I've been trying to create a RegExp that makes sure a sure has entered at least one word and at least one space. I tried to use this:

/\w+\s+/

But that makes sure that there is a word AFTER a space. I just want to make sure there is both in a string. They don't need to be in the order of the above RegExp.

How can I make the RegExp work, but without matching the order?

I've been trying to create a RegExp that makes sure a sure has entered at least one word and at least one space. I tried to use this:

/\w+\s+/

But that makes sure that there is a word AFTER a space. I just want to make sure there is both in a string. They don't need to be in the order of the above RegExp.

How can I make the RegExp work, but without matching the order?

• javascript
• regex

• could you post some possible and impossible matches? – Avinash Raj Commented Nov 19, 2014 at 17:31
• It SHOULD match "Hello, dude." It should NOT match "Hello,dude". – Ian Hazzard Commented Nov 19, 2014 at 17:33
• There are a lot of answers to my question. However, I do not want to simply pick the one that is upvoted. Please give a detailed explanation of why your regex works, and maybe why mine doesn't. – Ian Hazzard Commented Nov 19, 2014 at 17:44

3 Answers 3

/(?=.*?\w)(?=.*?\s)/

?= means "look-ahead", and .* means "any number of characters"

So "find any number of characters then a \w", "find any number of characters and a \s"

Another thing to note about how this works, look-aheads are "non-matching", making it so that this can match in any order.

You have two things:

• Is there a word character?
• Is there a space?

Two things.

• str.match(/\w/)
• str.match(/\s/)

So why are you trying to do them as one step?

if( str.match(/\w/) && str.match(/\s/))

There are a lot of answers to my question. However, I do not want to simply pick the one that is upvoted. Please give a detailed explanation of why your regex works, and maybe why mine doesn't.

My answer provides the simplest solution. It is very clear to anyone reading it that we are checking "if it has a word character, and if it contains a space character". It is also very easy to expand on, such as if you want to add another check.

zyklus' answer (/(?=.*?\w)(?=.*?\s)/) is the fastest when speed-tested on a 50Kb string of input. In more mon cases (ie. 100 character at most), this speed difference will be practically non-existent. It is twice as fast as my answer, but "2 * very small number = very small number". It's easy enough to add new test cases (just add another (?=.*something) block) but is less humanly-obvious as to what it does.

Jacob's answer ((\w+.*\s+)|(\s+.*\w+)) does quite literally what you asked, checking first if there is a word character and then a space character, then checks the other way around before failing. It works, however it is slower. Furthermore, if you decide to add a new test case, you'd get something like (\w+.*\s+.*\d+)|(\w+.*\d+.*\s)|(\s+.*\w+.*\d+)|(\s+.*\d+.*\w+)|(\d+.*\w+.*\s+)|‌​(\d+.*\s+.*\w+). It only gets worse if you add a fourth test (24 arrangements to check) and is unreadably ugly. Do not use this answer.

Other answers are variants of existing ones.

If you need to do it in one RegEx for some reason:

(\w+.*\s+)|(\s+.*\w+)

Can be handy if you're working with a library that only enables you to use a single regular expression.