我有一个相当简单的查询,它希望将另一个表中的所有结果合并到一个json对象中。
的MySQL
SELECT s.id, s.uid, s.sexnumber, s.rating, s.sextime, s.diary, GROUP_CONCAT(p.pid,',') as positions, GROUP_CONCAT(w.id,':',w.who) as who, GROUP_CONCAT('{',
'\"lat\":',l.lat,
',\"lon\":',l.lon,
',\"house\":',l.house,
',\"house_number\":',l.house_number,
',\"road\":',l.road,
',\"supermarket\":',l.supermarket,
',\"city\":',l.city,
',\"city_district\":',l.city_district,
',\"country\":',l.country,
',\"country_code\":',l.country_code,
',\"county\":',l.county,
',\"neighbourhood\":',l.neighbourhood,
',\"pedestrian\":',l.pedestrian,
',\"place_of_worship\":',l.place_of_worship,
',\"postcode\":',l.postcode,
',\"state\":',l.state,
',\"suburb\":',l.suburb,
'}'
) as location, GROUP_CONCAT(ww.id,':',ww.name) as place
FROM users u join sex s
on s.uid = u.uid
LEFT OUTER JOIN whos ws
ON s.id = ws.sid
LEFT OUTER JOIN who w
ON w.id = ws.wid
LEFT OUTER JOIN locations l
ON l.sid = s.id
LEFT OUTER JOIN wheresex whs
ON whs.sid = s.id
LEFT OUTER JOIN wherewhere ww
ON whs.wid = ww.id
LEFT OUTER JOIN positions p
ON s.id = p.sid
WHERE u.sessionCheck = '%s'
GROUP BY s.id
ORDER BY s.sextime DESC;
如果location
中的任何GROUP_CONCAT
结果为NULL
,则整个条目将返回NULL
。
如何让NULL
的个别结果以这种方式返回以及其他任何值的结果?
答案 0 :(得分:2)
我必须承认我以前没有尝试过执行这种查询,但是我看不出合并会怎么做呢?
SELECT s.id, s.uid, s.sexnumber, s.rating, s.sextime, s.diary, GROUP_CONCAT(p.pid,',') as positions, GROUP_CONCAT(w.id,':',w.who) as who, GROUP_CONCAT('{',
'\"lat\":',coalesce(l.lat,'Unknown'),
',\"lon\":',coalesce(l.lon,'Unknown'),
',\"house\":',coalesce(l.house,'Unknown'),
',\"house_number\":',coalesce(l.house_number,'Unknown'),
',\"road\":',coalesce(l.road,'Unknown'),
',\"supermarket\":',coalesce(l.supermarket,'Unknown'),
',\"city\":',coalesce(l.city,'Unknown'),
',\"city_district\":',coalesce(l.city_district,'Unknown'),
',\"country\":',coalesce(l.country,'Unknown'),
',\"country_code\":',coalesce(l.country_code,'Unknown'),
',\"county\":',coalesce(l.county,'Unknown'),
',\"neighbourhood\":',coalesce(l.neighbourhood,'Unknown'),
',\"pedestrian\":'coalesce(,l.pedestrian,'Unknown'),
',\"place_of_worship\":',coalesce(l.place_of_worship,'Unknown'),
',\"postcode\":',coalesce(l.postcode,'Unknown'),
',\"state\":',coalesce(l.state,'Unknown'),
',\"suburb\":',coalesce(l.suburb,'Unknown'),
'}'
) as location, GROUP_CONCAT(ww.id,':',ww.name) as place
编辑:我不确定我是否正确理解您的评论,但如果您想在合并返回的值周围添加引号,则可以使用concat function,如下所示:
'\"lat\":',concat('"',coalesce(l.lat,'Unknown'),'"')
答案 1 :(得分:1)
您可能正在寻找Concat
而不是Group_Concat
CONCAT(w.id,':',w.who)
和
相同 CONCAT('{',
'\"lat\":',l.lat,
',\"lon\":',l.lo