获取与单词之一匹配的引号之间的所有文本

Question

a:5:{s:12:"SubmissionID";s:0:"";s:13:"NotesClientID";s:4:"1891";s:5:"Field";a:5:
{i:0;s:7:"8/19/19";i:1;s:0:"";i:4;s:0:"";i:5;s:160:"client is dismissed due to no contact with me or with his assigned Recovery
Coach";i:7;s:0:"";}s:10:"SecurityID";s:40:"a31b7ea7f9191465525d0ac9a6358ba7b3d3e0fc";s:13:"action_doSave";s:4:"Save";}

在上面的字符串中，预期的输出是“客户因没有与我或他指定的康复教练联系而被解雇”

这包含 dismissed 字。

同样，正则表达式应该选择字符串中包含被解雇/出院/服从之一的字符串。

我已经在https://rubular.com/r/QMFAHjUgaYdjpM尝试过这个

Answer 1

您可以使用以下正则表达式匹配所需的文本。

/.*?"(?=(?:[^"]*\bdismissed\b[^"]*"))\K[^"]*(?=")/m

Start your engine!

regex101¹ 提供了有关正则表达式每个元素的详细信息。

如果变量str 包含问题中作为示例给出的文本，则可以这样写：

str.scan(/.*?"(?=(?:[^"]*\bdismissed\b[^"]*"))\K[^"]*(?=")/m)
  #=> ["client is dismissed due to no contact with me or with his assigned Recovery\nCoach"]

[String#scan] 返回一个包含正则表达式的每个匹配项的数组。这里有一个匹配项。

Ruby 的正则表达式引擎执行以下操作。

/
.*?                : match 0+ characters, lazily
"                  : match a double-quote
(?=                : begin a positive lookahead to assert that
                     'dismissed' appears before the next double-quote
  (?:              : begin a non-capture group
    [^"]*          : match 0+ characters other than double-quotes
    \bdismissed\b  : match 'dismissed'
    [^"]*          : match 0+ characters other than double-quotes
    "              : match a double-quote
  )                : end non-capture group
)                  : end positive lookahead
\K                 : forget all matched so far and reset match
[^"]*              : match 0+ characters other than double-quotes
(?=")              : assert the next character is a double-quote
/m                 : multiline mode to match line terminators

---

附录：我错过了问题中提到的目标词可能是“dismissed”、“discharged”或“amenable”中的任何一个。与其修改我的答案，不如说在哪里

\bdismissed\b

出现在应该替换为的正则表达式中

\b(?:dismissed|discharged|amenable)\b

^{1.移动光标查看详细说明。}