How to get integers from a specific string using regexp?

Question

I have some sample text

# HELP aaasd asdads
# TYPE ASDA dasdas
goodmetric_total{"camel-1"} 777.0
# HELP qqq www
# TYPE eee rrr
badmetric_total{"camel-1"} 888.0

I will need to get numbers from a specific string. Using String.format, I will substitute the values I need. For example goodmetric_total.

How do I write a regexp to get only a numerical value? At the moment, I have solutions for searching string first.

^goodmetric.+\d+

And already in this line look for numbers.

I think you can do all this in one operation. Thanks!

Do you mean you need to get 777.0 or 777.0 and 1? from goodmetric_total{"camel-1"} 777.0? — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Jan 11, 2022 at 9:20
@WiktorStribiżew Sorry. I only need to get a value like 777 — Evgeniy Zelenskiy
– Evgeniy Zelenskiy, Commented Jan 11, 2022 at 9:22
Then you could use a non-regex approach, or a regex one with a pattern similar to ^goodmetric.+ (\d+)(?:\.0)?$ (demo) (where you capture the int part of the number into Group 1). — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Jan 11, 2022 at 9:24
Why not just use the String#split() method? String value=""; if (line.trim().startsWith("goodmetric_total")) { value = line.split("\\s+")[1]; }. — DevilsHnd - 退した
– DevilsHnd - 退した, Commented Jan 11, 2022 at 9:57

Wiktor Stribiżew · Accepted Answer · 2022-01-11 09:30:24Z

1

You can use a regex approach with

(?m)^goodmetric.*\h(\d+)(?:\.0+)?$

See the regex demo. Details:

(?m)^ - start of a line
goodmetric - a string
.* - any zero or more chars other than line break chars, as many as possible
\h - a horizontal whitespace
(\d+) - Group 1: one or more digits
(?:\.0+)? - an optional sequence of a . and one or more 0 chars
$ - end of a line (due to (?m) modifier = Pattern.MULTILINE).

See the Java demo:

String s = "# HELP aaasd asdads\n# TYPE ASDA dasdas\ngoodmetric_total{\"camel-1\"} 777.0\n# HELP qqq www\n# TYPE eee rrr\nbadmetric_total{\"camel-1\"} 888.0";
Pattern pattern = Pattern.compile("^goodmetric.*\\h(\\d+)(?:\\.0+)?$", Pattern.MULTILINE);
Matcher matcher = pattern.matcher(s);
while (matcher.find()){
    System.out.println(matcher.group(1)); 
}

answered Jan 11, 2022 at 9:30

Wiktor Stribiżew

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1 Comment

Evgeniy Zelenskiy Over a year ago

I am very grateful to you for the explanation!

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