I have an issue with some homework, so basically we were asked to create a bash script that takes a variable greater than 2 and gives back the same number of Fibonacci sequence numbers, i.o. if i were to give 5 it would print:
0 1 1 2 3
I have done some research and have come up with a unique idea i havent really seen anywhere online, (keep in mind i have some experience in python) but i have run into a problem, so basically i use a for expression in bash and 2 variables to calculate the sequence, i first manually set them to be 0 and 1 (the first and second numbers in the sequence) and then i add them together while changing their values,
So here is the code :
#!/bin/bash
a=0
b=1
for i in $(seq 1 $1);
do
if ["$a" -gt "$b"]
then
b=($a +$b)
echo "$b"
else
a=($a +$b)
echo $a
fi
done
I have many issues and expect to be completely out of context but i hope you get the general idea and can help guide me through the problem :/
It seems like i want something to be treated as a number but its treated as text... Not sure though, any help is highly appreciated
You were very close, some small syntactic changes and new initial values make it work.
Notes:
whitespace is important in bash, especially with special characters [ ] ;
also, calculation with integer number can be forced using let
Nice idea to use a comparison to do the calculation with just two variables :)
#!/bin/bash -
a=1
b=0
echo $a
for i in $(seq 1 $1);
do
if [ $a -gt $b ] ; then
let b=($a +$b)
echo $b
else
let a=($a +$b)
echo $a
fi
done
there is an arithmetic context in bash.
slight re-write can be
$ a=0; b=1;
for i in {1..9};
do c=$((a+b));
echo $c;
if ((a>b)); then b=$c; else a=$c; fi;
done
1
2
3
5
8
13
21
34
55
You need a space
if [ "$a" -gt "$b" ]
and you can do arithmetic evaluation using
b=$((a+b))
once the script runs, verify the result is what you are expecting (i.e. does it print 0?)
Use double parens to evaluate arithmetic expressions. Instead of:
if [ "$a" -gt "$b" ]
write:
if ((a > b))
and instead of:
($a + $b)
write:
$((a + b))
You can also write the for header this way:
for ((i = 0; i < $1; i++))
But this is not required in this case. Your loop then becomes:
for ((i = 0; i < $1; i++)); do
if ((a > b)); then
b=$((a + b))
echo "$b"
else
a=$((a + b))
echo "$a"
fi
done
#!/bin/bash
a=0
b=1
echo $a
c=$1
for i in $(seq 1 $((c-1)));
do
if [ $a -gt $b ]; then
b=$((a+b))
echo $b
else
a=$((a+b))
echo $a
fi
done
Related
I'm unable to get numeric comparisons working:
echo "enter two numbers";
read a b;
echo "a=$a";
echo "b=$b";
if [ $a \> $b ];
then
echo "a is greater than b";
else
echo "b is greater than a";
fi;
The problem is that it compares the number from the first digit on, i.e., 9 is bigger than 10, but 1 is greater than 09.
How can I convert the numbers into a type to do a true comparison?
In Bash, you should do your check in an arithmetic context:
if (( a > b )); then
...
fi
For POSIX shells that don't support (()), you can use -lt and -gt.
if [ "$a" -gt "$b" ]; then
...
fi
You can get a full list of comparison operators with help test or man test.
Like this:
#!/bin/bash
a=2462620
b=2462620
if [ "$a" -eq "$b" ]; then
echo "They're equal";
fi
Integers can be compared with these operators:
-eq # Equal
-ne # Not equal
-lt # Less than
-le # Less than or equal
-gt # Greater than
-ge # Greater than or equal
See this cheatsheet.
There is also one nice thing some people might not know about:
echo $(( a < b ? a : b ))
This code will print the smallest number out of a and b
In Bash I prefer doing this as it addresses itself more as a conditional operation unlike using (( )) which is more of arithmetic.
[[ n -gt m ]]
Unless I do complex stuff like
(( (n + 1) > m ))
But everyone just has their own preferences. Sad thing is that some people impose their unofficial standards.
You can also do this:
[[ 'n + 1' -gt m ]]
Which allows you to add something else which you could do with [[ ]] besides arithmetic stuff.
