I am trying to print a filename which contains decimal point numbers...
say
L2.3stop.txt
I have variables defined as :
z1=2.3
z2=3.4
z3=7.8
z4=8.9
and so on
In a for loop i runs from 1 to 5
Inside the loop if I do
temp=`echo z$i`
and then I print the file name using
echo L${temp}stop.txt
it just prints
Lz1stop.txt
Lz2stop.txt
etc..
How can I print the desired filename....
I also tried using
echo L$((z$i))stop.txt
but this only works when z1, z2, z3 etc are integers and not floating numbers....
I hope this helps:
z1=2.3
z2=3.4
z3=7.8
z4=8.9
for i in {1..4};do
echo L$(eval "echo \$z"$i)stop.txt
done
or this should work too:
for i in {1..4};do
echo L$(echo $(echo "\$z$i"))stop.txt
done
outputs:
L2.3stop.txt
L3.4stop.txt
L7.8stop.txt
L8.9stop.txt
This is not portable, but in bash you can do:
name=z$i
echo ${!name}
Best (using an array):
z=( 2.3 3.4 7.8 8.9 9.8 ) # Added fifth value
for x in "${z[#]}"
do
echo L${x}stop.txt
done
Less good (using indirect expansion — a bash feature):
z1=2.3
z2=3.4
z3=7.8
z4=8.9
z5=9.8 # Added fifth value
for i in {1..5}
do
v=z$i
echo L${!v}stop.txt
done
Both produce:
L2.3stop.txt
L3.4stop.txt
L7.8stop.txt
L8.9stop.txt
L9.8stop.txt
Both avoid using eval which is not too harmful in this context but is dangerous (and difficult to use correctly) in general.
If you have defined variable z, you can do this:
for v in ${!z*}; do echo L${!v}stop.txt; done
or:
for v in z{1..4}; do echo L${!v}stop.txt; done
Related
echo 3+3
How can I evaluate such expressions in Bash, in this case to 6?
echo $(( 3+3 ))
expr is the standard way, but it only handles integers.
bash has a couple of extensions, which only handle integers as well:
$((3+3)) returns 6
((3+3)) used in conditionals, returns 0 for true (non-zero) and 1 for false
let 3+3 same as (( ))
let and (( )) can be used to assign values, e.g.
let a=3+3
((a=3+3))
for floating point you can use bc
echo 3+3 | bc
in shells such as zsh/ksh, you can use floats for maths. If you need more maths power, use tools like bc/awk/dc
eg
var=$(echo "scale=2;3.4+43.1" | bc)
var=$(awk 'BEGIN{print 3.4*43.1}')
looking at what you are trying to do
awk '{printf "%.2f\n",$0/59.5}' ball_dropping_times >bull_velocities
You can make use of the expr command as:
expr 3 + 3
To store the result into a variable you can do:
sum=$(expr 3 + 3)
or
sum=`expr 3 + 3`
Lots of ways - most portable is to use the expr command:
expr 3 + 3
I believe the ((3+3)) method is the most rapid as it's interpreted by the shell rather than an external binary.
time a large loop using all suggested methods for the most efficient.
Solved thanks to Dennis, an example of BC-use:
$ cat calc_velo.sh
#!/bin/bash
for i in `cat ball_dropping_times`
do
echo "scale=20; $i / 59.5" | bc
done > ball_velocities
My understanding of math processing involves floating point processing.
Using bashj (https://sourceforge.net/projects/bashj/) you can call a java method (with floating point processing, cos(), sin(), log(), exp()...) using simply
bashj +eval "3+3"
bashj +eval "3.5*5.5"
or in a bashj script, java calls of this kind:
#!/usr/bin/bashj
EXPR="3.0*6.0"
echo $EXPR "=" u.doubleEval($EXPR)
FUNCTIONX="3*x*x+cos(x)+1"
X=3.0
FX=u.doubleEval($FUNCTIONX,$X)
echo "x="$X " => f(x)=" $FUNCTIONX "=" $FX
Note the interesting speed : ~ 10 msec per call (the answer is provided by a JVM server).
Note also that u.doubleEval(1/2) will provide 0.5 (floating point) instead of 0 (integer)
One use case that might be useful in this regard is, if one of your operand itself is a bash command then try this.
echo $(( `date +%s\`+10 )) or even echo $(( `date +%s\`+(60*60) ))
In my case I was trying to get Unixtime 10 seconds and hour later than current time respectively.
The grub2 shell aims to be a minimalistic bash like shell.
But how can I increment a variable in grub2?
In bash I would do:
var=$((var+1))
or
((var=var+1))
In grub2 I get a syntax error on these calls. How can I achieve this in the grub2 shell?
Grub2 does not have builtin arithmetic support. You need to add Lua support if you want that, see this answer for details.
Based on this answer (as already linked by other answer), the following appears to work with GRUB's regexp command (allows incrementing from any number 0-5, add more <from>,<to> pairs as needed):
num=0
incr="" ; for x in 0,1 1,2 2,3 3,4 4,5 5,6 ; do
regexp --set=1:incr "${num},([0-9]+)" "${x}"
if [ "$incr" != "" ] ; then
echo "$num incremented to $incr"
num=$incr
break
fi
done
Decrementing similarly works (just flipping two the regular expression parts):
num=6
decr="" ; for x in 0,1 1,2 2,3 3,4 4,5 5,6 ; do
regexp --set=1:decr "([0-9]+),${num}" "${x}"
if [ "$decr" != "" ] ; then
echo "$num decremented to $decr"
num=$decr
break
fi
done
I'm working with an existing script which was written a bit messily. Setting up a loop with all of the spaghetti code could make a bigger headache than I want to deal with in the near term. Maybe when I have more time I can clean it up but for now, I'm just looking for a simple fix.
The script deals with virtual disks on a xen server. It reads multipath output and asks if particular LUNs should be formatted in any way based on specific criteria. However, rather than taking that disk path and inserting it, already formatted, into a configuration file, it simply presents every line in the format
'phy:/dev/mapper/UUID,xvd?,w',
UUID, of course, is an actual UUID.
The script actually presents each of the found LUNs in this format expecting the user to copy and paste them into the config file replacing each ? with a letter in sequence. This is tedious at best.
There are several ways to increment a number in bash. Among others:
var=$((var+1))
((var+=1))
((var++))
Is there a way to do the same with characters which doesn't involve looping over the entire alphabet such that I could easily "increment" the disk assignment from xvda to xvdb, etc?
To do an "increment" on a letter, define the function:
incr() { LC_CTYPE=C printf "\\$(printf '%03o' "$(($(printf '%d' "'$1")+1))")"; }
Now, observe:
$ echo $(incr a)
b
$ echo $(incr b)
c
$ echo $(incr c)
d
Because, this increments up through ASCII, incr z becomes {.
How it works
The first step is to convert a letter to its ASCII numeric value. For example, a is 97:
$ printf '%d' "'a"
97
The next step is to increment that:
$ echo "$((97+1))"
98
Or:
$ echo "$(($(printf '%d' "'a")+1))"
98
The last step is convert the new incremented number back to a letter:
$ LC_CTYPE=C printf "\\$(printf '%03o' "98")"
b
Or:
$ LC_CTYPE=C printf "\\$(printf '%03o' "$(($(printf '%d' "'a")+1))")"
b
Alternative
With bash, we can define an associative array to hold the next character:
$ declare -A Incr; last=a; for next in {b..z}; do Incr[$last]=$next; last=$next; done; Incr[z]=a
Or, if you prefer code spread out over multiple lines:
declare -A Incr
last=a
for next in {b..z}
do
Incr[$last]=$next
last=$next
done
Incr[z]=a
With this array, characters can be incremented via:
$ echo "${Incr[a]}"
b
$ echo "${Incr[b]}"
c
$ echo "${Incr[c]}"
d
In this version, the increment of z loops back to a:
$ echo "${Incr[z]}"
a
How about an array with entries A-Z assigned to indexes 1-26?
IFS=':' read -r -a alpharray <<< ":A:B:C:D:E:F:G:H:I:J:K:L:M:N:O:P:Q:R:S:T:U:V:W:X:Y:Z"
This has 1=A, 2=B, etc. If you want 0=A, 1=B, and so on, remove the first colon.
IFS=':' read -r -a alpharray <<< "A:B:C:D:E:F:G:H:I:J:K:L:M:N:O:P:Q:R:S:T:U:V:W:X:Y:Z"
Then later, where you actually need the letter;
var=$((var+1))
'phy:/dev/mapper/UUID,xvd${alpharray[$var]},w',
The only problem is that if you end up running past 26 letters, you'll start getting blanks returned from the array.
Use a Bash 4 Range
You can use a Bash 4 feature that lets you specify a range within a sequence expression. For example:
for letter in {a..z}; do
echo "phy:/dev/mapper/UUID,xvd${letter},w"
done
See also Ranges in the Bash Wiki.
Here's a function that will return the next letter in the range a-z. An input of 'z' returns 'a'.
nextl(){
((num=(36#$(printf '%c' $1)-9) % 26+97));
printf '%b\n' '\x'$(printf "%x" $num);
}
It treats the first letter of the input as a base 36 integer, subtracts 9, and returns the character whose ordinal number is 'a' plus that value mod 26.
