Java Regex, match pattern, pair of words

Question

i am using regex to check correctness of the string in my application. I want to check if string has a following pattern: x=y&a=b&... x,y,a,b etc. can be empty.

Example of correct strings:

abc=def&gef=cda&pdf=cdf
=&gef=def
abc=&gef=def
=abc&gef=def

Example of incorrect strings:

abc=def&gef=cda&
abc=def&gef==cda&
abc=defgef=cda&abc=gda

This is my code showing current solution:

    String pattern = "[[a-zA-Z0-9]*[=]{1}[a-zA-Z0-9]*[&]{1}]*";
    if(!Pattern.matches(pattern, s)){
        throw new IllegalArgumentException(s);
    }

This solution is bad because it accepts strings like:

abc=def&gef=def&

Can anyone help me with correct pattern?

No, string is containing only letters + numbers or is empty

Pan Wolodyjowsky
– Pan Wolodyjowsky

2017-05-21 19:46:57 +00:00
Commented May 21, 2017 at 19:46 — Pan Wolodyjowsky
– Pan Wolodyjowsky, Commented May 21, 2017 at 19:46
=&= would be correct

Pan Wolodyjowsky
– Pan Wolodyjowsky

2017-05-21 19:48:17 +00:00
Commented May 21, 2017 at 19:48 — Pan Wolodyjowsky
– Pan Wolodyjowsky, Commented May 21, 2017 at 19:48

Graham · Accepted Answer · 2017-09-24 14:21:14Z

3

You may use the following regex:

^[a-zA-Z0-9]*=[a-zA-Z0-9]*(?:&[a-zA-Z0-9]*=[a-zA-Z0-9]*)*$

See the regex demo

When used with matches(), the ^ and $ anchors may be omitted.

Details:

^ - start of string
[a-zA-Z0-9]* - 0+ alphanumeric chars (may be replaced with \p{Alnum})
= - a = symbol
[a-zA-Z0-9]* - 0+ alphanumeric chars
= - a = symbol
(?: - start of a non-capturing group matching sequences of...
- & - a & symbol
- [a-zA-Z0-9]*=[a-zA-Z0-9]* - same as above
)* - ... zero or more occurrences
$ - end of string

NOTE: If you want to make the pattern more generic, you may match any char other than = and & with a [^&=] pattern that would replace a more restrictive [a-zA-Z0-9] pattern:

^[^=&]*=[^=&]*(?:&[^=&]*=[^=&]*)*$

See this regex demo

edited Sep 24, 2017 at 14:21

Graham

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answered May 21, 2017 at 19:48

Wiktor Stribiżew

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Comments

Dawood ibn Kareem · Accepted Answer · 2017-05-21 19:49:20Z

1

I believe you want this.

([a-zA-Z0-9]*=[a-zA-Z0-9]*&)*[a-zA-Z0-9]*=[a-zA-Z0-9]*

This matches any number of repetitions like x=y, with a & after each one; followed by one repetition like x=y without the following &.

answered May 21, 2017 at 19:49

Dawood ibn Kareem

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1 Comment

Wiktor Stribiżew Over a year ago

Quantifying the pattern start causes more redundant backtracking than when quantifying the pattern end.

degant · Accepted Answer · 2017-05-21 19:58:16Z

1

Here you go:

^\w*=\w*(?:&(?:\w*=\w*))*$

^ is the starting anchor
(\w*=\w*) is to represent parameters like abc=def
- \w matches a word character [a-zA-Z0-9_]
- \w* represents 0 or more characters
& represents tha actual ampersand literal
(&(\w*=\w*))* matches any subsequents parameters like &b=d etc.
$ represents the ending anchor

Regex101 Demo

EDIT: Made all groups non-capturing.

Note: As @WiktorStribiżew has pointed out in the comments, \w will match _ as well, so above regex should be modified to exclude underscores if they are to be avoided in the pattern, i.e [A-Za-z0-9]

edited May 21, 2017 at 19:58

answered May 21, 2017 at 19:48

degant

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2 Comments

Wiktor Stribiżew Over a year ago

\w also matches _ symbols.

degant Over a year ago

Valid point @WiktorStribiżew, would have to change all \w to [A-Za-z0-9] instead. Thanks!

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