I want to create a select menu in it, like this:
echo "Choose your option:"
1) Factorial Calculation
2) Addition Calculator
3) Quit
And I have some shell scripts;
Factorial
./fact.sh
#!/bin/bash
fact=1
#taking input from user
echo -e "enter a number"
read n
#if enter value less than 0
if [ $n -le 0 ] ; then
echo "invalid number"
exit
fi
#factorial logic
if [ $n -gt 0 ] ; then
for((i=$n;i>=1;i--))
do
fact=`expr $fact \* $i`
done
fi
echo "The factorial of $n is $fact"
Addition
./add.sh
#!/bin/bash
#function to add two numbers
add()
{
x=$1
y=$2
echo -e "Number entered by u are: $x and $y"
echo "sum of $1 and $2 is `expr $x + $y` "
}
# main script
echo "enter first number"
read first
echo "enter second number"
read sec
#calling function
add $first $sec
echo "end of the script"
I have to create a menu, how should I proceed?
You can use select.
For your example:
select option in Factorial Addition Quit; do
case "$option" in
"Factorial")
echo "Factorial"
break ;;
"Addition")
echo "Addition"
break ;;
"Quit") exit ;;
esac
done
Related
I am trying to make a simple menu-driven calculator script. I am trying to make it such that after selecting A (add) or B (subtract) from the menu, it'll call the function and display the corresponding message when:
The parameters entered when function called is greater than 3
No parameters are entered when the function is called
The operator entered when the function is called is neither "+" or "-"
Right now it is doing the parameter check when I call the script ./mycalc.sh
executing the script
Am not sure how to make it check parameters after the function is called?
#!/bin/bash
display() {
echo "Calculator Menu"
echo "Please select an option between add, subtract or exit"
echo "A. Add"
echo "B. Subtract"
echo "C. Exit"
}
#initialize choice n
choice=n
if [[ $# -ne 3 ]]
then echo " You have not entered 3 parameters"
exit 1
fi
if [ $# -eq 0 ]
then
echo " You have not entered any parameters, please input 3. "
fi
if [[ $2 != [+-] ]]
then
echo " Please enter an add or subtract operator."
exit 1
fi
add() {
echo " The sum of $one + $three equals $(( $one $op $three ))"
}
subtract () {
echo " The difference of $one - $three equals $(( $one $op $three )) "
}
while [ $choice != 'C' ]
do display
read choice
if [ $choice = 'A' ]
then
read -p "Please enter two operands and the operator '+': " one op three
add $one $op $three
elif [ $choice = 'B' ]
then
read -p " Please enter two operands and the operator '-': " one op three
subtract $one $op $three
elif [ $choice = 'C' ]
then
echo "Thank you for using this program. The program will now exit."
fi
done
sleep 3
exit 0
if [ $# > 3 ]
then
echo " You need to input 3 parameters. "
fi
Within [...], the > is simply redirection. You'll find an empty file named 3 in your current directory.
Since this is an error condition for your script, you'll want to exit:
if [[ $# -ne 3 ]]; then
echo "usage: $0 operand1 operator operand2" >&2
exit 1
fi
And to test the operator, there are many ways.
case, but it's a bit verbose
case $2 in
'+'|'-') : ;;
*) echo "Operator must be + or -" >&2
exit 1
;;
esac
Within [[...]] the == and != operators are pattern matching operators
if [[ $2 != [+-] ]]; then
echo "Operator must be + or -" >&2
exit 1
fi
I am trying to make a simple calculator. I am sure if you even just glanced at this code you will see what I am trying to do.
Enter a number, then choose an operand, then vim should print out a table up to 15 with your number and operand...
Maybe this method is silly, trying to nest a load of loops in nested if statements. But I am new to bash.
The error is 'Unexpected token near else' line 24 but I feel there is a fundamental issue with the nests I do not understand.
Here is current code.
