I have the below script which is trying to get the pass percentage through shell script.
Script:
n= "$pass"/"$total";
"$n" * = 100;
echo "$n"
Output:
/tmp/jenkins3601870177535319162.sh: line 45: 20/25*=: No such file or
directory 20/25
I'm not sure the above calculation is correct. But I just have variable $pass which is having pass test case count and $total variable which had total test case count. Just wanna get percentage of the passed test cases using shell.
You have two primary problems: (1) there cannot be any spaces on either side of the '=' sign; and (2) shell uses integer math so 20/25 will equal 0.
The POSIX arithmetic operator is ((...)) where your expression goes within ((...)). Also, within ((...)) there is no need to derefernce the variable name by preceding it with a '$'.
To assign the result of the expression to a variable you precede the operator by the '$', e.g. n=$((pass/total)).
To get around the fact that shell uses integer math, you can multiply pass by 100 before you divide and at least get a whole-number percentage. For example:
#!/bin/sh
pass=20
total=25
n=$(((pass * 100)/total))
printf "pass percentage: %d\n" "$n"
If you run the script you will then get:
$ sh percent.sh
pass percentage: 80
Where 80% is what would be the percent result for 20/25. Look things over and let me know if you have further questions.
Related
I am trying to update a variable within a for loop, sequentially adding to the variable, and then printing it to a series of file names.
Actually, I want to sequentially update a series of file, from a tmp file distTmp.RST to sequentially dist1.RST, dist2.RST, etc..
The original distTmp.RST contains a line called "WWW". I want to replace the string with values called 21.5 in dist1.RST, 22.5 in dist2.RST, 23.5 in dist3.RST, etc...
My code is as follows:
#!/bin/bash
wm=1
wM=70
wS=1
F=20.5
for W in {${wm}..${wM}..${wS}}; do
F=$(${F} + 1 | bc)
echo ${F}
sed "s/WWW/"${F}"/g" distTmp.RST > dist${W}.RST
done
echo ${F}
========
But I am getting error message as follows:
change.sh: line 13: 20.5 + 1 | bc: syntax error: invalid arithmetic operator (error token is ".5 + 1 | bc")
Kindly suggest me a solution to the same.
Kindly suggest me a solution to the same.
This might do what you wanted. Using a c-style for loop.
#!/usr/bin/env bash
wm=1
wM=70
wS=1
F=20.5
for ((w=wm;w<=wM;w+=wS)); do
f=$(bc <<< "$F + $w")
echo "$f"
sed "s/WWW/$f/g" distTmp.RST > "dist${w}.RST"
done
The error from your script might be because the order of expansion. brace expansion expansion happens before Variable does.
See Shell Expansion
Use
F=$(echo ${F} + 1 | bc)
instead of F=$((${F} + 1 | bc)). The doubled-up parentheses are what caused your error. Double parentheses weren't in the original code, but I get a different error saying 20.5: command not found with the original code, so I tried doubling the parentheses and get the error in the question. Apparently, floating point numbers aren't supported by $(()) arithmetic evaluation expressions in Bash.
I'm reading a file with multiple columns, and dumping 2 columns of the file into into 2 different arrays. Now based on a condition, I need to get the difference between 2 values retrieved from the array. So my code looks like this -
if [ condition ]; then
VAL = (( ${local[$x]} - ${local[$y]} ))
fi
The thing is, while I'm able to echo and see both values ${local[$x]} and ${local[$y]}, the subtraction operation gives me a syntax error. I understand it's failing because the values currently held within the array involve floating point decimal values - like 3456712.126758, and the assignment throws errors with the decimal part. I understand arithmetic operations are not a strong point with the bash shell as floating point numbers are considered strings hence the issue.
Could you please help getting the right format please?
Should I do something like this
VAL= awk '{ print ${local[$x]} - ${local[$y]} }'
or
VAL=echo ${local[$x]} - ${local[$y]} | bc -l
I'm sure the syntax above is wrong, kindly help with the syntax, I need it assigned the subtracted result assigned to the field VAL.
Not only floating points, but also the spacing would lead to syntax errors. Bash variable assignments must have no spaces, as in val=x, and not val = x.
Uppercase variable names are reserved for environment variables, and it is recommended to use lowercase instead for your own variables. (Oh, and local is also a reserved word.)
Your assignment wouldn't work with proper spacing, either: the arithmetic expression
var=(( ${vals[$x]} - ${vals[$y]} )) # syntax error near unexpected token `('
is just evaluating its contents, but not returning anything. You could use the part after the = as a condition. To make it return something, you need arithmetic expansion (note the extra $):
var=$(( ${vals[$x]} - ${vals[$y]} )) # works for integers
^
In an arithmetic context, you don't even need to prepend $ to your variables:
var=$(( vals[x] - vals[y] ))
works just as well. Exception: in associative arrays, you still have to do it for indices:
$(( vals[$x] ))
And finally, as you noticed, this all doesn't work for floating point numbers. Instead of piping to bc, you can also use a here string and avoid spawning a subshell:
$ vals=(1.1 2.2)
$ x=0
$ y=1
$ echo $(( local[x] - local[y] )) # No '$' needed for variable expansion
bash: 1.1: syntax error: invalid arithmetic operator (error token is ".1") # But :(
$ bc -l <<< "local[x] - local[y]" # Requires '$' - these expand to nothing
0
$ bc -l <<< "${local[x]} - ${local[y]}" # Works!