Use Jot
While the Bash range option uses built-ins, you can also use a utility like the BSD jot utility. This is available on macOS by default, but your mileage may vary on Linux systems. For example, you'll need to install athena-jot on Debian.
More Loops
One trick here is to pre-populate a Bash array and then use an index variable to grab your desired output from the array. For example:
letters=( "" $(jot -w %c 26 a) )
for idx in 1 26; do
echo ${letters[$idx]}
done
A Loop-Free Alternative
Note that you don't have to increment the counter in a loop. You can do it other ways, too. Consider the following, which will increment any letter passed to the function without having to prepopulate an array:
increment_var () {
local new_var=$(jot -nw %c 2 "$1" | tail -1)
if [[ "$new_var" == "{" ]]; then
echo "Error: You can't increment past 'z'" >&2
exit 1
fi
echo -n "$new_var"
}
var="c"
var=$(increment_var "$var")
echo "$var"
This is probably closer to what the OP wants, but it certainly seems more complex and less elegant than the original loop recommended elsewhere. However, your mileage may vary, and it's good to have options!
I am new to bash scripting. Now, the question is self explanatory. I want to print the values 2^0, 2^1, 2^2, 2^3, 2^4, 2^5 using loop in bash.
I tried ..
for i in {0...5}; do echo 2^$i; done
result:
2^{0...5}
please suggest a solution
This is the correct form:
for i in {0..5}; do echo $((2**i)); done
Where { .. } is the range operator and $(( )) the arithmetical evaluation operator.
Note that the power operator in Bash is written ** and not ^.
Edit: Roberto Reale's answer is much better for this purpose, as it's shorter and simpler.
2^$i; won't work unfortunately, as bash doesn't support that operator.
You could try adding this
pow()
{
echo $(( ${1:?} ** ${2:?} ))
}
to your code (using it like pow [base] [exponent] (without the brackets of course)).
(Bare in mind I havent't tested this, so this may not work as expected)
Source
You could try this code also,
$ for i in {0..5}; do awk -v var=$i 'BEGIN{print 2^var}'; done
1
2
4
8
16
32
echo 3+3
How can I evaluate such expressions in Bash, in this case to 6?
echo $(( 3+3 ))
expr is the standard way, but it only handles integers.
bash has a couple of extensions, which only handle integers as well:
$((3+3)) returns 6
((3+3)) used in conditionals, returns 0 for true (non-zero) and 1 for false
let 3+3 same as (( ))
let and (( )) can be used to assign values, e.g.
let a=3+3
((a=3+3))
for floating point you can use bc
echo 3+3 | bc
in shells such as zsh/ksh, you can use floats for maths. If you need more maths power, use tools like bc/awk/dc
eg
var=$(echo "scale=2;3.4+43.1" | bc)
var=$(awk 'BEGIN{print 3.4*43.1}')
looking at what you are trying to do
awk '{printf "%.2f\n",$0/59.5}' ball_dropping_times >bull_velocities
You can make use of the expr command as:
expr 3 + 3
To store the result into a variable you can do:
sum=$(expr 3 + 3)
or
sum=`expr 3 + 3`
Lots of ways - most portable is to use the expr command:
expr 3 + 3
I believe the ((3+3)) method is the most rapid as it's interpreted by the shell rather than an external binary.
time a large loop using all suggested methods for the most efficient.
Solved thanks to Dennis, an example of BC-use:
$ cat calc_velo.sh
#!/bin/bash
for i in `cat ball_dropping_times`
do
echo "scale=20; $i / 59.5" | bc
done > ball_velocities
My understanding of math processing involves floating point processing.
Using bashj (https://sourceforge.net/projects/bashj/) you can call a java method (with floating point processing, cos(), sin(), log(), exp()...) using simply
bashj +eval "3+3"
bashj +eval "3.5*5.5"
or in a bashj script, java calls of this kind:
#!/usr/bin/bashj
EXPR="3.0*6.0"
echo $EXPR "=" u.doubleEval($EXPR)
FUNCTIONX="3*x*x+cos(x)+1"
X=3.0
FX=u.doubleEval($FUNCTIONX,$X)
echo "x="$X " => f(x)=" $FUNCTIONX "=" $FX
Note the interesting speed : ~ 10 msec per call (the answer is provided by a JVM server).
Note also that u.doubleEval(1/2) will provide 0.5 (floating point) instead of 0 (integer)
One use case that might be useful in this regard is, if one of your operand itself is a bash command then try this.
echo $(( `date +%s\`+10 )) or even echo $(( `date +%s\`+(60*60) ))
In my case I was trying to get Unixtime 10 seconds and hour later than current time respectively.