#!/bin/bash
choice=6
read -p "Enter a number bruv" num
#choose operand.
echo "Now choose an operand comrade"
#choices
echo "1. *"
echo "2. +"
echo "3. -"
echo "4. /"
echo "5. ^"
echo -n "Please choose [1,2,3,4,5]"
while [ $choice -eq 6 ]; do
read choice
if [ $choice -eq 1 ] ; then
for((i=0;i<=15;i++))
do
echo -n "$i * $num = $[ $i * $num ] "
echo " "
else
if [ $choice -eq 2 ] ; then
for((i=0;i<=15;i++))
do
echo -n "$i + $num = $[ $i + $num ] "
echo " "
else
if [ $choice -eq 3 ] ; then
for((i=0;i<=15;i++))
do
echo -n "$i - $num = $[ $i - $num ] "
echo " "
else
if [ $choice -eq 4 ] ; then
for((i=0;i<=15;i++))
do
echo -n "$i / $num = $[ $i / $num ] "
echo " "
else
if [ $choice -eq 5 ] ; then
for((i=0;i<=15;i++))
do
echo -n "$i ^$num = $[ $i ^$num ] "
echo " "
else echo "Please choose between 1 and 5!!!"
echo "1. *"
echo "2. +"
echo "3. -"
echo "4. /"
echo "5. ^"
echo -n "Please choose [1,2,3,4,5]"
fi
fi
fi
fi
fi
done
Would it be better to implement this?
# !/bin/bash
# Take user Input
echo "Enter number : "
read a
# Input type of operation
echo "Enter Choice :"
echo "1. Addition"
echo "2. Subtraction"
echo "3. Multiplication"
echo "4. Division"
echo "5. Power"
read ch
# Switch Case to perform
# calulator operations
case $ch in
1)res=`for((i=0;i<=15;i++))
do
echo -n "$i - $num = $[ $i - $num ] "
echo " "`
;;
2)res=`for((i=0;i<=15;i++))
do
echo -n "$i - $num = $[ $i - $num ] "
echo " "`
;;
3)res=`for((i=0;i<=15;i++))
do
echo -n "$i - $num = $[ $i - $num ] "
echo " "`
;;
4)res=`for((i=0;i<=15;i++))
do
echo -n "$i - $num = $[ $i - $num ] "
echo " "c`
;;
esac
echo "Result : $res" ```
Here is a solution, using a function:
#! /bin/bash
ITER_MAX=15
show_values() # n op
{
local n=$1 op=$2
for ((i=0; i<=ITER_MAX; i++)); do
((i>0)) && echo -n " ; "
echo -n "$i $op $n = $((i $op n))"
done
echo
}
# Take user Input
read -p "Enter number : " a
# Input type of operation
echo "Enter Choice (Ctrl+C to stop):"
PS3=">> "
select ch in Addition Subtraction Multiplication Division Power ; do
case "$ch" in
Add*) op="+" ;;
Sub*) op="-" ;;
Mul*) op="*" ;;
Div*) op="/" ;;
Pow*) op="**" ;;
*) echo "Bad choice, abort" >&2 ; break ;;
esac
show_values "$a" "$op"
done
Some explanations:
(( )) is arithmetic evaluation and $(( )) is arithmetic expansion
((i>0)) && echo -n " ; " is equivalent to if ((i>0)); then echo -n " ; " ; fi
read -p "Enter number : " a is equivalent to echo -n "Enter number : " ; read a
about select, see help select in your bash terminal.
Use https://www.shellcheck.net/
Line 20:
for((i=0;i<=15;i++))
^-- SC1009: The mentioned syntax error was in this for loop.
^-- SC1073: Couldn't parse this arithmetic for condition. Fix to allow more checks.
Line 21:
do
^-- SC1061: Couldn't find 'done' for this 'do'.
Line 24:
else
^-- SC1062: Expected 'done' matching previously mentioned 'do'.
^-- SC1072: Unexpected keyword/token. Fix any mentioned problems and try again.
An alternate take:
read -p "Enter a number and an operator: " -a a
for n in {1..15}; do printf "%+10s\n" $((${a[#]} $n)); done
-p tells read to supply a prompt. -a reads into an array (named a here).
{1..15} is built-in bash sequence syntax.
%+10s tells printf to space-pad/right justify out to 10 characters.
${a[#]} is replaced with all the elements of the array - the number and then operator.
$(( ... )) does arithmetic processing and replaces itself with the result (as a string).
So if you enter "10 *" then $((${a[#]} $n)) processes 10 * 1 the first time, so printf "%+10s\n" $((${a[#]} $n)) outputs " 10" with a trailing newline character. Then the loop replaces the 1 with the next number on each iteration, up to 15. This also allows other operators, such as % for modulus, but will crash if something invalid is given.
