Regex for partial match of word?

Question

Is it possible to perform the following match?

^do something$

.. where matches are possible for a partially typed 'something'.

For example:

^do s$
^do so$
^do som$
^do some$
^do somet$
^do someth$
^do somethi$
^do somethin$
^do something$

Is it possible to do this with regex without having to string a bunch of logical ORs?

[`^do s(?:o(?:m(?:e(?:t(?:h(?:i(?:ng?)?)?)?)?)?)?)?$`](https://regex101.com/r/jO6hB4/1) - not so nice. — Wiktor Stribiżew, Mar 09 '16 at 13:04
If a word is partially typed, there could be number of possibilities for trailing characters. Regex works with character patterns. — , Mar 09 '16 at 13:04
There is no number of possibilities as I explicitly define `$` at the end. — Zhro, Mar 09 '16 at 13:05
@Zhro: Really, just use `^do s\w*$`. Or `^do s[a-z]*$`. Isn't that enough? If you think it is not, please update your question with the *real* data and problem. — Wiktor Stribiżew, Mar 09 '16 at 13:22

score 0 · Answer 1 · answered Mar 09 '16 at 13:30

0

There is no need to use regex :

function isMatch(input)
{
   var key="do something";
   return key.startsWith(input) && input.length <= key.length;
}

answered Mar 09 '16 at 13:30

nAviD

score 0 · Accepted Answer · answered Mar 09 '16 at 14:29

0

If you just want to match partially "something", you could do:

\^do\s+((?:something|somethin|somethi|someth|somet|some|som|so|s))\$

answered Mar 09 '16 at 14:29

Quinn

2 Answers2