Regex to add string to beginning of every word based on condition

问题

I have a string which looks like this

someString = "#3Hello there! How many #4candies did you sell today? Do have any #4candies left?"

lookupDict = {"Hello there": "#3", "candies": "#4"}

Now I want to replace every terms in the string someString with #0 which are not in the dictionary lookupDict. I can't split by a space " " since this will make certain terms like Hello there appear as two different words Hello and there and that would never match my condition.

Now I know to apply basic regex that would add a #0 in front of every word. For example something like

let regex = /(\b\w+\b)/g;

someString = someString.replace(regex, '#0$1'));

But that would blindly add #0 to every term and won't lookup in the dictionary lookupDict.

Is there any way I can combine the regex with a lookup in the dictionary and assign the #0 accordingly? Basically the end result would something like

someString = "#3Hello there! #0How #0many #4candies #0did #0you #0sell #0today? #0Do #0have #0any #4candies #0left?"

Note: Spaces can be considered as word boundries here.

回答1:

You may use the following logic:

Build an array of substrings you need to skip that are concatenated values and keys of the associative array
Sort the items by length in the descending order since the word boundaries might not work well with phrases containing whitespace
Compile a regex pattern that will consist of two alternatives: the first will match the array items (escaped for use in a regex pattern) enclosed with a capturing group, and the other will match the rest of the "words"
When a match is found, check if Group 1 matched. If group 1 matches, just return the match value, else, add #0 to the match value.

Here is the implementation:

let someString = "#3Hello there! How many #4candies did you sell today? Do have any #4candies left? #0how #0much";
const lookupDict = {"Hello there": "#3", "candies": "#4", "how": "#0", "much": "#0"};
let patternDict = [];                             // Substrings to skip
for (var key in lookupDict) {                     
  patternDict.push( `${lookupDict[key]}${key}` ); // Values + keys
}
patternDict.sort(function(a, b){                  // Sorting by length, descending
  return b.length - a.length;
});
var rx = new RegExp("(?:^|\\W)(" + patternDict.map(function(m) { // Building the final pattern
    return m.replace(/[-\/\\^$*+?.()|[\]{}]/g, '\\$&');}
  ).join("|") + ")(?!\\w)|\\S+", "gi");
// rx = /(?:^|\W)(#3Hello there|#4candies|#0much|#0how)(?!\w)|\S+/gi
someString = someString.replace(rx, (x, y) => y ? x : `#0${x}` );
console.log(someString);
// => #3Hello there! #0How #0many #4candies #0did #0you #0sell #0today? #0Do #0have #0any #4candies #0left? #0how #0much

The regex will look like

/(?:^|\W)(#3Hello there|#4candies|#0much|#0how)(?!\w)|\S+/gi

See the regex demo (PHP option chosen to highlight groups green).

Details

(?:^|\W) - a non-capturing group matching either start of string (^) or (|) any non-word char (=a char other than an ASCII letter, digit or _)
(#3Hello there|#4candies|#0much|#0how) - Capturing group 1 matching any of the lookupDict concatenated value+keys
(?!\w) - a negative lookahead that fails the match if, immediately to the right of the current location, there is a word char
| - or
\S+ - 1+ non-whitespace chars.

回答2:

With this way, there is no worry for lookupDict key length or anything else:

let someString =
  "#3Hello there! How many #4candies did you sell today? #3Hello there! Do have any #4candies left?#3Hello there! #7John Doe! some other text with having #7John Doe person again";

const lookupDict = { "Hello there": "#3", candies: "#4", "John Doe": "#7" };

Object.keys(lookupDict).map((key, i) => {
  const regex = new RegExp(key, "g");
  someString = someString.replace(regex, lookupDict[key]); // replace each key to the value: Hello world => #3
});

someString = someString.replace(/ /gi, " #0"); // replace each space

Object.keys(lookupDict).map((key, i) => {
  const regex = new RegExp(lookupDict[key] + lookupDict[key], "g");
  someString = someString.replace(regex, `${lookupDict[key]}${key}`); // role back the value to key+value
});

someString = someString.replace(/#0#/gi, "#"); // replace #0 for each lookupDict key value

console.log(someString, '<TheResult/>');

回答3:

You can pass a function to .replace as second parameter and check the matching token in dictionary

I've changed regex to not include # in results

Hello there is problematic, how long can a single term be? max 2 words?

someString = "#3Hello there! How many #4candies did you sell today? Do have any #4candies left?"

let regex = /(?<!#)(\b\w+\b)/g;

someString = someString.replace(regex, x => {
// check x in dict
	return `#0${x}`
});
console.log(someString)

来源：https://stackoverflow.com/questions/60365460/regex-to-add-string-to-beginning-of-every-word-based-on-condition

标签

javascript

regex

replace