I need to do an operation but something is wrong in my code in bash
I have 4 variables, km1, km2, km3, km4.
I want to sum the 4 variables except when the value is "CLOSED"
3.200
CLOSED
1.800
0.600
When I do the following sum, there is an error...I thing my variables are not numeric, any help? How can I force them to be numeric and then do the sum?
let km=$km1+$km3+$km4
echo $km
./sum.sh: line 41: let: km=3.200: syntax error: invalid arithmetic operator (error token is ".200")
km1=3.200
km2=CLOSED
km3=1.800
km4=0.600
total=`LC_ALL=C echo "$km1 $km2 $km3 $km4"|awk '{sum += $1+$2+$3+$4}END {print sum}'`
Not that good with awk but i think the above can help. total the is sum of all vars
There are 2 issues with you code. The first one is that you are trying to work with values other than integers. Bash only does integers. You can round up the values to integers using bc (An arbitrary precision calculator language). The second issue is that you are trying to do math on strings. So consider the code below:
#!/bin/bash
km1=3.200;
km2="CLOSED";
km3=1.800;
km4=0.600;
km1=$(echo "$km1/1" | bc)
km3=$(echo "$km3/1" | bc)
km4=$(echo "$km4/1" | bc)
array=($km1 $km2 $km3 $km4)
for i in ${array[#]}; do
case $i in
*[0-9]*)
(( result+=$i ))
esac
done
echo $result
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 $(( `date +%s\`+(60*60) ))
In my case I was trying to get Unixtime 10 seconds and hour later than current time respectively.
I'm a bit stuck with something. I have a for loop like this:
#!/bin/bash
for i in {10..15}
do
I want to obtain the last digit of the number, so if i is 12, I want to get 2. I'm having difficulties with the syntax though. I've read that I should convert it into a character array, but when I do something like:
j=${i[#]}
echo $j
I don't get 1 0 1 1 1 2 and so on...I get 10, 11, 12...How do I get the numbers to be split up so I can get the last one of i, when I don't always know how many digits will make up i (ex. it may be 1, or 10, or a 100, etc.)?
Trick is to treat $i like a string.
for i in {10..15}; do j="${i: -1}"; echo $j; done
Of course, you do not need to assign to a variable if you don't want to:
for i in {10..15}; do echo "${i: -1}"; done
This answer which uses GNU shell parameter expansion is the most sensible method, I guess.
However, you can also use the double parenthesis construct which allows C-style manipulation of variables in Bash.
for i in {10..15}
do
(( j = i % 10 )) # modulo 10 always gives the ones' digit
echo $j
done
This awk command could solve your problem:
awk '{print substr($0,length,1)}' test_file
I'm assuming that the numbers are saved in a file test_file
If you want to use for loop:
for i in `cat test_1`
do
echo $i |tail -c 2
done
When running this part of my bash script am getting an error
Script
value=0
for (( t=0; t <= 4; t++ ))
do
d1=${filedates[$t]}
d2=${filedates[$t+1]}
((diff_sec=d2-d1))
SEC=$diff_sec
compare=$((${SEC}/(60*60*24)))
value=$((value+compare))
done
Output
jad.sh: line 28: ((: 10#2014-01-09: value too great for base (error token is "09")
jad.sh: line 30: /(60*60*24): syntax error: operand expected (error token is "/(60*60*24)")
d1 and d2 are dates in that form 2014-01-09 and 2014-01-10
Any solution please?
Prepend the string "10#" to the front of your variables. That forces bash to treat them as decimal, even though the leading zero would normally make them octal.
What are d1 and d2? Are they dates or seconds?
Generally, this error occurs if you are trying to do arithmetic with numbers containing a zero-prefix e.g. 09.
Example:
$ echo $((09+1))
-bash: 09: value too great for base (error token is "09")
In order to perform arithmetic with 0-prefixed numbers you need to tell bash to use base-10 by specifying 10#:
$ echo $((10#09+1))
10
As others have said, the error results from Bash interpreting digit sequences with leading zeros as octal numbers. If you have control over the process creating the date values and you're using date, you can prefix the output format string with a hyphen to remove leading zero padding.
Without prefixing date format with hyphen:
$ (( $(date --date='9:00' +%H) > 10 )) && echo true || echo oops
-bash: ((: 09: value too great for base (error token is "09")
oops
With prefixing date format with hyphen:
$ (( $(date --date='9:00' +%-H) > 10 )) && echo true || echo oops
true
From the date man page:
By default, date pads numeric fields with zeroes. The following
optional flags may follow '%':
- (hyphen) do not pad the field
d1 and d2 are dates in that form 2014-01-09 and 2014-01-10
and then
((diff_sec=d2-d1))
What do you expect to get? ((diffsec=2014-01-09-2014-01-10)) ??
