how can I skip a string if it contains some word by using regex

Question

I want to skip the string if it contains 'have' after Vehicles word str 1 = "3 Vehicles have already been added", str 2 = "13 Vehicles" Regex I m using = ([0-9]+)[\s]*Vehicles[\s]*[^have]

output needed is "13 Vehicles"(only) but "3 Vehicles" is also matches..that is not right

Thanks in advance

Provide a compilable code, please. – Nikolas Charalambidis Aug 18 '20 at 07:37 — Nikolas Charalambidis, Aug 18 '20 at 07:37
Do you also want to match "vehicles" (lower-case v)? – NomadMaker Aug 18 '20 at 07:46 — NomadMaker, Aug 18 '20 at 07:46
that I am using in my regex "(?i)" in advance – Jenny Aug 18 '20 at 07:54 — Jenny, Aug 18 '20 at 07:54

The fourth bird · Accepted Answer · 2020-08-18T07:41:52.120

0

You could use a negative lookahead to assert that have is not directly at the right.

The negated character class [^have] matches a single char which is not any of the listed chars.

[0-9]+\h+Vehicles(?!\h+have\b)

Regex demo

In Java

String regex = "[0-9]+\\h+Vehicles(?!\\h+have\\b)";

edited Aug 18 '20 at 07:41

answered Aug 18 '20 at 07:35

The fourth bird

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how can I skip a string if it contains some word by using regex

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