value less than 99999999.9999

Question

Pattern regex = Pattern.compile("^\\d{0,8}\\.\\d{0,4}$");

is working, but if I enter value e.g 5000, it should work. Basically valid value should be equal or less than "99999999.9999".

If value is "123456789" it is invalid.

Decimal point is not mandatory.

Please help.

So include the dot and everything after it in a `(...)?` group. — Marko Topolnik, Nov 22 '12 at 13:57
99999999.88888888 is lower but has more digits after the decimal point. — jlordo, Nov 22 '12 at 13:57
Perhaps related: http://stackoverflow.com/questions/6349161/how-to-generate-a-regular-expression-at-runtime-to-match-a-numeric-range — aioobe, Nov 22 '12 at 13:57
Never use regex for working with numbers.. regex is for strings.. — Damien Overeem, Nov 22 '12 at 14:30
While it is possible to write a regular expression to do this, it's absolutely horrible. Truly awful. **DO NOT DO IT THIS WAY.** There's a lovely static `Double` method for parsing which will be far more satisfactory. — Donal Fellows, Nov 22 '12 at 15:08

score 4 · Answer 1 · answered Nov 22 '12 at 13:58

4

Since every floating-point number has many different representations (think 100, 100.0, 1e2, etc), I'd suggest parsing the number into a double, and then using a numeric comparison to establish whether it's within the desired range.

answered Nov 22 '12 at 13:58

NPE

486,780
108
951
1,012

John Woo · Answer 2 · 2012-11-22T14:03:09.210

3

make the decimal part optional

^\d{0,8}(\.\d{0,4})?$

but i think you mean until only 12345678 but if not

^\d{0,9}(\.\d{0,4})?$

but i'm proposing to have atleast 1 number after a decimal so if the user tries to enter 123. it will fail

^\d{0,9}(\.\d{1,4})?$

edited Nov 22 '12 at 14:03

answered Nov 22 '12 at 13:57

John Woo

258,903
69
498
492

replace that part `(\.\d{0,4})?` with `(\.\d+)?` to match numbers with more digits after the decimal point but are still smaller from the mathematical aspect – jlordo Nov 22 '12 at 14:00

value less than 99999999.9999

2 Answers2