java - 如何删除不平衡/不成对的双引号(在 Java 中)

Question

想到这里把这个比较聪明的问题分享给大家。我正在尝试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，我还没有找到可行的解决方案。 我无法从字符串中删除未配对/未配对的双引号。

示例输入

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

输出应该是

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

这个问题听起来类似于 Using Java remove unbalanced/unpartnered parenthesis

到目前为止，这是我的代码(它不会删除所有未配对的双引号)

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，它可能可以在单个正则表达式中完成。
粗略定义了一个分隔符的概念，可以‘偏向’
这些规则以获得更好的结果。
这完全取决于规定的规则。此正则表达式考虑了
三种可能的情况；

1. 有效对
2. 无效对(有偏差)
3. 无效单

它也不会解析超出行尾的“”。但它确实做了很多
行合并为一个字符串。要更改它，请在您看到它的地方删除 \n。

---

全局上下文 - 原始查找正则表达式
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

扩展的原始正则表达式:

(?:                            // 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 测试用例(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"
'