This first removes all non-letter characters, folds to lowercase, then splits the input, doing all the work in a single line: Spaces are initially left in the input so the split will still work. By removing the rubbish characters before splitting, you avoid having to loop through the elements.
That regex “\\s*,\\s*” means: \s* any number of whitespace characters a comma \s* any number of whitespace characters which will split on commas and consume any spaces either side
Square brackets are meant for character class, and you’re actually trying to match any one of: s, |, s (again), e, a, s (again), o and n. Use parentheses instead for grouping: or non-capturing group: Note: Non-capture groups tell the engine that it doesn’t need to store the match, while the other one (capturing group does). For small stuff, either works, for ‘heavy duty’ stuff, you … Read more
I agree that by definition your regex does not miss any UUID. However it may be useful to note that if you are searching especially for Microsoft’s Globally Unique Identifiers (GUIDs), there are five equivalent string representations for a GUID:
Actually, your second attempt works for me in bash 3 and 4: However, to talk theory for a second, it all has to do with how bash interprets the expression as a whole. As long as the regular-expression characters themselves aren’t quoted, the rest of the expression can be quoted without side-effects: but perhaps the … Read more
To remove a property from an object (mutating the object), you can do it like this: Demo For anyone interested in reading more about it, Stack Overflow user kangax has written an incredibly in-depth blog post about the delete statement on their blog, Understanding delete. It is highly recommended. If you’d like a new object with all the keys of the original … Read more
Using regular expressions is probably the best way. You can see a bunch of tests here (taken from chromium) Here’s the example of regular expresion that accepts unicode: But keep in mind that one should not rely only upon JavaScript validation. JavaScript can easily be disabled. This should be validated on the server side as well. Here’s an example of … Read more
Use to make the letter optional. {1} is redundant. (Of course you could also write [A-Z]{0,1} which would mean the same, but that’s what the ? is there for.) You could improve your regex to And, since in most regex dialects, \d is the same as [0-9]: But: do you really need 11 separate capturing groups? And if so, why don’t you capture the … Read more
Here’s yet another way (using a negative look-ahead): Note: There’s only one capturing expression: ([a-z0-9]+).
Don’t use String.replaceAll in this case – that’s specified in terms of regular expressions, which means you’d need even more escaping. This should be fine: Note that the backslashes are doubled due to being in Java string literals – so the actual strings involved here are “single backslash” and “double backslash” – not double and quadruple. replace works on simple … Read more