bash：从文本字符串中解析数字

Question

I'm writing a little bash script that scan a list of text lines, each of which has the format:

num1 num2 num3 filename

For each line, I only want to parse out the first numerical token. This is my code:

printf "input line: %s\n" "${line}"
let number="${line//^[0-9]+/}"
printf "regexp parsed %s\n" "${number}"

Well, it does parse out the first number in the line, but also outputs an error message:

input line: 11531          1008      16   12555    310b /usr/bin/gresource
./statistics.sh: line 21: let: number=11531           1008      16   12555    310b /usr/bin/gresource: syntax error in expression (error token is "1008          16   12555    310b /usr/bin/gresource")
regexp parsed 11531

Why do I get this error message? How can I apply the regexp $[0-9]+ on $line without getting the error?

Answer 1

Parameter expansions expect patterns, not regular expressions. Further, your attempt would remove the number rather than capturing it. What's really happening is that let is converting the entire line to a number by commenting on, but ignoring, the non-numeric part of the line. (That is, it only "works" because the line actually starts with a number.)

Consider the following, using the extended pattern equivalent to the regular expression [0-9]+ . Note that your regular expression, treated as a pattern, doesn't match anything.

$ echo "$line"
11531          1008      16   12555    310b /usr/bin/gresource
$ echo "${line//^[0-9]+/}"
11531          1008      16   12555    310b /usr/bin/gresource
$ shopt -s extglob
$ echo "${line/+([0-9])}"
          1008      16   12555    310b /usr/bin/gresource

Use a regular expression match.

[[ $line =~ [0-9]+ ]] && number=${BASH_REMATCH[0]}

Answer 2

If the lines are all that format, use cut , since there'd be no need to parse for numbers:

cut -d ' ' -f 1 <<< 'num1 num2 num3 filename'

Output:

num1

---

For an input file do:

cut -d ' ' -f 1  inputfile.txt