If you want error checking -
while [[ ! "${a[*]}" =~ ^[0-9][0-9]*\ ([*/+-]|\*\*)$ ]]
do read -p "Enter a, integer and an operator (one of: + - * / **) " -a a
done
for n in {1..15}; do printf "%+10s\n" $((${a[#]} $n)); done
If you want floating point math, try bc:
while [[ ! "${a[*]}" =~ ^[0-9]+.?[0-9]*\ ([*/+-]|\*\*)$ ]]
do read -p "Enter a, integer and an operator (one of: + - * / **) " -a a
done
for n in {1..15}; do printf "%+15s\n" $(bc -l <<< "scale=3;${a[#]} $n"); done
Personally, I'd put the inputs on the command line. Less futzy, though it might require quoting the * operator, depending on how you run it and what's in the directory.
#!/bin/bash
me=${0##*/}
use="
use: $me {number} {operator} [iterations] [scale]
Numbers may include one decimal point.
Operators must be one of: + - * / **
"
while getopts "n:o:i:s:" arg
do case $arg in
n) n="$OPTARG" ;;
o) o="$OPTARG" ;;
i) i="$OPTARG" ;;
s) s="$OPTARG" ;;
*) printf "%s\n" "Invalid argument '$arg' $use";
exit 1;;
esac
done
[[ -n "$n" && -n "$o" ]] || { echo "$use"; exit 1; }
[[ "$n" =~ ^[0-9]+.?[0-9]*$ ]] || { printf "%s\n" "Invalid number '$n' $use"; exit 1; }
[[ "$o" =~ ^([*/^+-]|\*\*)$ ]] || { printf "%s\n" "Invalid operator '$o' $use"; exit 1; }
[[ "${i:=15}" =~ ^[0-9]+.?[0-9.]$ ]] || { printf "%s\n" "Invalid number '$i' $use"; exit 1; }
[[ "${s:=3}" =~ ^[0-9]+$ ]] || { printf "%s\n" "Invalid scale '$s' (must be an integer) $use"; exit 1; }
c=1
while ((c < i))
do printf "%+15s\n" $(bc -l <<< "scale=$s; $n $o $c")
((c++))
done
I commented the things I have problem.
and also, is there any other way I can exit my loop without using the exit command?.................
#!/bin/bash
while [ "$done" != "true" ] #this don't work
do
echo "Please enter one of the following options"
echo "1. Move empty files"
echo "2. Check file size"
echo "3. Which file is newer"
echo "4. File check rwx"
echo "5. Exit".
echo -n "Enter Choice: "
read scale # starting from this part for checking if user only inputs numbers 1-5 not working
if ! [[ "$scale" =~ ^[0-6]+$ ]]
then
echo "Invalid Input"
fi #up to this part is not working
read -r answer
case "$answer" in
1) ./move_empty
exit 55
;;
2) ./file_size
exit 0
;;
3) ;;
;;
4)
;;
5) done="true";;
esac
done
There are few things Which should be avoided
while [ "$done" != "true" ] --> while [ $done -ne 1 ] /* -ne is not equal to */
case "$answer" in --> case $answer in
Complete Code
var=0
flag=1
while [ $var -ne 1 ]
do
echo "Please enter one of the following options"
echo "1. Move empty files"
echo "2. Check file size"
echo "3. Which file is newer"
echo "4. File check rwx"
echo "5. Exit".
echo -n "Enter Choice: "
read scale
if [ $scale -gt 0 -a $scale -lt 6 ]
then
echo "valid Input, you can procees for switch "
else
echo "invalid input.. go again & give correct one "
flag=0
fi
if [ $flag -eq 1 ]
then
read -r answer
case $answer in
1) ./move_empty
#exit 55
break
;;
2) ./file_size
#exit 0
break
;;
3)
exit 0
;;
4)
;;
5) done="true";;
esac
fi
done
To come out from loop, without using exit command, use break. Here is the answer from man 1 bash
break [n]
Exit from within a for, while, until, or select loop.
If n is specified, break n levels. n must be ≥ 1. If n
is greater than the number of enclosing loops, all
enclosing loops are exited. The return value is 0
unless n is not greater than or equal to 1.
