check string with multiple length patterns

Question

I need to check a string with a regex.

the valid strings are for example: ABC0001 or A00023.

There are three parts I have to check:

So the string should start with a word signs: [a-zA-Z]{1,}
Then there are min one zero: [0]{1,}
The third part is a number \\d+

The whole string may not be longer than 8 signs.

What I tried so far:

String NR_PATTERN = "^([a-zA-Z]{1,}[0]{1,}\\d+){3,8}$";

The problem is, that the regex do not accepts the string: KDS0234

And the trouble you have with this expression is (other than the fact that you can have arbitrary length strings by inserting zeros or additional letters and that [0] is the same as bare-bones 0)? — dhke
– dhke, Commented Jul 8, 2016 at 8:38
"^([a-zA-Z]{1,}[0]{1,}\\d+){3,8}$"; mean ([a-zA-Z]{1,}[0]{1,}\\d+) appear 3-8 times, not whole length is 3-8 — Antony Dao
– Antony Dao, Commented Jul 8, 2016 at 8:44

Moishe Lipsker · Accepted Answer · 2016-07-08 08:43:33Z

3

You could try with a positive lookahead in the beginning:

(?=^.{3,8}$)([a-zA-Z]{1,}[0]{1,}\\d+)

The positive lookahead:

(?=^.{3,8}$)

will look ahead ?= and confirm that from the beginnning ^ to the end $ of the string, there is any character matched by . between 3 and 8 times {3,8}.

answered Jul 8, 2016 at 8:43

Moishe Lipsker

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raphaëλ · Accepted Answer · 2016-07-08 08:41:37Z

1

(…){3,8} means you want to … sequence to be repeated between 3 and 8 times, not that the string is 8 characters

You could just use ( in know you said you need a regex, but perhaps this will work for you)

String NR_PATTERN = "[a-zA-Z]+0+\\d+";
String s = "ABC0001";
boolean match  = s.length() < 8 &&  s.matches(NR_PATTERN);

answered Jul 8, 2016 at 8:41

raphaëλ

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Comments

Maria Ivanova · Accepted Answer · 2016-07-08 08:56:42Z

1

You can check the length condition with a lookahead and then match your pattern:

\b(?=[a-zA-Z0-9]{3,8})([A-Za-z]+0[0-9]+)\b

edited Jul 8, 2016 at 8:56

answered Jul 8, 2016 at 8:42

Maria Ivanova

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Jan · Accepted Answer · 2016-07-08 10:30:58Z

1

You could go for:

\b(?=[A-Z]+0+\d+)\w{3,8}\b

See a demo on regex101.com, in Java use double backslashes.

This will match:

ABC0001 A00023 and ABCD001

But not

B123456 or 0001B

edited Jul 8, 2016 at 10:30

answered Jul 8, 2016 at 9:14

Jan

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1 Comment

DVarga Over a year ago

In the OP: "The whole string may not be longer than 8 signs."

Antony Dao · Accepted Answer · 2016-07-08 08:53:25Z

0

(?=^.{3,8}$)([a-zA-Z]+[0]+\d+)

Hope this help

answered Jul 8, 2016 at 8:53

Antony Dao

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2 Comments

abarisone Over a year ago

Please edit with more information. Code-only and "try this" answers are discouraged, because they contain no searchable content, and don't explain why someone should "try this". We make an effort here to be a resource for knowledge.

Toby Speight Over a year ago

Although this code may be help to solve the problem, providing additional context regarding why and/or how it answers the question would significantly improve its long-term value. Please edit your answer to add some explanation.

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