The bracket stuff (e.g., [[ $a -gt $b ]] or (( $a > $b )) ) isn't enough if you want to use float numbers as well; it would report a syntax error. If you want to compare float numbers or float number to integer, you can use (( $(bc <<< "...") )).
For example,
a=2.00
b=1
if (( $(bc <<<"$a > $b") )); then
echo "a is greater than b"
else
echo "a is not greater than b"
fi
You can include more than one comparison in the if statement. For example,
a=2.
b=1
c=1.0000
if (( $(bc <<<"$b == $c && $b < $a") )); then
echo "b is equal to c but less than a"
else
echo "b is either not equal to c and/or not less than a"
fi
That's helpful if you want to check if a numeric variable (integer or not) is within a numeric range.
One-line solution.
a=2
b=1
[[ ${a} -gt ${b} ]] && echo "true" || echo "false"
gt reference: https://www.gnu.org/software/bash/manual/html_node/Bash-Conditional-Expressions.html
&& reference: https://www.gnu.org/software/bash/manual/html_node/Shell-Arithmetic.html
[[...]] construct reference: https://www.gnu.org/software/bash/manual/bash.html#index-_005b_005b
${} reference: https://pubs.opengroup.org/onlinepubs/9699919799/utilities/V3_chap02.html#tag_18_06_02 (2.6.2)
The format for parameter expansion is as follows:
${expression}
where expression consists of all characters until the matching '}'.
Any '}' escaped by a or within a quoted string, and
characters in embedded arithmetic expansions, command substitutions,
and variable expansions, shall not be examined in determining the
matching '}'.
The simplest form for parameter expansion is:
${parameter}
This code can also compare floats. It is using AWK (it is not pure Bash). However, this shouldn't be a problem, as AWK is a standard POSIX command that is most likely shipped by default with your operating system.
$ awk 'BEGIN {return_code=(-1.2345 == -1.2345) ? 0 : 1; exit} END {exit return_code}'
$ echo $?
0
$ awk 'BEGIN {return_code=(-1.2345 >= -1.2345) ? 0 : 1; exit} END {exit return_code}'
$ echo $?
0
$ awk 'BEGIN {return_code=(-1.2345 < -1.2345) ? 0 : 1; exit} END {exit return_code}'
$ echo $?
1
$ awk 'BEGIN {return_code=(-1.2345 < 2) ? 0 : 1; exit} END {exit return_code}'
$ echo $?
0
$ awk 'BEGIN {return_code=(-1.2345 > 2) ? 0 : 1; exit} END {exit return_code}'
$ echo $?
To make it shorter for use, use this function:
compare_nums()
{
# Function to compare two numbers (float or integers) by using AWK.
# The function will not print anything, but it will return 0 (if the comparison is true) or 1
# (if the comparison is false) exit codes, so it can be used directly in shell one liners.
#############
### Usage ###
### Note that you have to enclose the comparison operator in quotes.
#############
# compare_nums 1 ">" 2 # returns false
# compare_nums 1.23 "<=" 2 # returns true
# compare_nums -1.238 "<=" -2 # returns false
#############################################
num1=$1
op=$2
num2=$3
E_BADARGS=65
# Make sure that the provided numbers are actually numbers.
if ! [[ $num1 =~ ^-?[0-9]+([.][0-9]+)?$ ]]; then >&2 echo "$num1 is not a number"; return $E_BADARGS; fi
if ! [[ $num2 =~ ^-?[0-9]+([.][0-9]+)?$ ]]; then >&2 echo "$num2 is not a number"; return $E_BADARGS; fi
# If you want to print the exit code as well (instead of only returning it), uncomment
# the awk line below and comment the uncommented one which is two lines below.
#awk 'BEGIN {print return_code=('$num1' '$op' '$num2') ? 0 : 1; exit} END {exit return_code}'
awk 'BEGIN {return_code=('$num1' '$op' '$num2') ? 0 : 1; exit} END {exit return_code}'
return_code=$?
