如何使用正则表达式限制位数但允许前面有零

Question

I want to limit the matches to 7 digits at most, but at the same time I want to allow zeroes in front of the number. Right now it's allowing numbers such as 0, 1, 12, 123, 1234, 12345, 123456, 1234567 which is expected. This is the problem, I also want to find a match for numbers such as 0001234567 but not 12345678. This is the regex I'm using:

/^[1-9][0-9]{0,6}$|^0$/g

Answer 1

Try this /^0*[0-9]{7}$/

• 0001234567 - pass
• 1234567 - pass
• 12345678 - error
• 12345 - error

Answer 2

This should do it:

/^0*[1-9][0-9]{0,6}$/

It allows any number of leading zeroes.

Answer 3

As you als want to match 0, you might also use an alternation to match 1+ more zeroes as well

^(?:0*[1-9]\d{0,6}|0+)$

Regex demo

You might also use parseInt and check for the string length

 const limitNumbers = s => parseInt(s, 10).toString().length < 8; const strings = [ "0", "1", "12", "123", "1234", "12345", "123456", "1234567", "0001234567", "12345678", ]; strings.forEach(s => console.log(`${s}: ${limitNumbers(s)}`))