mysql 匹配json 数组字段中值出现的次数
精确匹配
SELECT COUNT(id) FROM Fund f WHERE JSON_EXTRACT(f.manager,’$[0]’) = ‘段鹏’
select count(1) from Fund where JSON_CONTAINS(manager->’ [ ∗ ] ′ , ′ " 段 鹏 " ′ , ′ [*]', '"段鹏"', ' [∗]′,′"段鹏"′,′’)
模糊匹配
SELECT count(1) FROM Fund WHERE JSON_EXTRACT(manager,’$[*]’) LIKE “%段鹏%”;
[1]: http://meta./questions/5020/mathjax-basic-tutorial-and-quick-reference[2]: https://mermaidjs.github.io/[3]: https://mermaidjs.github.io/[4]: http://adrai.github.io/flowchart.js/
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