I have a string 36K total,TOO_MANY_NEWLINE_IN_TEXT_tABLE. From this string i need only 36.
is that possible?
4 Answers
As described in https://www.gnu.org/software/bash/manual/bashref.html#Shell-Parameter-Expansion, you can write
echo "${foo:0:2}"
to output the first two characters of a variable foo.
Comments
You could also try this:
echo "36K total,TOO_MANY_NEWLINE_IN_TEXT_tABLE" | grep -oEi '([0-9])+'
2 Comments
melpomene
Why use the
-i flag?
Ravi Gehlot
@melpomene man grep. I guess you don't need to ignore case for numbers. I thought it would be nice to include -i, in case, he uses it to capture values other integers.
To extract the leading number (of variable length) from a string in pure bash (without any external tools), you can use the =~ operator (extended regex match) inside the conditional expression [[ .. ]]:
$ line='36K total,TOO_MANY_NEWLINE_IN_TEXT_tABLE'
$ [[ $line =~ [0-9]* ]] && echo "$BASH_REMATCH"
36
Comments
You have a delimiter; the letter K separates the first desired substring, 36, from the rest of the string.
$ str="36K total,TOO_MANY_NEWLINE_IN_TEXT_tABLE"
$ IFS=K read num rest <<< "$str"
$ echo "$num"
36
echo '36K total,TOO_MANY_NEWLINE_IN_TEXT_tABLE' | sed 's/[^0-9].*//'?