我正在嘗試創建一個函式，它將在兩個引數（rewrite和redirect）之間獲取一個字串。我無法繞過它。
我有一個看起來像這樣的字串：
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;
我想要以下輸出：
/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
獲取兩個引數之間的所有字串的正確正則運算式或方法是什么？
uj5u.com熱心網友回復：
嘗試：
\brewrite \^(.*?)\s redirect;
查看在線演示
\brewrite \^- 在左側的單詞邊界和右側的文字“^”之間的字面上“重寫”；(.*?)- 匹配（惰性）0 個字符；\s redirect;- 字面意思是“重定向”左側的 1 個空白字符和右側的分號。
查看將列印的在線 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
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