I keep getting unexpected End of file error while running a if else statement
#! /bin/bash
echo -e "1: Proband\n 2: mincount\n Enter an option:"
read promin
echo $promin
if ($promin == 1) then
echo -e "Enter the proband file name\n"
read proband_file
echo "$proband_file"
endif
if ($promin == 2) then
echo -e "enter the min count number\n"
read mincount
echo "$mincount mincount"
endif
I tried fi instead of elseif too. But i still get the same error. Can someone help me fix that?
This is how you write an if-statement in bash:
if - then - fi
if [ conditional expression ]
then
statement1
statement2
fi
if - then - else - fi
If [ conditional expression ]
then
statement1
statement2
else
statement3
statement4
fi
if - then - elif - else - fi
If [ conditional expression1 ]
then
statement1
statement2
elif [ conditional expression2 ]
then
statement3
statement4
else
statement5
fi
Example of a conditional expression:
#!/bin/bash
count=100
if [ $count -eq 100 ]
then
echo "Count is 100"
fi
IMPROVED
The if is syntax is not correct. In the if there should be a program (bash internal or external) run, which returns an exit code. If it is 0 then if is true, otherwise it is false. You can use grep or any other utility, like test or /usr/bin/[. But bash has a built-in test and [.
So [ "$var" -eq 1 ] returns 0 if $var equals 1, or return 1 if $var not equals 1.
In your case I would suggest to use case instead of if-then-elif-else-fi notation.
case $x in
1) something;;
2) other;;
*) echo "Error"; exit 1;;
easc
Or even use select. Example:
#!/bin/bash
PS3="Enter an option: "
select promin in "Proband" "mincount";do [ -n "$promin" ] && break; done
echo $promin
case "$promin" in
Proband) read -p "Enter the proband file name: " proband_file; echo "$proband_file";;
mincount) read -p "Enter the min count number: " mincount; echo "$mincount mincount";;
*) echo Error; exit 1;;
esac
This will print the "Enter an option: " prompt and wait until a proper answer is presented (1 or 2 or ^D - to finish the input).
1) Proband
2) mincount
Enter an option: _
Then it checks the answer in the case part. Meanwhile $promin contains the string, $REPLY contains the entered answer. It also can be used in case.
I just changed your code and I think it works now.
I think the problem is you should fi instead of endif...
#!/bin/sh
echo "1: Proband\n2: mincount\nEnter an option:"
read promin
echo $promin
if [ $promin -eq "1" ]
then
echo "Enter the proband file name\n"
read proband_file
echo "$proband_file"
elif [ $promin -eq "2" ]
then
echo "enter the min count number\n"
read mincount
echo "$mincount mincount"
fi
#! /bin/bash
echo -e "1: Proband\n2: mincount\nEnter an option:"
read promin
echo $promin
if (($promin == 1)); then
echo -e "Enter the proband file name\n"
read proband_file
echo "$proband_file"
elif (($promin == 2)); then
echo -e "Enter the min count number\n"
read mincount
echo "$mincount mincount"
fi
I don't know if you need an if-else statement or two if statemnts. The above has an if-else.
If you need two if statements, then insert a line of code below the "echo "$proband_file"" line with the text:
fi
Then replace the line "elif (($promin == 2)); then" with the following code:
if (($promin == 2)); then
When I write parameters they should be checked in a case and then added in a list if u type -c in front the parameter should be in uppercase and if u type -4 the parameter should appear 4 times
#!/bin/bash
function forloop {
for each in naam
do
echo $naam zegt je dat...
echo ""
echo
let "x+=1"
done
}
function helper {
echo "use names only."
echo "if u type -c before a word then it wil be shown in capital"
echo "if u type -(number) before a word the word will be shown (number)times "
}
naam=null
x=0
while [ $x -le 4 ]
do
case $1 in
-c) echo "${2 ^^}";shift ;;
help) helper ;shift;naam=helper;;
-4) $1x-1
exit ;shift;;
esac
naam=" $1"
shift
forloop
done
exit
#!/bin/bash
if [ -z "$1" ]
then
echo "use names only."
echo "if you type -c before a word, then it will be shown in capitals"
echo "if you type -NUMBER before a word, the word will be shown NUMBER times"
exit
fi
while [ $1 ]
do
case $1 in
-c) shift; <<<$1 tr '[:lower:]' '[:upper:]';;
-[0-9]*)for ((count=$1; count++; )) do echo $2; done; shift;;
*) echo $1
esac
shift
done