Regex with form As123456-89

Question

I need regex for the following text validation:

must start with two alpabetic characters, six digits, minus and two digits represent the year, for example: Ew142356-19

I have tried:

 Console.WriteLine(new Regex("[0-9]{2}[a-zA-Z]{6}[\\-]{1}[1-9]{1}[0-9]{1}").IsMatch("Be123456-19"));

but it prins False

You swapped `[0-9]` and `[a-zA-Z]` patterns. Besides, you need to use anchors, `^` and `$`. Use `^[a-zA-Z]{2}[0-9]{6}-[0-9]{2}$`. See https://ideone.com/q2V40C — Wiktor Stribiżew, Jan 14 '19 at 13:16

Superluminal · Accepted Answer · 2019-01-14T13:23:24.280

2

Try this one:

^[a-zA-Z]{2}\d{6}-[1-9]\d$

Notes:

There is no need to do something like [1-9]{1}. [1-9] already means that this is one element.

Use Anchors: ^ means the start of the string and $ means the end of it. If you dont use them you could match something aBe123456-19

\d is equivalent to [0-9]

There is no need to create a set for the - sign

edited Jan 14 '19 at 13:23

answered Jan 14 '19 at 13:18

Superluminal

[\d is less efficient than \[0-9\]](https://stackoverflow.com/questions/16621738/d-is-less-efficient-than-0-9) – JohnyL Jan 14 '19 at 17:02
@JohnyL Thanks. Dont knew about it. – Superluminal Jan 15 '19 at 09:05

1 Answers1