Regex match for beginning of multiple words in string

Question

In Javascript i want to be able to match strings that begin with a certain phrase. However, I want it to be able to match the start of any word in the phrase, not just the beginning of the phrase.

For example:

Phrase: "This is the best"

Need to Match: "th"

Result: Matches Th and th

EDIT: \b works great however it proposes another issue:

It will also match characters after foreign ones. For example if my string is "Männ", and i search for "n", it will match the n after Mä...Any ideas?

Answer 1

"This is the best moth".match(/\bth/gi);

or with a variable for your string

var string = "This is the best moth";
alert(string.match(/\bth/gi));

\b in a regex is a word boundary so \bth will only match a th that at the beginning of a word.

gi is for a global match (look for all occurrences) and case insensitive

(I threw moth in there to as a reminder to check that it is not matched)

jsFiddle example

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Edit:

So, the above only returns the part that you match (th). If you want to return the entire words, you have to match the entire word.

This is where things get tricky fast. First with no HTML entity letter:

string.match(/\bth[^\b]*?\b/gi);

Example

To match the entire word go from the word boundary \b grab the th followed by non word boundaries [^\b] until you get to another word boundary \b. The * means you want to look for 0 or more of the previous (non word boundaries) the ? mark means that this is a lazy match. In other words it doesn't expand to as big as would be possible, but stops at the first opportunity.

If you have HTML entity characters like ä (ä) things get complicated really fast, and you have to use whitespace or whitespace and a set of defined characters that may be at word boundaries.

string.match(/\sth[^\s]*|^th[^\s]*/gi);

Example with HTML entities.

Since we're not using word boundaries, we have to take care of the beginning of the string separately (|^).

The above will capture the white space at the beginning of words. Using \b will not capture white space, since \b has no width.