return $return_code
}
$ compare_nums -1.2345 ">=" -1.2345 && echo true || echo false
true
$ compare_nums -1.2345 ">=" 23 && echo true || echo false
false
If you have floats, you can write a function and then use that. For example,
#!/bin/bash
function float_gt() {
perl -e "{if($1>$2){print 1} else {print 0}}"
}
x=3.14
y=5.20
if [ $(float_gt $x $y) == 1 ] ; then
echo "do stuff with x"
else
echo "do stuff with y"
fi
I solved this by using a small function to convert version strings to plain integer values that can be compared:
function versionToInt() {
local IFS=.
parts=($1)
let val=1000000*parts[0]+1000*parts[1]+parts[2]
echo $val
}
This makes two important assumptions:
The input is a "normal SemVer string"
Each part is between 0-999
For example
versionToInt 12.34.56 # --> 12034056
versionToInt 1.2.3 # --> 1002003
Example testing whether npm command meets the minimum requirement...
NPM_ACTUAL=$(versionToInt $(npm --version)) # Capture npm version
NPM_REQUIRED=$(versionToInt 4.3.0) # Desired version
if [ $NPM_ACTUAL \< $NPM_REQUIRED ]; then
echo "Please update to npm#latest"
exit 1
fi
Just adding to all the above answers:
If you have more than one expression in single if statement, you can do something like this:
if (( $a % 2 == 0 )) && (( $b % 2 != 0));
then
echo "What you want to do"
fi
Hope this helps!
I have to write a script in bash where I have to provide 10 numbers to the table. Then script have to write the content out and arithmetic mean of even numbers. I did like 90% of the script, but I can't find out how to extract information about quantity of even numbers that is needed for arithmetic mean.
Here is my code:
echo "Provide data:"
i=0
for (( i = 0 ; i < 10; i++ ))
do
echo "Provide $[$i+1] number:"
read x
if [ "$x" = "" ]
then
break
else
table[$i]=$x
fi
done
echo "Provided data: ${table[*]}"
result=0
for (( i = 0 ; i < 10; i++))
do
res=$[${table[i]}%2]
if [ $res -eq 0 ]
then
echo "Number ${table[i]} is even"
result=$[$result+${table[$i]}]
fi
done
echo "SUM:$[$result]"
ignoring data input adding only odd inputs can look like :
$ cat c.sh
#!/bin/bash
declare -A xDarray
sum=0
xDarray[0 1]=1
xDarray[0 2]=3
xDarray[1 0]=2
xDarray[2 0]=4
for var in ${xDarray[#]}
do
if [ $(( $var & 1 )) == 0 ] ; then
echo $var is even
i+=1
tab[$i]=$var
sum=$(( $sum + $var))
fi
done
var=$(echo ${tab[#]} | sed 's/ / + /g' )
echo $var = $sum
result in
$ ./c.sh
2 is even
4 is even
2 + 4 = 6
$
whatever the number of data is used it would work
I let you work around your data input
Here are some suggested modifications for your script, syntax is simple.
#!/bin/bash
arr=()
for (( i=1;i<=10;i++ )); do
number=''
while [[ ! $number =~ ^[0-9]+$ ]]; do
printf "Please enter number $i:\n"
read number
done
arr+=($number)
done
printf "\nProvided numbers:"
printf " %d" "${arr[#]}"
printf "\nEven numbers:"
s=0
n=0
for x in "${arr[#]}"; do
if ! (( x % 2 )); then
printf " %d" "$x"
s=$(( s + x ))
(( n++ ))
fi
done
m=$(( s / n ))
printf "\nMean of the %d even numbers: %d / %d = %d\n" "$n" "$s" "$n" "$m"
Use an array arr to hold the input numbers, declare with arr=(), append numbers with arr+=($x), we refer only ${arr[#]} for all the items and we avoid any other complex array references, indices etc.
Every input number is tested against regular expression ^[0-9]+$ which means one or more digits (and only digits) with the =~ operator, and if this is not true, we prompt again for the same i-th input number.
Also we prefer printf for printing.
The last loop is the standard array loop, where we use again the arithmetic expansion syntax to find the even numbers, to add them to the sum and get the mean of them (result is an integer).
If you want to print a decimal result, e.g. with 2 floating points, you could use bc and printf "%f" like this:
m=$( bc <<< "scale=2; $s/$n" )
printf "%.2f" "$m"
I have this Bash script and I had a problem in line 16.
How can I take the previous result of line 15 and add
it to the variable in line 16?
#!/bin/bash
num=0
metab=0
for ((i=1; i<=2; i++)); do
for j in `ls output-$i-*`; do
echo "$j"
metab=$(cat $j|grep EndBuffer|awk '{sum+=$2} END { print sum/120}') (line15)
num= $num + $metab (line16)
done
echo "$num"
done
For integers:
Use arithmetic expansion: $((EXPR))
num=$((num1 + num2))
num=$(($num1 + $num2)) # Also works
num=$((num1 + 2 + 3)) # ...
num=$[num1+num2] # Old, deprecated arithmetic expression syntax
Using the external expr utility. Note that this is only needed for really old systems.
num=`expr $num1 + $num2` # Whitespace for expr is important
For floating point:
Bash doesn't directly support this, but there are a couple of external tools you can use:
num=$(awk "BEGIN {print $num1+$num2; exit}")
num=$(python -c "print $num1+$num2")
num=$(perl -e "print $num1+$num2")
num=$(echo $num1 + $num2 | bc) # Whitespace for echo is important
You can also use scientific notation (for example, 2.5e+2).
Common pitfalls:
When setting a variable, you cannot have whitespace on either side of =, otherwise it will force the shell to interpret the first word as the name of the application to run (for example, num= or num)
num= 1 num =2
bc and expr expect each number and operator as a separate argument, so whitespace is important. They cannot process arguments like 3+ +4.
num=`expr $num1+ $num2`
Use the $(( )) arithmetic expansion.
num=$(( $num + $metab ))
See Chapter 13. Arithmetic Expansion for more information.
There are a thousand and one ways to do it. Here's one using dc (a reverse Polish desk calculator which supports unlimited precision arithmetic):
dc <<<"$num1 $num2 + p"
But if that's too bash-y for you (or portability matters) you could say
echo $num1 $num2 + p | dc
But maybe you're one of those people who thinks RPN is icky and weird; don't worry! bc is here for you:
bc <<< "$num1 + $num2"
echo $num1 + $num2 | bc
That said, there are some unrelated improvements you could be making to your script:
#!/bin/bash
num=0
metab=0
for ((i=1; i<=2; i++)); do
for j in output-$i-* ; do # 'for' can glob directly, no need to ls
echo "$j"
# 'grep' can read files, no need to use 'cat'
metab=$(grep EndBuffer "$j" | awk '{sum+=$2} END { print sum/120}')
num=$(( $num + $metab ))
done
echo "$num"
done
As described in Bash FAQ 022, Bash does not natively support floating point numbers. If you need to sum floating point numbers the use of an external tool (like bc or dc) is required.
In this case the solution would be
num=$(dc <<<"$num $metab + p")
To add accumulate possibly-floating-point numbers into num.
In Bash,
num=5
x=6
(( num += x ))
echo $num # ==> 11
Note that Bash can only handle integer arithmetic, so if your AWK command returns a fraction, then you'll want to redesign: here's your code rewritten a bit to do all math in AWK.
num=0
for ((i=1; i<=2; i++)); do
for j in output-$i-*; do
echo "$j"
num=$(
awk -v n="$num" '
/EndBuffer/ {sum += $2}
END {print n + (sum/120)}
' "$j"
)
done
echo "$num"
done
I always forget the syntax so I come to Google Search, but then I never find the one I'm familiar with :P. This is the cleanest to me and more true to what I'd expect in other languages.
i=0
((i++))
echo $i;
I really like this method as well. There is less clutter:
count=$[count+1]
#!/bin/bash
read X
read Y
echo "$(($X+$Y))"
You should declare metab as integer and then use arithmetic evaluation
declare -i metab num
...
num+=metab
...
For more information, see 6.5 Shell Arithmetic.
Use the shell built-in let. It is similar to (( expr )):
A=1
B=1
let "C = $A + $B"
echo $C # C == 2
Source: Bash let builtin command
Another portable POSIX compliant way to do in Bash, which can be defined as a function in .bashrc for all the arithmetic operators of convenience.
addNumbers () {
local IFS='+'
printf "%s\n" "$(( $* ))"
}
and just call it in command-line as,
addNumbers 1 2 3 4 5 100
115
The idea is to use the Input-Field-Separator(IFS), a special variable in Bash used for word splitting after expansion and to split lines into words. The function changes the value locally to use word-splitting character as the sum operator +.
