Java regex confusing

Question

i want parse string like this

  0.1142     0.0000     0.0000     0.0004     0.0000     0.0000    2299/2299        MakeRequest   [23]

I want get ol doubles from this string. I am using next pattern

.+ ([0-9\\.]+) [\\p{Space}]+ ([0-9\\.]+) [\\p{Space}]+ ([0-9\\.]+) [\\p{Space}]+ ([0-9\\.]+) [\\p{Space}]+ ([0-9\\.]+) [\\p{Space}]+ ([0-9\\.]+) [\\p{Space}]+ ([0-9\\/])+ [\\p{Space}] "\\.(MakeRequest) .+

And this string is parsed well.
But i cann't parse next string!

    20.1600     0.0001     0.0000     0.0053     0.0001     0.0000  383248/383248      MakeRequest   [22]

This string are very similar! But first is parsed, second not:-(

Update Easily pattern

.+ ([0-9\\.]+) .+ ([0-9\\.]+) .+([0-9\\.]+) .+ ([0-9\\.]+) .+ ([0-9\\.]+) .+ ([0-9\\.]+) .+ ([0-9\\/])+ .+\\.(MakeRequest) .+

Don't works too

I'm using

java.util.regex.Pattern

and

java.util.regex.Matcher

Are there supposed to be two + symbols between the last two groupings? [\\p{Space}]+ + "\\.(MakeRequest) .+? — Hunter McMillen
– Hunter McMillen, Commented Apr 20, 2012 at 13:06

Johan Soderberg · Accepted Answer · 2012-04-20 13:08:52Z

4

([0-9\.]+) .+ ([0-9\/])+

Matches SPACE, ONE OR MORE CHARACTERS, SPACE (total of 3 characters)

Second string has only 2 spaces.

0.0000 383248/383248

answered Apr 20, 2012 at 13:08

Johan Soderberg

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1 Comment

Johan Soderberg Over a year ago

Between the numbers that is, which is where it fails (didn't check if it fails anywhere else too).

Marko Topolnik · Accepted Answer · 2012-04-20 13:09:44Z

2

What would be wrong with code like the following:

  final Matcher m = Pattern.compile("\\d+\\.\\d+").matcher(s); // s is your line
  while (m.find()) System.out.println(Double.parseDouble(m.group()));

answered Apr 20, 2012 at 13:09

Marko Topolnik

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Piotr Kochański · Accepted Answer · 2012-04-20 13:10:50Z

1

This is pattern to get all doubles

[0-9]*\.[0-9]*

answered Apr 20, 2012 at 13:10

Piotr Kochański

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