How to make Regular expression into non-greedy?

The non-greedy regex modifiers are like their greedy counter-parts but with a ? immediately following them:

*  - zero or more
*? - zero or more (non-greedy)
+  - one or more
+? - one or more (non-greedy)
?  - zero or one
?? - zero or one (non-greedy)

I believe it would be like this

takedata.match(/(\[.+\])/g);

the g at the end means global, so it doesn't stop at the first match.

You are right that greediness is an issue:

--A--Z--A--Z--
  ^^^^^^^^^^
     A.*Z

If you want to match both A--Z, you'd have to use A.*?Z (the ? makes the * "reluctant", or lazy).

There are sometimes better ways to do this, though, e.g.

A[^Z]*+Z

This uses negated character class and possessive quantifier, to reduce backtracking, and is likely to be more efficient.

In your case, the regex would be:

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

Unfortunately Javascript regex doesn't support possessive quantifier, so you'd just have to do with:

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

See also

• regular-expressions.info/Repetition
  • See: An Alternative to Laziness
    • Possessive quantifiers
  • Flavors comparison

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Quick summary

*   Zero or more, greedy
*?  Zero or more, reluctant
*+  Zero or more, possessive

+   One or more, greedy
+?  One or more, reluctant
++  One or more, possessive

?   Zero or one, greedy
??  Zero or one, reluctant
?+  Zero or one, possessive

Note that the reluctant and possessive quantifiers are also applicable to the finite repetition {n,m} constructs.

Examples in Java:

System.out.println("aAoZbAoZc".replaceAll("A.*Z", "!"));  // prints "a!c"
System.out.println("aAoZbAoZc".replaceAll("A.*?Z", "!")); // prints "a!b!c"

System.out.println("xxxxxx".replaceAll("x{3,5}", "Y"));  // prints "Yx"
System.out.println("xxxxxx".replaceAll("x{3,5}?", "Y")); // prints "YY"