Remember the IFS is changed locally and does not take effect on the default IFS behaviour outside the function scope. An excerpt from the man bash page,
The shell treats each character of IFS as a delimiter, and splits the results of the other expansions into words on these characters. If IFS is unset, or its value is exactly , the default, then sequences of , , and at the beginning and end of the results of the previous expansions are ignored, and any sequence of IFS characters not at the beginning or end serves to delimit words.
The "$(( $* ))" represents the list of arguments passed to be split by + and later the sum value is output using the printf function. The function can be extended to add scope for other arithmetic operations also.
#!/usr/bin/bash
#integer numbers
#===============#
num1=30
num2=5
echo $(( num1 + num2 ))
echo $(( num1-num2 ))
echo $(( num1*num2 ))
echo $(( num1/num2 ))
echo $(( num1%num2 ))
read -p "Enter first number : " a
read -p "Enter second number : " b
# we can store the result
result=$(( a+b ))
echo sum of $a \& $b is $result # \ is used to espace &
#decimal numbers
#bash only support integers so we have to delegate to a tool such as bc
#==============#
num2=3.4
num1=534.3
echo $num1+$num2 | bc
echo $num1-$num2 | bc
echo $num1*$num2 |bc
echo "scale=20;$num1/$num2" | bc
echo $num1%$num2 | bc
# we can store the result
#result=$( ( echo $num1+$num2 ) | bc )
result=$( echo $num1+$num2 | bc )
echo result is $result
##Bonus##
#Calling built in methods of bc
num=27
echo "scale=2;sqrt($num)" | bc -l # bc provides support for calculating square root
echo "scale=2;$num^3" | bc -l # calculate power
#!/bin/bash
num=0
metab=0
for ((i=1; i<=2; i++)); do
for j in `ls output-$i-*`; do
echo "$j"
metab=$(cat $j|grep EndBuffer|awk '{sum+=$2} END { print sum/120}') (line15)
let num=num+metab (line 16)
done
echo "$num"
done
Works on MacOS. bc is a command line calculator
#!/bin/bash
sum=0
for (( i=1; i<=5; i++ )); do
sum=$(echo "$sum + 1.1" | bc) # bc: if you want to use decimal
done
echo "Total: $sum"
I'm unable to get numeric comparisons working:
echo "enter two numbers";
read a b;
echo "a=$a";
echo "b=$b";
if [ $a \> $b ];
then
echo "a is greater than b";
else
echo "b is greater than a";
fi;
The problem is that it compares the number from the first digit on, i.e., 9 is bigger than 10, but 1 is greater than 09.
How can I convert the numbers into a type to do a true comparison?
In Bash, you should do your check in an arithmetic context:
if (( a > b )); then
...
fi
For POSIX shells that don't support (()), you can use -lt and -gt.
if [ "$a" -gt "$b" ]; then
...
fi
You can get a full list of comparison operators with help test or man test.
Like this:
#!/bin/bash
a=2462620
b=2462620
if [ "$a" -eq "$b" ]; then
echo "They're equal";
fi
Integers can be compared with these operators:
-eq # Equal
-ne # Not equal
-lt # Less than
-le # Less than or equal
-gt # Greater than
-ge # Greater than or equal
See this cheatsheet.
There is also one nice thing some people might not know about:
echo $(( a < b ? a : b ))
This code will print the smallest number out of a and b
In Bash I prefer doing this as it addresses itself more as a conditional operation unlike using (( )) which is more of arithmetic.
[[ n -gt m ]]
Unless I do complex stuff like
(( (n + 1) > m ))
But everyone just has their own preferences. Sad thing is that some people impose their unofficial standards.
You can also do this:
[[ 'n + 1' -gt m ]]
Which allows you to add something else which you could do with [[ ]] besides arithmetic stuff.
The bracket stuff (e.g., [[ $a -gt $b ]] or (( $a > $b )) ) isn't enough if you want to use float numbers as well; it would report a syntax error. If you want to compare float numbers or float number to integer, you can use (( $(bc <<< "...") )).
