从字符串中提取第一个数字

Question

I want to extract the first number from a given string. The number is a float but I only need the integers before the decimal.

example:

string1="something34521.32somethingmore3241"

Output I want is 34521

What is the easiest way to do this in bash?

Thanks!

Answer 1

So just for simplicity I'll post what I was actually looking for when I got here.

echo $string1 | sed 's@^[^0-9]*\([0-9]\+\).*@\1@'

Simply skip whatever is not a number from beginning of the string ^[^0-9]* , then match the number \([0-9]\+\) , then the rest of the string .* and print only the matched number \1 .

I sometimes like to use @ instead of the classical / in the sed replace expressions just for better visibility within all those slashes and backslashes. Handy when working with paths as you don't need to escape slashes. Not that it matters in this case. Just sayin'.

Answer 2

This sed 1 liner will do the job I think:

str="something34521.32somethingmore3241"
echo $str | sed -r 's/^([^.]+).*$/\1/; s/^[^0-9]*([0-9]+).*$/\1/'

OUTPUT

 34521

Answer 3

You said you have a string and you want to extract the number, i assume you have other stuff as well.

$ echo $string
test.doc_23.001
$ [[ $string =~ [^0-9]*([0-9]+)\.[0-9]+ ]]
$ echo "${BASH_REMATCH[1]}"
23

$ foo=2.3
$ [[ $string =~ [^0-9]*([0-9]+)\.[0-9]+ ]]
$ echo "${BASH_REMATCH[1]}"
2

$ string1="something34521.32somethingmore3241"
$ [[ $string1 =~ [^0-9]*([0-9]+)\.[0-9]+ ]]
$ echo "${BASH_REMATCH[1]}"
34521

Answer 4

There's the awk solution:

echo $string1 | awk -F'[^0-9]+' '{ print $2 }'

Use -F'[^0-9,]+' to set the field separator to anything that is not a digit (0-9) or a comma, then print it.