从文本中删除 bb_quotes ([quote..] [/quote]) 的 Javascript

Question

目标是使用 Javascript (vanilla) 删除介于 [quote][/quote] 和 [quote=something][/quote] 之间的所有文本(包括)(不区分大小写).最好也删除双空格，如果它们在删除引号后存在。我按如下方式尝试了这个 Javascript，即:

t.replace(/\[quote.*\](.*?)\[\/quote\]/gi,'')

，但我没有得到正确的结果。正确的做法是什么？

var t='Starting [QUOTE]this should be ignored hello[/quote] it. This is not quote and [quote=frank]HELLO quotes[/quote] Marky Mark 84WD. Last [quote=irene]try.[/quote]';

console.log(t.replace(/\[quote.*\](.*?)\[\/quote\]/gi,''));

//Current result: Starting.
//Expected result: Starting it. This is not quote and Marky Mark 84WD. Last

var t='[Quote]this should be ignored hello[/quote]. This is not quote and [quote=frank]HELLO quote[/quote] Marky Mark 84WD.';

console.log(t.replace(/\[quote.*\](.*?)\[\/quote\]/gi,''));

//Current result: Marky Mark 84WD.
//Expected result: . This is not quote and Marky Mark 84WD.

Answer 1

可以使用/\[(quote)[^\]]*](.*?)\[\/\1\]/gi来实现过滤:

var t='Starting [QUOTE]this should be ignored hello[/quote] it. This is not quote and [quote=frank]HELLO quotes[/quote] Marky Mark 84WD. Last [quote=irene]try.[/quote]';

console.log(t.replace(/\[(quote)[^\]]*](.*?)\[\/\1\]/gi,''));

var t='[Quote]this should be ignored hello[/quote]. This is not quote and [quote=frank]HELLO quote[/quote] Marky Mark 84WD.';

console.log(t.replace(/\[(quote)[^\]]*](.*?)\[\/\1\]/gi,''));