-1.1
With awk:
awk -v a=${a} -v b=${b} 'BEGIN{print a - b}'
With bc:
echo "${a} - ${b}" | bc -l
See also other options here.
I have a number 2014061200 and I am trying to extract the one's digit of this number (0) and set it to a variable $STR, so that $STR would equal 0, because that is in the one's place of the above number.
I made a shell script showing the following:
$NUM=2014061200
$STR= $NUM | tail -c 2
echo $STR
However when I do this I get a blank for $STR instead of the expected result of 0. I mean when I type
echo $NUM | tail -c 2, I do get the output of 0, but how do I get this into a variable?
Thanks
Use parameter expansion:
num=2014061200
last=${num: -1}
-1 tells bash to extract one character from the right.
Here's another variation using parameter expansion where ${#num}-1 is the length of the string less 1 and ${num:position:length} is a substring expression:
${num:(${#num}-1):1}
Why can't I do this simple arithmetic operation and store it in a variable in bash shell? I've been struggling with this and playing around with () and $ symbols but no luck.
read t
let r=$(5/9)*$($t-32)
I get a: let: r=*: syntax error: operand expected (error token is "*")
When you are using the let statement, you don't need the dollar-sign, but single-quote the expression instead to keep the shell preprocessor from messing with your operators. Note that bash does not seem to be able to handle numbers which aren't integers, so the (5/9) expression will always be zero. Try the second let statement.
read -p 'Temp in Fahrenheit (no decimals): ' t
# let r='(5/9)*(t-32)' -- this doesn't work
let r='5*(t-32)/9'
echo "Centigrade: $r"
Try that :
read -p 'Type integer temp (Fahrenheit) >>> ' int
echo "$(( 5 * ( int - 32 ) / 9 )) Celcius"
I have a bash scripts which an argument enclosed with double quotes, which creates a shape-file of map within the given boundries, e.g.
$ export_map "0 0 100 100"
Within the script, there are two select statements:
select ENCODING in UTF8 WIN1252 WIN1255 ISO-8859-8;
...
select NAV_SELECT in Included Excluded;
Naturally, these two statements require the input to enter a number as an input. This can by bypassed by piping the numbers, followed by a newline, to the script.
In order to save time, I would like to have a script that would create 8 maps - for each combination of ENCODING (4 options) and NAV_SELECT (2 options).
I have written another bash script, create_map, to server as a wrapper:
#!/bin/bash
for nav in 1 2 3 4;
do
for enc in 1 2;
do
printf "$nav\n$enc\n" | /bin/bash -c "./export_map.sh \"0 0 100 100\""
done
done
**This works (thanks, Brian!), but I can't find a way to have the numeric argument "0 0 100 100" being passed from outside the outer script. **
Basically, I'm looking for way to accept an argument within double quotes to a wrapper bash script, and pass it - with the double quotes - to an inner script.
CLARIFICATIONS:
export_map is the main script, being called from create_map 8 times.
Any ideas?
Thanks,
Adam
If I understand your problem correctly (which I'm not sure about; see my comment), you should probably add another \n to your printf; printf does not add a trailing newline by default the way that echo does. This will ensure that the second value will be read properly by the select command which I'm assuming appears in export_map.sh.
printf "$nav\n$enc\n" | /bin/bash -c "./export_map.sh \"100 200 300 400\""
Also, I don't think that you need to add the /bin/bash -c and quote marks. The following should be sufficient, unless I'm missing something:
printf "$nav\n$enc\n" | ./export_map.sh "100 200 300 400"
edit Thanks for the clarification. In order to pass an argument from your wrapper script, into the inner script, keeping it as a single argument, you can pass in "$1", where the quotes indicate that you want to keep this grouped as one argument, and $1 is the first parameter to your wrapper script. If you want to pass all parameters from your outer script in to your inner script, each being kept as a single parameter, you can use "$#" instead.
#!/bin/bash
for nav in 1 2 3 4;
do
for enc in 1 2;
do
printf "$nav\n$enc\n" | ./export_map.sh "$1"
done
done
Here's a quick example of how "$#" works. First, inner.bash:
#!/bin/bash
for str in "$#"
do
echo $str
done
outer.bash:
#!/bin/bash
./inner.bash "$#"
And invoking it:
$ ./outer.bash "foo bar" baz "quux zot"
foo bar
baz
quux zot