You need to convert the dates to seconds first:
d1=$( date -d "${filedates[$t]}" +%s )
d2=$( date -d "${filedates[$t+1]}" +%s )
(( compare = (d2 - d1) / (60*60*24) ))
(( value += compare ))
Posting some tips here related to the title of this question, but not directly related to the details of the original question. I realize that's a bit controversial action on Stack Overflow, however these related questions:
convert octal to decimal in bash [duplicate]
Value too great for base (error token is "08") [duplicate]
point to this one, and yet they are closed and hence, I could not post this answer there. Therefore, this seemed like a logical place (at least to me) to post this information that may help others in a similar situation, especially new-to-BaSH programmers.
An alternative approach to ensuring a number is treated as a 10-base integer is to use printf. This command instructs printf to treat $num as an integer and round it to 0 decimal places.
num="$(printf "%.0f" "$num")"
Or, if you want to also ensure there are no non-numeric characters in the string, you can do this:
num="$(printf "%.0f" "${num//[!0-9]/}")"
Both commands will strip out leading zeroes and round decimal values to the nearest whole number. Note the first (simpler) solution works with negative numbers, but the second does not (it will always return absolute value).
Note that printf rounds down, meaning .01 to 0.5 is rounded down to 0, while .51 to .99 is rounded up to 1. Basically, the difference between rounding up versus down in this case is that printf rounds down 0.5 and any below. I mention this because 0.5 rounded up is a more common practice.
Now, addressing the OP's specific scenario.... Combining printf with awk allows arithmetic expressions not possible with printf alone.
This
compare=$((${SEC}/(606024)))
could be alternatively be expressed as
compare=$(awk -v sec=$SEC 'BEGIN { print int(sec/(60*60*24))}')
or
compare="$(printf "%.0f" "$(awk "BEGIN { print ( $SEC / ( 60 * 60 * 24 ) ) }")")"
Meanwhile,
value=$((value+compare))
Could be calculated as
value="$(printf "%.0f" "$(awk "BEGIN { print ( $value + $compare ) }")")"
You don't need the $ and the {} in an arithmetic expansion expression. It should look like this:
compare=$((SEC/(60*60*24)))
For 'mm' and 'dd' values in dates, I use this trick:
mm="1${date:5,2}" # where 5 is the offset to mm in the date
let mm=$mm-100 # turn 108 into 8, and 109 into 9
In a text file, test.txt, I have the next information:
sl-gs5 desconnected Wed Oct 10 08:00:01 EDT 2012 1001
I want to extract the hour of the event by the next command line:
hour=$(grep -n sl-gs5 test.txt | tail -1 | cut -d' ' -f6 | awk -F ":" '{print $1}')
and I got "08". When I try to add 1,
14 echo $((hour+1))
I receive the next error message:
./test2.sh: line 14: 08: value too great for base (error token is "08")
If variables in Bash are untyped, why?
See ARITHMETIC EVALUATION in man bash:
Constants with a leading 0 are interpreted as octal numbers.
You can remove the leading zero by parameter expansion:
hour=${hour#0}
or force base-10 interpretation:
$((10#$hour + 1))
what I'd call a hack, but given that you're only processing hour values, you can do
hour=08
echo $(( ${hour#0} +1 ))
9
hour=10
echo $(( ${hour#0} +1))
11
with little risk.
IHTH.
You could also use bc
hour=8
result=$(echo "$hour + 1" | bc)
echo $result
9
Here's an easy way, albeit not the prettiest way to get an int value for a string.
hour=`expr $hour + 0`
Example
bash-3.2$ hour="08"
bash-3.2$ hour=`expr $hour + 0`
bash-3.2$ echo $hour
8
In Short: In order to deal with "Leading Zero" numbers (any 0 digit that comes before the first non-zero) in bash
- Use bc An arbitrary precision calculator language
Example:
a="000001"
b=$(echo $a | bc)
echo $b
Output: 1
From Bash manual:
"bc is a language that supports arbitrary precision numbers with interactive execution
of statements. There are some similarities in the syntax to the C programming lan-
guage. A standard math library is available by command line option. If requested, the
math library is defined before processing any files. bc starts by processing code from
all the files listed on the command line in the order listed. After all files have
been processed, bc reads from the standard input. All code is executed as it is read.
(If a file contains a command to halt the processor, bc will never read from the standard input.)"
Since hours are always positive, and always 2 digits, you can set a 1 in front of it and subtract 100:
echo $((1$hour+1-100))
which is equivalent to
echo $((1$hour-99))
Be sure to comment such gymnastics. :)
The leading 0 is leading to bash trying to interpret your number as an octal number, but octal numbers are 0-7, and 8 is thus an invalid token.
If I were you, I would add some logic to remove a leading 0, add one, and re-add the leading 0 if the result is < 10.
How about sed?
hour=`echo $hour|sed -e "s/^0*//g"`
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.