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Hello guys i want to write a regex to check if a value has only numbers but 0 not to be in first position . for example the value 10 is correct but the value 01 is wrong.

so far i have this mystr.matches("[123456789]+")

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  • mystr.matches("[1-9]+") Commented Feb 24, 2013 at 16:52
  • 1
    Is 0 a correct number? Commented Feb 24, 2013 at 16:54
  • @LuiggiMendoza I'm asking about 0, just 0 Commented Feb 24, 2013 at 16:56
  • What about floating point numbers, like 10.5? Commented Feb 24, 2013 at 16:59

5 Answers 5

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3

Try this:

mystr.matches("^[1-9]\\d*$")
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3 Comments

You need to escape the \ character by using \\.
Thanks, changed it, apparently I have to put in three \ characters in for it to show two.
matches will return true only if entire string matches used regex so there is no need for ^ and $ :). Also try using "{}" key (or Ctrl+K) next time while creating answer to format selected code and you wont have to use tricks like \\\ to get two backslashes :)
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Proposed solution:

mystr.matches("[1-9]\\d*")

Explanation:

  • [1-9] in the beginning to check if the first digit is between 1 and 9.
  • \\d* to look for any digit (form 0 to 9).
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4 Comments

how about 1, 2, 3, 4, 5, 6, 7, 8, 9 ? Should be \\d* no?
when i put 2 i have problem
@user1884030 answer updated. By the way, is 0 a legal input?
This will match anything with a number (1-9) in it.
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All answers here will validate numbers like 1, 2, 3, ..., 10, 11, ... but in case you want to also include simple 0 but not 00, 01 and so on you can use 

mystr.matches("0|[1-9][0-9]*")
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2

I think that one should do the job: "([1-9]+[0-9]*)"

Cheers!

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1 Comment

you don't need + in your regex since you have * after [0-9]. Try to avoid catastrophic backtracking
1

You can also use this regex

^(?!0)\d+$
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