Regular Expression: first character alphabet second onward numeric

Question

This is my string, which always starts with a alphabet followed by 3 digits, then - (hyphen), followed by 3 digits, then alphabet, then 4 digits

I123-123S1234N1234

I have come up with a regular expression, which seems to be very long! How can I write a better expression for this?

[a-zA-Z][0-9][0-9][0-9][-][0-9][0-9][0-9][a-zA-Z].........goes on

Note

Shall support case insensitive
Hyphen, S and N always constant

If it's always that fixed format, I think I would split it up using substring rather than regex. — Hans Kilian, Apr 09 '19 at 06:15

CertainPerformance · Answer 1 · 2019-04-09T05:42:16.493

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First use the case-insensitive flag, and then when repeating tokens, use {<number of times to repeat>}` instead of repeating yourself:

(?i)[a-z][0-9]{3}-[0-9]{3}[a-z][0-9]{4}

If you want to match the trailing N1234 as well, then put the final [a-z][0-9]{4} in a group, and repeat it twice:

(?i)[a-z][0-9]{3}-[0-9]{3}(?:[a-z][0-9]{4}){2}

https://regex101.com/r/5ra1eU/1

If the S and N are constant, then don't use character sets to match them, just match the plain characters:

(?i)[a-z][0-9]{3}-[0-9]{3}S[0-9]{4}N[0-9]{4}

https://regex101.com/r/5ra1eU/2

edited Apr 09 '19 at 05:42

answered Apr 09 '19 at 05:36

CertainPerformance

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\d is also a lot shorter than [0-9] – bdebaere Apr 09 '19 at 05:37
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@fubo It does with case insensitive flag. Which he mentions to use. – bdebaere Apr 09 '19 at 05:38
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what is (?i) here – kudlatiger Apr 09 '19 at 05:38
@bdebaere Better to avoid `\d` when you don't *want* to possibly match [other numeric characters](https://stackoverflow.com/questions/16621738/d-is-less-efficient-than-0-9) - it's nice to be as specific as possible – CertainPerformance Apr 09 '19 at 05:39
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@kudlatiger `(?i)` is the case-insensitive flag – CertainPerformance Apr 09 '19 at 05:39
Hyphen, R and N always constant. it won't change at those positions – kudlatiger Apr 09 '19 at 05:39
@kudlatiger Look at the regex101 link - it seems to work just fine – CertainPerformance Apr 09 '19 at 05:41
Let us [continue this discussion in chat](https://chat.stackoverflow.com/rooms/191496/discussion-between-kudlatiger-and-certainperformance). – kudlatiger Apr 09 '19 at 05:43

Michał Turczyn · Accepted Answer · 2019-04-09T06:00:10.897

3

Try this:

var testString = "I123-123S1234N1234";
var pattern = @"[A-Za-z]\d{3}-\d{3}[sS]\d{4}[nN]\d{4}";
var isMatch = Regex.Match(testString, pattern).Success;

Pattern used: [A-Za-z]\d{3}-\d{3}[sS]\d{4}[nN]\d{4}

Explanation:

[A-Za-z] - match any letter

\d{3} - match three digits

[sS] - match S or s literally

\d{4} - match four digits

[nN] - match N or n literally

\d{4} - match four digits

edited Apr 09 '19 at 06:00

answered Apr 09 '19 at 05:43

Michał Turczyn

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will mark as answer! But it shall support for small letters too! [a-z] – kudlatiger Apr 09 '19 at 05:51
There is a regex option that makes it case insensitive, thus making it more simple. – Cleptus Apr 09 '19 at 06:15
@bradbury9 Yes, you are right, but I prefer using flags only when needed badly – Michał Turczyn Apr 09 '19 at 06:17

Frenchy · Answer 3 · 2019-04-09T05:56:04.113

2

if you are sure to have capital letter and S and N constant:

[A-Z]\d{3}-\d{3}S\d{4}N\d{4}

[A-Z] - match any uppercase letter

\d{3} - match 3 digits

S - match S literally

\d{4} - match 4 digits

N - match N literally

\d{4} - match 4 digits

if you want upper and lower case:

[a-zA-Z]\d{3}-\d{3}[Ss]\d{4}[Nn]\d{4}

or

(?i)[a-z]\d{3}-\d{3}s\d{4}n\d{4}

edited Apr 09 '19 at 05:56

answered Apr 09 '19 at 05:42

Frenchy

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score 0 · Answer 4 · answered Apr 09 '19 at 05:40

0

This should work:

[a-zA-Z][0-9]{3}-[0-9]{4}[a-zA-Z][0-9]{4}

answered Apr 09 '19 at 05:40

Nico Schreiner

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Regular Expression: first character alphabet second onward numeric

4 Answers4