Regex match multiple same expression multiple times

Question

I have got this string {bgRed Please run a task, {red a list has been provided below} , I need to do a string replace to remove the braces and also the first word.

So below I would want to remove {bgRed and {red and then the trailing brace which I can do separate.

I have managed to create this regex, but it is only matching {bgRed and not {red , can someone lend a hand?

/^\\{.+?(?=\\s)/gm

Answer 1

Note you are using ^ anchor at the start and that makes your pattern only match at the start of a line (mind also the m modifier). .+?(?=\\s|$) is too cumbersome, you want to match any 1+ chars up to the first whitespace or end of string, use {\\S+ (or {\\S* if you plan to match { without any non-whitespace chars after it).

You may use

s = s.replace(/{\S*|}/g, '')

You may trim the outcome to get rid of resulting leading/trailing spaces:

s = s.replace(/{\S*|}/g, '').trim()

See the regex demo and the regex graph :

Details

• {\\S* - { char followed with 0 or more non-whitespace characters
• | - or
• } - a } char.

Answer 2

If the goal is go to from

"{bgRed Please run a task, {red a list has been provided below}"

to

"Please run a task, a list has been provided below"

a regex with two capture groups seems simplest:

 const original = "{bgRed Please run a task, {red a list has been provided below}"; const rex = /\\{\\w+ ([^{]+)\\{\\w+ ([^}]+)}/g; const result = original.replace(rex, "$1$2"); console.log(result); 

\\{\\w+ ([^{]+)\\{\\w+ ([^}]+)} is:

• \\{ - a literal {
• \\w+ - one or more word characters ("bgRed")
• a literal space
• ([^{]+) one or more characters that aren't { , captured to group 1
• \\{ - another literal {
• \\w+ - one or more word characters ("red")
• ([^}]+) - one or more characters that aren't } , captured to group 2
• } - a literal }

The replacement uses $1 and $2 to swap in the capture group contents.