined and are not shown
 
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