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?

CodePudding user response：

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));
}