I have a log line that I am pulling from bash like:
There are 5 apples and 7 oranges
And I want to get the 5 and 7 into bash variables and am having problems doing so.
awk '{ print $3, $6 }' won't work due to the fact that their position might change, if there are bananas. Bonus points if the digits can be associated with their fruits.
Thanks in advance
and to get the fruits:
echo 'There are 5 apples and 7 oranges' | grep -o -E '[[:digit:]]+ [[:alpha:]]+'
A bash-only method, tested with bash v4:
str=" There are 5 apples, 12 bananas and 7 oranges"
i=0
while [[ "$str" =~ ([0-9]+)" "([a-z]+)(.*) ]] ; do
num[$i]=${BASH_REMATCH[1]}
type[$i]=${BASH_REMATCH[2]}
str=${BASH_REMATCH[3]}
(( i++ ))
done
for (( i=0; i<${#num[#]}; i++ )); do
printf "%4d\t%s\t%s\n" $i ${num[$i]} ${type[$i]}
done
outputs
0 5 apples
1 12 bananas
2 7 oranges
Grep will do this:
echo 'There are 5 apples and 7 oranges' | grep -o -E '[[:digit:]]+'
Slightly more compact version:
echo 'There are 5 apples and 7 oranges' | egrep -o '[0-9]+ \w+'
Note however that \w is synonymous with [:alnum:], and will thus match "appl3s" as well, for example.
Well, shell is a strange language, and you can do nice things. For example:
a="there are 7 apples"
for i in $a; do
case $i in
[0-9]*)
value=$i
expectname=1
;;
*)
test -n "$expectname" && name=$i
expectname=
;;
esac
done
echo $value
echo $name
If you have more than one occurrence, you can fill an array, or a bash map.
Related
I need a bash script to find the sum of the absolute value of integers separated by spaces. For instance, if the input is:
1 2 -3
the script should print 6 to standard output
I have:
while read x ; do echo $(( ${x// /+} )) ; done
which gives me
0
Without over complicated things, how would I include an absolute value of each x in that statement so the output would be:
6
With Barmar's idea:
echo "1 2 -3" | tr -d - | tr ' ' '+' | bc -l
Output:
6
You have almost done it, but the -s must have been removed from the line read:
while read x; do x=${x//-}; echo $(( ${x// /+} )); done
POSIX friendly implementation without running a loop and without spawning a sub-shell:
#!/usr/bin/env sh
abssum() {
IFS='-'
set -- $*
IFS=' '
set -- $*
IFS=+
printf %d\\n $(($*))
}
abssum 1 2 -3
Result:
6
I'm working on a Bash script which that takes in two integers and outputs all the numbers in between the two. It would look something like this:
Input:
bash testScript 3 10
3
4
5
6
7
8
9
10
This is some code that I wrote that I thought would work but I haven't had much luck getting it to work yet.
read myvar
read myvar2
while [ $myvar -le myvar2 ]
do
echo $myvar
myvar=$(($myvar+1))
//timer in-between numbers
sleep .5
done
Bash supports c style for loops using a double parenthesis construct:
$ for ((x=3; x<=10; x++)); { echo $x; }
3
4
5
6
7
8
9
10
Or, brace expansion:
$ for i in {3..6}; do echo $i; done
3
4
5
6
Problem with brace expansion is you need to use eval to use variables...
A common GNU utility for this is seq but it is not POSIX, so may not be on every *nix. If you want to write a similar function in Bash, it would look like this:
my_seq ()
# function similar to seq but with integers
# my_seq [first [incr]] last
{
incr=1
start=1
if [[ $# -gt 2 ]]; then
start=$1
incr=$2
end=$3
elif [[ $# -gt 1 ]]; then
start=$1
end=$2
else
end=$1
fi
for ((x=start; x<=end; x+=incr)); { echo $x; }
}
Then call that with 1, 2 or 3 arguments:
$ my_seq 30 10 60
30
40
50
60
with brace expansion
$ echo {3..10} | tr ' ' '\n'
or for variables with eval
$ a=3; b=10; eval echo {$a..$b} | ...
or, if you have awk
$ awk -v s=3 -v e=10 'BEGIN{while(s<=e) print s++}'
Use positional parameters:
myvar=$1
myvar2=$2
while [ $myvar -le $myvar2 ]
do
echo $myvar
myvar=$(($myvar+1))
#timer in-between numbers
sleep .5
done
Output:
scottsmudger#ns207588:~ $ bash test 1 5
1
2
3
4
5
See http://www.tldp.org/LDP/abs/html/othertypesv.html
Your posted codes are not aligned properly, not sure if it's your actual codes. But only problem other than the alignment is you missed a $ for myvar2 in the while statment.
read myvar
read myvar2
while [ $myvar -le $myvar2 ]
do
echo $myvar
myvar=$(($myvar+1))
#//timer in-between numbers
sleep .5
done
I'm new to this and what I'm trying to do is create a simple adding script in bash.
I have to use a for loop. What I'm starting so far looks like this:
#!/bin/bash
sum=0
for num in {1..15}
do
echo $num
done
echo$sum
I need help with how to make the for loop show for example if I type:
add 4 -3 2 8
it will output as:
4
-3
2
8
=11
How would I make it so the $num only show what I typed in such as the '4 -3 2 8' and negative numbers?
You can use $# to get all parameters, and $(()) to do arithmetic.
sum=0
for num in $#
do
sum=$((sum + num))
done
echo $# = $sum
I'll retag your question as bash; d is not appropriate.
I'm trying to create an exit error to expand on the previous script now. if I type anything but a number now and what I have tried so far is:
sum=0
for num in "$#"
do
echo $num | grep -i [^0-9+-]
if ["$?" = 1] then
echo "Sorry, '$num' is not a number"
fi
sum=$((sum + num))
done
echo $sum
Example if I type in
add 1 2 3 four five
it would say
four
Sorry, 'four' is not a number
I have a file with many lines, like
1 jfkdajfd 1 2 3 5
2 fkldfjld
3 fdkfloaf 9 10
4 fldfldkf
5 fdskf;ak 12 1 4
I want to get all the numbers and put them in a column in a file, like
1
2
3
5
9
10
12
1
4
how can I achieve this?
thanks
In your case, it looks like you can do this:
awk '{for (i=3;i<=NF;++i) {print $i}}'
This is assuming that all the numbers you want to print occur in column 3 or after.
cat file | while read line
do
for i in $(echo ${line})
do
isnumeric=$(echo ${i} | grep -q [0-9]; echo ${?})
if [ ${isnumeric} -eq 0 ]
then
echo ${i} >> outfile
fi
done
done
not bulletproof and not as elegant as the previously given solutions, but it shows what is being used for determining if this is a numeric or not.
while read num alpha rest; do
[[ "$rest" ]] && printf "%s\n" $rest # <-- variable is unquoted
done < filename
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