CODEWITHSUNDEEP
javascriptregex
Muhammad Umer
Muhammad Umer

Reputation: 18097

REGEX for finding the word that doesn't start with space or end with space

consider the following string:

"abc123 123 123abc abc123abc"

Now i want to select 123 that is not by itself. So all digit but digit \s\d+\s+

I've tried ton of stuff but no..

Upvotes: 2

Views: 517

Answers (5)

wumpz
wumpz

Reputation: 9131

So here is a short solution to your problem:

\B123|123\B

Regular expression visualization

Debuggex Demo

\B matches non word boundaries. So before or after 123 a part of a word must be found.

This will match all 123 without the one that is standing alone.

Edit 1:

If 123 stands for a sequence of digits and abc for a sequence of letters you could try

[a-zA-Z]\d+|\d+[a-zA-Z]

Regular expression visualization

Debuggex Demo

Unfortunately for this solution you have to create matching groups to retrieve the digits. This is the regexp then

[a-zA-Z](\d+)|(\d+)[a-zA-Z]

Upvotes: 4

dnl-blkv
dnl-blkv

Reputation: 2128

My suggestion in your case would be:

[^\s\d]+\d+[^\s\d]*|[^\s\d]*\d+[^\s\d]+

Regular expression visualization

Debuggex Demo

Detailed explanation:

[^\s\d]+ // Represents one or more non-whitespace and non-digit character
\d+ // Represents one or more digit (your required sequence)
[^\s\d]* // Represents zero or more non-whitespace and non-digit characters
| // Represents logical OR operation
[^\s\d]* // Zero or more non-WS and non-digit
\d+ // Your sequence
[^\s\d]+ // One or more non-WS and non-digit

NOTE: [^\s\d]* entries are used for capturing the whole group in case of abc123abc.

UPD: In the current version, from the string abc13 124233 356abc abc12333abc my regex would match abc13, 356abc and abc12333abc.

Also tested with Rubular.

Upvotes: 1

xiaoboa
xiaoboa

Reputation: 1953

Probably you're looking for lookahead assertion and lookbehind assertion in regular expression.

[^\s]*(?<!\s)123(?!\s)[^\s]*

will match abc123abc only

Regular expression visualization

Debuggex Demo

But the bad news is that javascript regular expression doesn't support lookahead/lookbehind assertion

Upvotes: 0

ghoti
ghoti

Reputation: 46836

If what you're looking for is the string of digits as a distinct word, regexp can use word boundaries. For example:

$ echo "abc123 123 123abc abc123abc" | egrep -o '\<[[:digit:]]+\>'
123

Or perhaps a better example:

$ echo "abc123 234 345abc abc456abc" | egrep -o '\<[[:digit:]]+\>'
234

Now, I'm not a JavaScript guy. But if as @wumpz seems to suggest, JavaScript's regex parser uses \B in place of \< and \> then a regex of \B[[:digit:]]+\B would seem to do the trick, assuming JavaScript understands classes.

Upvotes: 0

Dormouse
Dormouse

Reputation: 5198

The following regex will work in the specific case:

/\w+123\w+|\w+123|123\w+/

As in:

"abc123 123 123abc abc123abc".match(/\w+123\w+|\w+123|123\w+/g);

Will output:

["abc123", "123abc", "abc123abc"]

Assuming that abc and 123 are both obfuscations then you'll need to alter 123 in the regex to target your relevant case.

Upvotes: 3

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