Javascript regular expressions modifier U [duplicate]

Question

I have a string

"{1} {2} {4} {abc} {abs}"

and the regular expression

/\{(.+)\}/g

In PHP regular expressions i can use modifier "U" (U modifier: Ungreedy. The match becomes lazy by default). If i use /\{(.+)\}/gU my response looks like this:

Array (5) [ "1", "2", "4", "abc", "abs" ]

In javascript no modifier U. Without this modifier my response looks like this:

Array (1) [ "1 2 4 abc abs" ]

How i can do this?

Menderes

.+?... There's no option to do it by default but you can always do it manually... – Supervision

Is expected result Array (5) [ "1", "2", "4", "abc", "abs" ] ? – Megaspore

@Megaspore Yes ....! – Brittney

Answer 1

One way is to make the + ungreedy by adding the ? modifier:

"{1} {2} {4} {abc} {abs}".match(/\{(.+?)\}/g)

Another way would be to replace . with "anything except closing brace":

"{1} {2} {4} {abc} {abs}".match(/\{([^}]+)\}/g)

Kessinger

Why did you escape { character in your pattern? I think that's useless. regex101.com/r/eW0dF3/1 – Brittney

@Shafizadeh: I copy/pasted from the question. Useless but also harmless. – Kessinger

Answer 2

You can remove all {s and explode the string per }. Something like this:

var str = "{1} {2} {4} {abc} {abs}";
var result = str.replace(/{|}$/g,"").split(/} ?/);
document.write(result);

Brittney

Answer 3

Try RegExp /([a-z]+|\d+)(?=\})/ig to match a-z case insensitive or digit characters followed by }

"{1} {2} {4} {abc} {abs}".match(/([a-z]+|\d+)(?=\})/ig)

console.log("{1} {2} {4} {abc} {abs}".match(/([a-z]+|\d+)(?=\})/ig))

Megaspore

I think your pattern could be simply this /(\w+)(?=})/g. You don't need to escape { character, Also what OP mentioned in his question is just an example. I guess there maybe are some values like these: 1a, ba3. – Brittney