For example,
a=2.00
b=1
if (( $(bc <<<"$a > $b") )); then
echo "a is greater than b"
else
echo "a is not greater than b"
fi
You can include more than one comparison in the if statement. For example,
a=2.
b=1
c=1.0000
if (( $(bc <<<"$b == $c && $b < $a") )); then
echo "b is equal to c but less than a"
else
echo "b is either not equal to c and/or not less than a"
fi
That's helpful if you want to check if a numeric variable (integer or not) is within a numeric range.
One-line solution.
a=2
b=1
[[ ${a} -gt ${b} ]] && echo "true" || echo "false"
gt reference: https://www.gnu.org/software/bash/manual/html_node/Bash-Conditional-Expressions.html
&& reference: https://www.gnu.org/software/bash/manual/html_node/Shell-Arithmetic.html
[[...]] construct reference: https://www.gnu.org/software/bash/manual/bash.html#index-_005b_005b
${} reference: https://pubs.opengroup.org/onlinepubs/9699919799/utilities/V3_chap02.html#tag_18_06_02 (2.6.2)
The format for parameter expansion is as follows:
${expression}
where expression consists of all characters until the matching '}'.
Any '}' escaped by a or within a quoted string, and
characters in embedded arithmetic expansions, command substitutions,
and variable expansions, shall not be examined in determining the
matching '}'.
The simplest form for parameter expansion is:
${parameter}
This code can also compare floats. It is using AWK (it is not pure Bash). However, this shouldn't be a problem, as AWK is a standard POSIX command that is most likely shipped by default with your operating system.
$ awk 'BEGIN {return_code=(-1.2345 == -1.2345) ? 0 : 1; exit} END {exit return_code}'
$ echo $?
0
$ awk 'BEGIN {return_code=(-1.2345 >= -1.2345) ? 0 : 1; exit} END {exit return_code}'
$ echo $?
0
$ awk 'BEGIN {return_code=(-1.2345 < -1.2345) ? 0 : 1; exit} END {exit return_code}'
$ echo $?
1
$ awk 'BEGIN {return_code=(-1.2345 < 2) ? 0 : 1; exit} END {exit return_code}'
$ echo $?
0
$ awk 'BEGIN {return_code=(-1.2345 > 2) ? 0 : 1; exit} END {exit return_code}'
$ echo $?
To make it shorter for use, use this function:
compare_nums()
{
# Function to compare two numbers (float or integers) by using AWK.
# The function will not print anything, but it will return 0 (if the comparison is true) or 1
# (if the comparison is false) exit codes, so it can be used directly in shell one liners.
#############
### Usage ###
### Note that you have to enclose the comparison operator in quotes.
#############
# compare_nums 1 ">" 2 # returns false
# compare_nums 1.23 "<=" 2 # returns true
# compare_nums -1.238 "<=" -2 # returns false
#############################################
num1=$1
op=$2
num2=$3
E_BADARGS=65
# Make sure that the provided numbers are actually numbers.
if ! [[ $num1 =~ ^-?[0-9]+([.][0-9]+)?$ ]]; then >&2 echo "$num1 is not a number"; return $E_BADARGS; fi
if ! [[ $num2 =~ ^-?[0-9]+([.][0-9]+)?$ ]]; then >&2 echo "$num2 is not a number"; return $E_BADARGS; fi
# If you want to print the exit code as well (instead of only returning it), uncomment
# the awk line below and comment the uncommented one which is two lines below.
#awk 'BEGIN {print return_code=('$num1' '$op' '$num2') ? 0 : 1; exit} END {exit return_code}'
awk 'BEGIN {return_code=('$num1' '$op' '$num2') ? 0 : 1; exit} END {exit return_code}'
return_code=$?
return $return_code
}
$ compare_nums -1.2345 ">=" -1.2345 && echo true || echo false
true
$ compare_nums -1.2345 ">=" 23 && echo true || echo false
false
If you have floats, you can write a function and then use that. For example,
#!/bin/bash
function float_gt() {
perl -e "{if($1>$2){print 1} else {print 0}}"
}
x=3.14
y=5.20
if [ $(float_gt $x $y) == 1 ] ; then
echo "do stuff with x"
else
echo "do stuff with y"
fi
I solved this by using a small function to convert version strings to plain integer values that can be compared:
function versionToInt() {
local IFS=.
parts=($1)
let val=1000000*parts[0]+1000*parts[1]+parts[2]
echo $val
}
This makes two important assumptions:
The input is a "normal SemVer string"
Each part is between 0-999
For example
versionToInt 12.34.56 # --> 12034056
versionToInt 1.2.3 # --> 1002003
Example testing whether npm command meets the minimum requirement...
