I currently want to make the first letter of a city name uppercase; but since in my country we have a lot of cities with a multiple words composed name (eg. New York), i am currently using this regex:
/\w\S*/g
so I am using this function
function capitalizeWords(myString) {
return myString.replace(/\w\S*/g, w => w.replace(/^\w/, c => c.toUpperCase()))
}
so I can change new yorkin New York.
But now I have a small issue, I have some cities like
sauze d'oulx
and I need to convert it to Sauze d'Oulx, excluding the d, or in any case the first character before the ' symbol.
What is the trick to exclude one char before a known one?
You can define all your exceptions as alternatives before matching and capturing your expected match:
const texts = ["new york", "sauze d'oulx"];
const capitalizeWords = (myString) =>
myString.replace(/\bd['’]|(\w)(\S*)/g, (m, g, h) => g ? g.toUpperCase() + h : m);
for (const text of texts) {
console.log(text, '->', capitalizeWords(text));
}The \bd['’]|(\w)(\S*) regex matches
\bd['’] - a d' or d’ after a non-word char or at the start of string| - or(\w)(\S*) - a word char captured into Group 1 (g in the code above) and then zero or more non-whitespaces into Group 2 (h in the code above).The (m, g, h) => g ? g.toUpperCase() + h : m replacement means that the replacement is the match itself if Group 1 did not match, else, the char matched with \w is capitalized.
Extending with di preposition:
const texts = ["new york", "sauze d'oulx", "bagno di romagna"];
const capitalizeWords = (myString) =>
myString.replace(/\bd(?:['’]|i\b)|(\w)(\S*)/g, (m, g, h) => g ? g.toUpperCase() + h : m);
for (const text of texts) {
console.log(text, '->', capitalizeWords(text));
}