Regex to test whether there are more than two digits in a string
As far as I know, \d{2,}
matches 2 or more consecutive digits, but I need to know whether there are any 2 digits in a string.
I would also appreciate some good links to password strength meters.
What I use now is
function passwordStrengthPercent(pwd,username)
{
var score = 0, special = /(.*[!,@,#,$,%,^,&,*,?,_,~,;,:,`,|,\\,\/,<,>,\{,\},\[,\],=,\+])/
if (pwd.length < 8 ) return 0
if (pwd.toLowerCase() == username.toLowerCase()) return -1
score += pwd.length * 4
score += ( checkRepetition(1,pwd).length - pwd.length )
score += ( checkRepetition(2,pwd).length - pwd.length )
score += ( checkRepetition(3,pwd).length - pwd.length )
score += ( checkRepetition(4,pwd).length - pwd.length )
if (pwd.match(/(.*[e].*[e].*[e])/)) score -= 15//most common letter in passswords?
if (pwd.match(/(.*[a].*[a]开发者_JAVA技巧.*[a])/)) score -= 15//most common letter in passswords?
if (pwd.match(/(.*[o].*[o].*[o])/)) score -= 10//most common letter in passswords?
if (pwd.match(/^\l+$/) || pwd.match(/^\d+$/) ) return score/2//there was w here in regexp
if (pwd.match(/(.*[0-9].*[0-9].*[0-9])/)) score += 10
//todo additional rules 11
if (pwd.match(special)) score += 15
if (pwd.match(/([a-z].*[A-Z])|([A-Z].*[a-z])/)) score += 15
if (pwd.match(/(w)/) && pwd.match(/(d)/)) score += 15
if (pwd.match(special) && pwd.match(/(d)/)) score += 10
if (pwd.match(special) && pwd.match(/(w)/)) score += 10
if ( score < 0 ) return 0
if ( score > 100 ) return 100
return score
}
You can try
^.*\d.*\d.*$
which will only match if (at least) two digits are included.
You can just use:
\d.*\d
That matches two numerals anywhere in the line, even with other characters between them.
check if there are Exactly two digits:
^[^\d]*\d[^\d]*\d[^\d]*$
see the test with grep:
kent$ echo "23
2 3
ax3x2x
aaaaa23
a222
2
3x3x4
3 4 5"|grep -P "^[^\d]*\d[^\d]*\d[^\d]*$"
23
2 3
ax3x2x
aaaaa23
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