How can I iterate through a simple range of ints using a for loop in ksh?
For example, my script currently does this...
for i in 1 2 3 4 5 6 7
do
#stuff
done
...but I'd like to extend the range way above 7. Is there a better syntax?
Curly brackets?
for i in {1..7}
do
#stuff
done
While loop?
while [[ $i -lt 1000 ]] ; do
# stuff
(( i += 1 ))
done
on OpenBSD, use jot:
for i in `jot 10`; do echo $i ; done;
ksh93, Bash and zsh all understand C-like for loop syntax:
for ((i=1; i<=9; i++))
do
echo $i
done
Unfortunately, while ksh and zsh understand the curly brace range syntax with constants and variables, Bash only handles constants (including Bash 4).
Using seq:
for i in $(seq 1 10)
do
echo $i
done
The following will work on AIX / Linux / Solaris ksh.
#!/bin/ksh
d=100
while (( $d < 200 ))
do
echo "hdisk$d"
(( d=$d+1 ))
done
Optionally if you wanted to pad to 5 places, i.e. 00100 .. 00199 you could begin with:
#!/bin/ksh
typeset -Z5 d
-Scott
Just a few examples I use in AIX because there is no range operator or seq, abusing perl instead.
Here's a for loop, using perl like seq:
for X in `perl -e 'print join(" ", 1..10)'` ; do something $X ; done
This is similar, but I prefer while read loops over for. No backticks or issues with spaces.
perl -le 'print "$_ " for 1..10;' | while read X ; do xargs -tn1 ls $X ; done
My fav, do bash-like shell globbing, in this case permutations with perl.
perl -le 'print for glob "e{n,nt,t}{0,1,2,3,4,5}"' | xargs -n1 rmdev -dl
Related
I have this Bash script and I had a problem in line 16.
How can I take the previous result of line 15 and add
it to the variable in line 16?
#!/bin/bash
num=0
metab=0
for ((i=1; i<=2; i++)); do
for j in `ls output-$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 can't seem to be able to increase the variable value by 1. I have looked at tutorialspoint's Unix / Linux Shell Programming tutorial but it only shows how to add together two variables.
I have tried the following methods but they don't work:
i=0
$i=$i+1 # doesn't work: command not found
echo "$i"
$i='expr $i+1' # doesn't work: command not found
echo "$i"
$i++ # doesn't work*, command not found
echo "$i"
How do I increment the value of a variable by 1?
You can use an arithmetic expansion like so:
i=$((i+1))
or declare i as an integer variable and use the += operator for incrementing its value.
declare -i i=0
i+=1
or use the (( construct.
((i++))
The way to use expr:
i=0
i=`expr $i + 1`
The way to use i++ (unless you're running with -e/-o errexit):
((i++)); echo $i;
Tested in gnu bash.
you can use bc as it can also do floats
var=$(echo "1+2"|bc)
These are the methods I know:
ichramm#NOTPARALLEL ~$ i=10; echo $i;
10
ichramm#NOTPARALLEL ~$ ((i+=1)); echo $i;
11
ichramm#NOTPARALLEL ~$ ((i=i+1)); echo $i;
12
ichramm#NOTPARALLEL ~$ i=`expr $i + 1`; echo $i;
13
Note the spaces in the last example, also note that's the only one that uses $i.
Please tell me what is wrong with the UNIX code below.
#!/bin/ksh
p=10
for i in $p
do
echo $i
done
i am expecting output as
1
2
3
.
.
.
but the output am getting is just 10
I need for loop not while loop.
in ksh
#!/bin/ksh
p=10
i=1
while ((i<=p)); do
echo $i
i=$((i+1))
done
or
#!/bin/ksh
# with for you can only do this
for i in 1 2 3 4 5 6 7 8 9 10; do
echo $i
done
in bash it works as expected
#!/bin/bash
p=10
for (( i=1; i<=p; i++ )); do
echo $i
done
there is a Linux command seqthat can be used for both ksh and bash. But it is a Linux command. So this will not work on Solaris or other Unix systems that don't have the progrtam seq installed.
# on Linux, bash or ksh
p=10
for i in $(seq $p); do
echo $i
done
The following uses only shell built-ins and therefore will work for all bash installations (e.g. on Solaris) but not for ksh
#!/bin/bash
p=10
for i in `eval echo {1..$p}`; do
echo $i
done
This complicated construct is necessary because of brace expansion occurrs before variable expansion
You have to assign a range. Otherwise the loop can't work. This should do it:
#!/bin/ksh
p=10
for i in {0..$p}
do
echo $i
done
#fedorqui: You are right, I absolutely missed that. When I do stuff like this in Bash (I don't know if it's the same for KornShell), I go like:
for ((i=0; i<$p; i++))
in UNIX KSH
#!/bin/ksh
while [ ${i:=1} -le 10 ]
do
echo "$i"
let i+=1
done
Why the output is {1..3} rather than 123 ?
#!/bin/sh
a=1
for i in {$a..3}
do
echo -n $i
done
If I change {$a..3} to $(echo {$a..3}), it does not work either.
Brace expansion is performed before parameter substitution. But since that isn't a valid brace expansion, it isn't expanded. Use seq instead.
Ignacio's answer is right.
Here are some other solutions!
You can use a c-style for-loop in bash:
for (( i=a; i<=3; i++ ))
Or you can use dangerous eval, but you have to be sure that $a variable can't be anything else but a number, especially if the user is able to change it:
for i in $(echo eval {$a..3})
Or while loop with a variable in pure sh:
i=$a
while [ "$i" -le 3 ]
do
echo -n $i
i=$(( i + 1 ))
done
I want to write a loop in Bourne shell which iterates a specific set of numbers. Normally I would use seq:
for i in `seq 1 10 15 20`
#do stuff
loop
But seemingly on this Solaris box seq does not exist. Can anyone help by providing another solution to iterating a list of numbers?
try
for i in 1 10 15 20
do
echo "do something with $i"
done
else if you have recent Solaris, there is bash 3 at least. for example this give range from 1 to 10 and 15 to 20
for i in {1..10} {15..20}
do
echo "$i"
done
OR use tool like nawk
for i in `nawk 'BEGIN{ for(i=1;i<=10;i++) print i}'`
do
echo $i
done
OR even the while loop
while [ "$s" -lt 10 ]; do s=`echo $s+1|bc`; echo $s; done
You can emulate seq with dc:
For instance:
seq 0 5 120
is rewritten as:
dc -e '0 5 120 1+stsisb[pli+dlt>a]salblax'
Another variation using bc:
for i in $(echo "for (i=0;i<=3;i++) i"|bc); do echo "$i"; done
For the Bourne shell, you'll probably have to use backticks, but avoid them if you can:
for i in `echo "for (i=0;i<=3;i++) i"|bc`; do echo "$i"; done
#!/bin/sh
for i in $(seq 1 10); do
echo $i
done
I find that this works, albeit ugly as sin:
for i in `echo X \n Y \n Z ` ...
for i in `seq 1 5 20`; do echo $i; done
Result:
5
10
15
20
$ man seq
SEQ(1) User Commands SEQ(1)
NAME
seq - print a sequence of numbers
SYNOPSIS
seq [OPTION]... LAST
seq [OPTION]... FIRST LAST
seq [OPTION]... FIRST INCREMENT LAST