NPM_ACTUAL=$(versionToInt $(npm --version)) # Capture npm version
NPM_REQUIRED=$(versionToInt 4.3.0) # Desired version
if [ $NPM_ACTUAL \< $NPM_REQUIRED ]; then
echo "Please update to npm#latest"
exit 1
fi
Just adding to all the above answers:
If you have more than one expression in single if statement, you can do something like this:
if (( $a % 2 == 0 )) && (( $b % 2 != 0));
then
echo "What you want to do"
fi
Hope this helps!
Here is my simple shell code. I want the result to be 2.Shell treats everything as a string.
How can i get this done ?
num=1
num=$(( $num + 1 ))
EDIT :
Complete code : Whats wrong in this if i want to print from 1 to 10 ?
#! /bin/bash
num=1
until test $num -eq 10
do
num=$(( $num + 1 ))
echo $num
done
In bash, you don't need to do anything special:
aix#aix:~$ num=1
aix#aix:~$ num=$(( $num + 1 ))
aix#aix:~$ echo $num
2
#tonio; please don't advocate using subshell (` ... or $( ... ) ) constructs when they're not needed (to keep confusion to the maximum, $(( ... )) is not a sub-shell construct). Sub-shells can make a tremendous performance hit even with rather trivial amounts of data. The same is true for every place where an external program is used to do somethign that could be done with a shel built-in.
Example:
num=1
time while [[ $num -lt 10000 ]]; do
num=$(( num+1 ))
done
echo $num
num=1
time while /bin/test $num -lt 10000; do
num=$( /bin/expr $num + 1 )
done
echo $num
Output (run in ksh on Linux):
real 0m0.04s
user 0m0.04s
sys 0m0.01s
10000
real 0m20.32s
user 0m2.23s
sys 0m2.92s
10000
...so run-time factor of 250, and CPU-time factor of 100. I admit the example I used was a exaggerated one, with explicitly requiring all built-ins to be bypassed, but I think the point was made: creating new processes is expenisve, avoid it when you can, and know your shell to recognise where new processes are created.
This might work for you:
num=1; ((num++)); echo $num
2
or
num=1; echo $((++num))
2
for loops
for num in {1..10}; do echo $num; done
or (in bash at least)
for ((num=1; num<=10; num++)) { echo $num; }
second loop more useful when more programming involved:
for (( num=1,mun=10; num<=10; num++,mun--)) { echo $num $mun; }
You just did:
$ num=1; num=$(( $num + 1 ));echo $num
2
Note: You don't need to quote variables inside $(( )). Also, you can just use $((num++))
You are not specifying which shell you are using, but the most concise form I know is this one (works at least in bash):
num=$[num+1]
If only incrementing by one and changing the variable itself rather than printing/assigning, then:
((num++))
Is a better/more elegant solution. See dogbane's answer for that.
If looping over the values, I would use this form instead:
for i in `seq 1 10`; do
echo $i
done
Use ((num++)) as shorthand for incrementing num.
$ num=1
$ ((num++))
$ echo $num
2
try this
$ num=1; num=`expr $num + 1`; echo $num;
EDIT:
More efficient would be:
num=$(( num + 1 ))
Thanks #Charles Duffy for your comment.
works for me
$ num=1
$ num=$(( $num + 1 ))
$ echo $num
2
What output do you get?
Read more about bash Arithmetic # tldp
EDIT
To do something 10 times in bash you can use (using brace-expansion}
$ for i in {1..10}; do echo $i; done
1
2
3
4
5
6
7
8
9
10
However, you cannot use variables between the {}. If this is the case, use seq instead.