This is a follow-up of Find multiple keywords within a dictionary.

My questions are...

1. The first is: I believe this matches words that are not whole. Like if short is in my dictionary it matches the word shortly. How would I stop this?

2. And the second not so important but would be nice is: How would I make it so it only matches once per content? So short doesn't get defined twice within the same content area.

Thanks!

This is a follow-up of Find multiple keywords within a dictionary.

My questions are...

1. The first is: I believe this matches words that are not whole. Like if short is in my dictionary it matches the word shortly. How would I stop this?

2. And the second not so important but would be nice is: How would I make it so it only matches once per content? So short doesn't get defined twice within the same content area.

Thanks!

• javascript
• regex
• search
• dictionary
• replace

1 Answer 1

I have implemented the following additional requirements:

• Do not match shortly when looking for short (because shortly is a different word)
• Use keys in the dictionary only once.
  Example input: key=foo, replacement=bar, content=foo foo.
  Output: bar foo (only the first foo is replaced).

Demo: http://jsfiddle/bhGE3/3/

Usage:

1. Define a dictionary. Each key will be used only once.
2. Define content. A new string will be created, based on this string.
3. Optionally, define a replacehandler function. This function is called at each match. The return value will be used to replace the matched phrase.
   The default replacehandler will return the dictionary's matching phrase. The function should take two arguments: key and dictionary.
4. Call replaceOnceUsingDictionary(dictionary, content, replacehandler)
5. Process the output, eg. show content to the user.

Code:

var dictionary = {
    "history": "war . ",
    "no": "in a",
    "nothing": "",
    "oops": "",
    "time": "while",
    "there": "We",
    "upon": "in",
    "was": "get involved"
};
var content = "Once upon a time... There was no history. Nothing. Oops";
content = replaceOnceUsingDictionary(dictionary, content, function(key, dictionary){
    return '_' + dictionary[key] + '_';
});
alert(content);
// End of implementation

/*
* @name        replaceOnceUsingDictionary
* @author      Rob W http://stackoverflow./users/938089/rob-w
* @description Replaces phrases in a string, based on keys in a given dictionary.
*               Each key is used only once, and the replacements are case-insensitive
* @param       Object dictionary  {key: phrase, ...}
* @param       String content
* @param       Function replacehandler
* @returns     Modified string
*/
function replaceOnceUsingDictionary(dictionary, content, replacehandler) {
    if (typeof replacehandler != "function") {
        // Default replacehandler function.
        replacehandler = function(key, dictionary){
            return dictionary[key];
        }
    }
    
    var patterns = [], // \b is used to mark boundaries "foo" doesn't match food
        patternHash = {},
        oldkey, key, index = 0,
        output = [];
    for (key in dictionary) {
        // Case-insensitivity:
        key = (oldkey = key).toLowerCase();
        dictionary[key] = dictionary[oldkey];
        
        // Sanitize the key, and push it in the list
        patterns.push('\\b(?:' + key.replace(/([[^$.|?*+(){}])/g, '\\$1') + ')\\b');
        
        // Add entry to hash variable, for an optimized backtracking at the next loop
        patternHash[key] = index++;
    }
    var pattern = new RegExp(patterns.join('|'), 'gi'),
        lastIndex = 0;

    // We should actually test using !== null, but for foolproofness,
    //  we also reject empty strings
    while (key = pattern.exec(content)) {
        // Case-insensitivity
        key = key[0].toLowerCase();

        // Add to output buffer
        output.push(content.substring(lastIndex, pattern.lastIndex - key.length));
        // The next line is the actual replacement method
        output.push(replacehandler(key, dictionary));

        // Update lastIndex variable
        lastIndex = pattern.lastIndex;

        // Don't match again by removing the matched word, create new pattern
        patterns[patternHash[key]] = '^';
        pattern = new RegExp(patterns.join('|'), 'gi');

        // IMPORTANT: Update lastIndex property. Otherwise, enjoy an infinite loop
        pattern.lastIndex = lastIndex;
    }
    output.push(content.substring(lastIndex, content.length));
    return output.join('');
}