Question

在一个表中，有一个params行，它有一些JSON编码数据 - 如下：

{"categories":211,"singleCatOrdering":"","menu-anchor_title":"","menu-anchor_css":"","menu_image":"","menu_text":1,"page_title":"","show_page_heading":0,"page_heading":"","pageclass_sfx":"","menu-meta_description":"","menu-meta_keywords":"","robots":"","secure":0}

我需要做的是在[]参数周围包裹categories - 看起来像这样：

{"categories":[211],"singleCatOrdering":"","menu-anchor_title":"","menu-anchor_css":"","menu_image":"","menu_text":1,"page_title":"","show_page_heading":0,"page_heading":"","pageclass_sfx":"","menu-meta_description":"","menu-meta_keywords":"","robots":"","secure":0}

我尝试了以下（和其他一些失败的）并且没有骰子：

UPDATE j17_menu SET params = REPLACE(params,'"categories":%,','"categories":[%],') WHERE component_id = 10021;

我是否可能使用通配符错误？任何朝着正确方向的推动都将是一个巨大的帮助。谢谢！

Answer 1

这个将独立于上下文工作

UPDATE j17_menu 
SET params=CONCAT (
    SUBSTR(params,1,LOCATE('"categories":',params)),
    '"categories":[',
    substr(
        params,
        LOCATE('"categories":',params)+13,
        LOCATE(',',params,LOCATE('"categories":',params))-LOCATE('"categories":',params)-13
    ),
    ']',
    substr(params,LOCATE(',',params,LOCATE('"categories":',params)))
)
WHERE component_id = 10021;

MySQL替换语法 - 包装特定的文本

1 个答案: