Regex to get string between curly braces

Try

/{(.*?)}/

That means, match any character between { and }, but don't be greedy - match the shortest string which ends with } (the ? stops * being greedy). The parentheses let you extract the matched portion.

Another way would be

/{([^}]*)}/

This matches any character except a } char (another way of not being greedy)

/\{([^}]+)\}/

/        - delimiter
\{       - opening literal brace escaped because it is a special character used for quantifiers eg {2,3}
(        - start capturing
[^}]     - character class consisting of
    ^    - not
    }    - a closing brace (no escaping necessary because special characters in a character class are different)
+        - one or more of the character class
)        - end capturing
\}       - the closing literal brace
/        - delimiter

If your string will always be of that format, a regex is overkill:

>>> var g='{getThis}';
>>> g.substring(1,g.length-1)
"getThis"

substring(1 means to start one character in (just past the first {) and ,g.length-1) means to take characters until (but not including) the character at the string length minus one. This works because the position is zero-based, i.e. g.length-1 is the last position.

For readers other than the original poster: If it has to be a regex, use /{([^}]*)}/ if you want to allow empty strings, or /{([^}]+)}/ if you want to only match when there is at least one character between the curly braces. Breakdown:

• /: start the regex pattern
  • {: a literal curly brace
    • (: start capturing
      • [: start defining a class of characters to capture
        • ^}: "anything other than }"
      • ]: OK, that's our whole class definition
      • *: any number of characters matching that class we just defined
    • ): done capturing
  • }: a literal curly brace must immediately follow what we captured
• /: end the regex pattern

Try this:

/[^{\}]+(?=})/g

For example

Welcome to RegExr v2.1 by #{gskinner.com},  #{ssd.sd} hosted by Media Temple!

will return gskinner.com, ssd.sd.