I am trying to build a regex in javascript to match a 9-digit number with these characteristics:
First 3 digits should not be ‘0’ ,
4th and 5th Digit should not be ‘0’,
Last 4 digits should not be ‘0000’,
First 3 digits should not be ‘666’,
Last 3 Digits Should not be greater than ‘899’
Can someone please help me out with this.
Here is my current regex:
/^666[^0]{3}[1-9]{2}0000$/
, but it’s not meeting the criteria
Try This regular expression it will work ^(?!(000)|(666))[0-9]{3}[1-9]{2}[0-9][0-8][0-9]{2}(?<!0000)$. See demo here
Try this ^[^06]{3}[^0]{2}[^0][1-8][^0]{2}$
Below is my explanation for it
First 3 digits should not be ‘0’
First 3 digits should not be ‘666’,-> ^[^06]{3}
4th and 5th Digit should not be ‘0’, -> [^0]{2}
Last 4 digits should not be ‘0000’, -> [^0]{4}
Last 3 Digits Should not be greater than ‘899’ -> [1-8][^0]{2}$
Last 3 Digits Should be greater than ‘899’, I do not pay attention to not
666 - not the same thing. Also, the ending [1-8][^0]{2}$ wrongfully excludes 0xx where x is anything and y00 where y is 0-8.Because all nine characters need to be a digit, you might use lookahead from the beginning to check that there are 9 digits followed by the end of the string, which will make the subsequent groups easier to manage. Then, you need to utilize character sets and negative lookahead. The first two conditions look to collapse together - the first five characters need to be other than 0:
/^(?=\d{9}$)(?!666)[^0]{5}(?!0000)\d[^9]/
const re = /^(?=\d{9}$)(?!666)[^0]{5}(?!0000).[^9]/;
`555555555
5555555555
055555555
555505555
666555555
555550000
555550001
555550900
555550953`
.split('\n')
.forEach(n => console.log(re.test(String(n))));Explanation:
555555555 true
5555555555 false; 10 digits, not 9
055555555 false: has 0 in first 5 digits
555505555 false: has 0 in first 5 digits
666555555 false: starts with 666
555550000 false: ends with 0000
555550001 true
555550900 false: sixth digit is a 9 (so last 3 digits are 9xx, which is greater than 899)
555550953 false: same as above
