Question

使用此查询获取json结果，如下面的数据结构，

SELECT c.* ,
    row_to_json(mg.*) as media_gallery, 
    row_to_json(m.*) as media,
    row_to_json(g.*) as gallery
  FROM comment c

  LEFT JOIN media m ON m.id = c.endpoint_id
  .... join mg g table

是否可以根据列值选择选择表可能就像这样的语法，如果为null则排除结果中的media_gallery，media和gallery

SELECT c.* ,
  CASE WHEN c.endpoint_code = 1
  THEN
    row_to_json(mg.*) as media_gallery, 
    row_to_json(m.*) as media,
  END

  CASE WHEN c.endpoint_code = 0 THEN
    row_to_json(g.*) as gallery
  END

  FROM comment c

  LEFT JOIN media m ON m.id = c.endpoint_id
  .... join mg g table

结果

  rows: 
[ { id: 7,
  endpoint_code: 0,
  endpoint_id: 27,
  status: 0,

  media_gallery: null,
  media: null,
  gallery: 
   { id: 27,
     status: 0,
     create_date: 
   ...

更新

我也尝试了这个，但没有工作

COALESCE( row_to_json(mg.*), null) AS media_gallery,

Answer 1

这是你想要的吗？

SELECT c.* ,
       (CASE WHEN c.endpoint_code = 1 THEN row_to_json(mg.*) END) as media_gallery,
       (CASE WHEN c.endpoint_code = 1 THEN row_to_json(m.*) END) as media,
       (CASE WHEN c.endpoint_code = 0 THEN row_to_json(g.*) END) as gallery
.  . .

if / else列值选择select table，在json中排除null

1 个答案: