我正在嘗試創建一個函數，它將在兩個參數（重寫和重定向）之間獲取一個字符串。我無法理解它。我有一個看起來像這樣的字符串：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獲取兩個參數之間的所有字符串的正確正則表達式或方法是什么？
1 回答
胡子哥哥
TA貢獻1825條經驗 獲得超6個贊
嘗試：
\brewrite \^(.*?)\s+redirect;
查看在線演示
\brewrite \^- 在左側的單詞邊界和右側的空格后跟文字“^”之間字面地“重寫”；(.*?)- 匹配（惰性）0+ 個字符；\s+redirect;- 在左側的 1+ 個空白字符和右側的分號之間“重定向”。
查看將打印的在線 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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