如何删除不平衡/未共享的双引号（在Java中）

Question

我想与大家分享这个相对聪明的问题。 我试图从字符串中删除不平衡/不成对的双引号。

我的工作正在进行中，我可能接近解决方案。但是，我还没有得到一个有效的解决方案。 我无法从字符串中删除未配对/未提交的双引号。

示例输入

string1=injunct! alter ego."
string2=successor "alter ego" single employer"  "proceeding "citation assets"

输出

string1=injunct! alter ego.
string2=successor "alter ego" single employer  proceeding "citation assets"

这个问题听起来很像 Using Java remove unbalanced/unpartnered parenthesis

到目前为止，这是我的代码（它不会删除所有无用的双引号）

private String removeUnattachedDoubleQuotes(String stringWithDoubleQuotes) {
    String firstPass = "";

    String openingQuotePattern = "\\\"[a-z0-9\\p{Punct}]";
    String closingQuotePattern = "[a-z0-9\\p{Punct}]\\\"";

    int doubleQuoteLevel = 0;
    for (int i = 0; i < stringWithDoubleQuotes.length() - 3; i++) {
        String c = stringWithDoubleQuotes.substring(i, i + 2);
        if (c.matches(openingQuotePattern)) {
            doubleQuoteLevel++;
            firstPass += c;
        }
        else if (c.matches(closingQuotePattern)) {
            if (doubleQuoteLevel > 0) {
                doubleQuoteLevel--;
                firstPass += c;
            }
        }
        else {
            firstPass += c;
        }
    }

    String secondPass = "";
    doubleQuoteLevel = 0;
    for (int i = firstPass.length() - 1; i >= 0; i--) {
        String c = stringWithDoubleQuotes.substring(i, i + 2);
        if (c.matches(closingQuotePattern)) {
            doubleQuoteLevel++;
            secondPass = c + secondPass;
        }
        else if (c.matches(openingQuotePattern)) {
            if (doubleQuoteLevel > 0) {
                doubleQuoteLevel--;
                secondPass = c + secondPass;
            }
        }
        else {
            secondPass = c + secondPass;
        }
    }

    String result = secondPass;

    return result;
}

Answer 1

如果没有嵌套，可以在单个正则表达式中完成 有一个大致定义的分界面的概念，并且有可能“偏向” 这些规则可以获得更好的结果 这一切都取决于规定的规则。这个正则表达式考虑到了 三种可能的情况按顺序排列;

1. 有效对
2. 无效配对（有偏见）
3. 无效单

它也不会在行尾之外解析“”。但它确实做了多个 行组合为单个字符串。要更改它，请删除您看到的\n。

---

全球背景 - 原始查找正则表达式
的缩短

(?:("[a-zA-Z0-9\p{Punct}][^"\n]*(?<=[a-zA-Z0-9\p{Punct}])")|(?<![a-zA-Z0-9\p{Punct}])"([^"\n]*)"(?![a-zA-Z0-9\p{Punct}])|")

替换分组

$1$2 or \1\2

扩展原始正则表达式：

(?:                            // Grouping
                                  // Try to line up a valid pair
   (                                 // Capt grp (1) start 
     "                               // "
      [a-zA-Z0-9\p{Punct}]              // 1 of [a-zA-Z0-9\p{Punct}]
      [^"\n]*                           // 0 or more non- [^"\n] characters
      (?<=[a-zA-Z0-9\p{Punct}])         // 1 of [a-zA-Z0-9\p{Punct}] behind us
     "                               // "
   )                                 // End capt grp (1)

  |                               // OR, try to line up an invalid pair
       (?<![a-zA-Z0-9\p{Punct}])     // Bias, not 1 of [a-zA-Z0-9\p{Punct}] behind us
     "                               // "
   (  [^"\n]*  )                        // Capt grp (2) - 0 or more non- [^"\n] characters
     "                               // "
       (?![a-zA-Z0-9\p{Punct}])      // Bias, not 1 of [a-zA-Z0-9\p{Punct}] ahead of us

  |                               // OR, this single " is considered invalid
     "                               // "
)                               // End Grouping

Perl testcase（没有Java）

$str = '
string1=injunct! alter ego."
string2=successor "alter ego" single employer "a" free" proceeding "citation assets"
';

print "\n'$str'\n";

$str =~ s
/
  (?:
     (
       "[a-zA-Z0-9\p{Punct}]
        [^"\n]*
        (?<=[a-zA-Z0-9\p{Punct}])
       "
     )
   |
       (?<![a-zA-Z0-9\p{Punct}])
       " 
     (  [^"\n]*  )
       " (?![a-zA-Z0-9\p{Punct}])
   |
       "
  )
/$1$2/xg;

print "\n'$str'\n";

输出

'
string1=injunct! alter ego."
string2=successor "alter ego" single employer "a" free" proceeding "citation assets"
'

'
string1=injunct! alter ego.
string2=successor "alter ego" single employer "a" free proceeding "citation assets"
'

Answer 2

您可以使用类似（Perl表示法）的内容：

s/("(?=\S)[^"]*(?<=\S)")|"/$1/g;

在Java中将是：

str.replaceAll("(\"(?=\\S)[^\"]*(?<=\\S)\")|\"", "$1");