在两个参数之间查找多个字符串

Question

我正在尝试创建一个函数，它将在两个参数（重写和重定向）之间获取一个字符串。我无法理解它。我有一个看起来像这样的字符串：add_header X-Robots-Tag "noindex, follow" always; rewrite ^/en/about/Administration/index.aspx /en/about/more-about/administration redirect; rewrite ^/en/about/Administration/supervisory-board/index.aspx /nl/over/meer-over/administration redirect; rewrite ^/en/about/Departments-sections-and-fields/index.aspx /en/about/more-about/department-divisions-and-fields redirect; rewrite ^/en/about/For-companies/index.aspx /en/about/more-about/for-companies redirect; rewrite ^/en/about/contact-information/index.aspx /en/about/more-about/contact-information redirect; rewrite ^/en/about/index.aspx /nl/over redirect;我想要以下输出：/en/about/Administration/index.aspx /en/about/more-about/administration /en/about/Administration/supervisory-board/index.aspx /nl/over/meer-over/administration /en/about/Departments-sections-and-fields/index.aspx /en/about/more-about/department-divisions-and-fields  /en/about/For-companies/index.aspx /en/about/more-about/for-companies/en/about/contact-information/index.aspx /en/about/more-about/contact-information/en/about/index.aspx /nl/over获取两个参数之间的所有字符串的正确正则表达式或方法是什么？

Answer 1

尝试：

\brewrite \^(.*?)\s+redirect;

查看在线演示

---

• \brewrite \^- 在左侧的单词边界和右侧的空格后跟文字“^”之间字面地“重写”；

• (.*?)- 匹配（惰性）0+ 个字符；

• \s+redirect;- 在左侧的 1+ 个空白字符和右侧的分号之间“重定向”。

---

查看将打印的在线 GO演示：

/en/about/Administration/index.aspx /en/about/more-about/administration

/en/about/Administration/supervisory-board/index.aspx /nl/over/meer-over/administration

/en/about/Departments-sections-and-fields/index.aspx /en/about/more-about/department-divisions-and-fields

/en/about/For-companies/index.aspx /en/about/more-about/for-companies

/en/about/contact-information/index.aspx /en/about/more-about/contact-information

/en/about/index.aspx /nl/over