有查询:" SELECT日期,COUNT(id)FROM引导WHERE userkey =' $ userkey' GROUP BY DAY(日期)"
我想在is_lead 1
中添加第二个示例:
[FIRST]
SELECT date, COUNT(id) FROM leads WHERE userkey = '$userkey' GROUP BY DAY(date)
[SECOND]
SELECT COUNT(price) from leads WHERE userkey = '$userkey' and is_lead = 1
联盟不起作用。
请帮助!
答案 0 :(得分:2)
或者你想这样做?
SELECT a.userkey,DATE, COUNT(id),b.countp
FROM leads a
LEFT JOIN (SELECT COUNT(price) AS countp
FROM leads
WHERE userkey = '$userkey' AND is_lead = 1) b
ON a.userkey=b.userkey
WHERE userkey = '$userkey'
GROUP BY DAY(DATE)