I want a regular expression that prevents symbols and only allows letters and numbers. The regex below works great, but it doesn't allow for spaces between words.

^[a-zA-Z0-9_]*$

For example, when using this regular expression "HelloWorld" is fine, but "Hello World" does not match.

How can I tweak it to allow spaces?

One possibility would be to just add the space into you character class, like acheong87 suggested, this depends on how strict you are on your pattern, because this would also allow a string starting with 5 spaces, or strings consisting only of spaces.

The other possibility is to define a pattern:

I will use \w this is in most regex flavours the same than [a-zA-Z0-9_] (in some it is Unicode based)

^\w+( \w+)*$

This will allow a series of at least one word and the words are divided by spaces.

^ Match the start of the string

\w+ Match a series of at least one word character

( \w+)* is a group that is repeated 0 or more times. In the group it expects a space followed by a series of at least one word character

$ matches the end of the string

This one worked for me

([\w ]+)

Try with:

^(\w+ ?)*$

Explanation:

\w             - alias for [a-zA-Z_0-9]
"whitespace"?  - allow whitespace after word, set is as optional

I assume you don't want leading/trailing space. This means you have to split the regex into "first character", "stuff in the middle" and "last character":

^[a-zA-Z0-9_][a-zA-Z0-9_ ]*[a-zA-Z0-9_]$

or if you use a perl-like syntax:

^\w[\w ]*\w$

Also: If you intentionally worded your regex that it also allows empty Strings, you have to make the entire thing optional:

^(\w[\w ]*\w)?$

If you want to only allow single space chars, it looks a bit different:

^((\w+ )*\w+)?$

This matches 0..n words followed by a single space, plus one word without space. And makes the entire thing optional to allow empty strings.

This regular expression

^\w+(\s\w+)*$

will only allow a single space between words and no leading or trailing spaces.

Below is the explanation of the regular expression:

1. ^ Assert position at start of the string
2. \w+ Match any word character [a-zA-Z0-9_]
   1. Quantifier: + Between one and unlimited times, as many times as possible, giving back as needed [greedy]
3. 1st Capturing group (\s\w+)*
   1. Quantifier: * Between zero and unlimited times, as many times as possible, giving back as needed [greedy]
   2. \s Match any white space character [\r\n\t\f ]
   3. \w+ Match any word character [a-zA-Z0-9_]
      1. Quantifier: + Between one and unlimited times, as many times as possible, giving back as needed [greedy]
4. $ Assert position at end of the string

This does not allow space in the beginning. But allowes spaces in between words. Also allows for special characters between words. A good regex for FirstName and LastName fields.

\w+.*$

For alphabets only:

^([a-zA-Z])+(\s)+[a-zA-Z]+$

For alphanumeric value and _:

^(\w)+(\s)+\w+$

Just add a space to end of your regex pattern as follows:

[a-zA-Z0-9_ ]

It was my regex: @"^(?=.{3,15}$)(?:(?:\p{L}|\p{N})[._()\[\]-]?)*$"

I just added ([\w ]+) at the end of my regex before *

@"^(?=.{3,15}$)(?:(?:\p{L}|\p{N})[._()\[\]-]?)([\w ]+)*$"

Now string is allowed to have spaces.

Try this: (Python version)

"(A-Za-z0-9 ){2, 25}"

change the upper limit based on your data set

Try with this one:

result = re.search(r"\w+( )\w+", text)

I find this one works well for a "FullName":

([a-z',.-]+( [a-z',.-]+)*){1,70}/

Had a good look at many of these supposed answers...

...and bupkis after scouring Stack Overflow as well as other sites for a regex that matches any string with no starting or trailing white-space and only a single space between strictly alpha character words.

^[a-zA-Z]+[(?<=\d\s]([a-zA-Z]+\s)*[a-zA-Z]+$

Thus easily modified to alphanumeric:

^[a-zA-Z0-9]+[(?<=\d\s]([a-zA-Z0-9]+\s)*[a-zA-Z0-9]+$

(This does not match single words but just use a switch/if-else with a simple ^[a-zA-Z0-9]+$ if you need to catch single words in addition.)

enjoy :D

try .*? to allow white spaces it worked for me