如何使用正则表达式查看带有引号内文本的字符串，但忽略括号内的任何内容？

Question

例如我的字符串是：

var str = 'Hello "Counts1 [ignore1] Counts2 [ignore2] Counts3 [ignore3] Count these too"';

如何在忽略括号内的字符的情况下获取引号内字符串中的所有内容？

例如要收集的正则表达式：“Counts1”、“Counts2”、“Counts3”、“Count these too”

到目前为止我只得到：

var regex = /".*?"/g;

它输出：

"Counts1 [ignore1] Counts2 [ignore2] Counts3 [ignore3]"

Answer 1

应该这样做：/(?<="|\] )[\w ]+(?="| \[)/g

正后视 (?<="|\] ) 确保 [\w ]+ 左侧有 " 或 ] 。
正面前瞻 (?="| \[) 确保 [\w ]+ 在右边有 " 或 [。

Demo

Answer 2

您还可以使用以下同样支持嵌套方括号的非正则表达式方法：

const extract = (str) => {
    let result = [], start = 0, i=0, level = 0, inquote=false;
    for (let i = 0; i < str.length; ++i) {
        switch (str[i]) {
            case '"':
                if (inquote && !level) {
                    if (start < i) {
                        result.push(str.substr(start, i - start));
                    }
                    start = i + 1;
                } else {
                    start = i+1;
                }
                inquote = !inquote;
                break;
            case '[':
                if (inquote) {
                  if (start < i && inquote && !level) {
                    result.push(str.substr(start, i - start));
                  }
                  ++level;
                }
                break;
            case ']':
                if (inquote) {
                    if (level > 0) {  
                        --level;
                    }
                    if (level == 0) {
                        start = i + 1;
                    }
                }
                break;
        }
    }
 
    if (start < i)
        result.push(str.substr(start, i - start));
   
    return result;
}
console.log(extract('Hello "Counts1 [ignore1] Counts2 [ignore2] Counts3 [ignore3] Count these too"'));
console.log(extract('Hello "Counts1 [[ignore1] this [2]] Counts2 [ignore2] Counts3 [ignore3] Count these too"'));
console.log(extract('"Counts1 [[ignore1] [2]] Counts2 [ignore2] Count these too" and [junk], and "maybe this, too [not this]"'));

如何使用正则表达式查看带有引号内文本的字符串，但忽略括号内的任何内容？

How do I look in a string with text inside quotations, but ignoring anything inside brackets using regex?

javascript

regex

square-bracket

quotation-marks