問題一：1、$array數組如下：
array (size=7)
1 => string '2018-09-21' (length=10)
2 => string '2018-09-22' (length=10)
3 => string '2018-09-23' (length=10)
4 => string '2018-09-24' (length=10)
5 => string '2018-09-25' (length=10)
6 => string '2018-09-26' (length=10)
7 => string '2018-09-27' (length=10)
2、$list數組如下：
array (size=3)
0 =>
array (size=2)
'riqi' => string '2018-09-21' (length=10)
'count' => int 2
1 =>
array (size=2)
'riqi' => string '2018-09-26' (length=10)
'count' => int 2
2 =>
array (size=2)
'riqi' => string '2018-09-27' (length=10)
'count' => int 3
求助：如何php實現以下效果
array (size=3)
0 =>
array (size=2)
'riqi' => string '2018-09-21' (length=10)
'count' => int 2
1 =>
array (size=2)
'riqi' => string '2018-09-22' (length=10)
'count' => int 0
2 =>
array (size=2)
'riqi' => string '2018-09-23' (length=10)
'count' => int 0
3 =>
array (size=2)
'riqi' => string '2018-09-24' (length=10)
'count' => int 0
4 =>
array (size=2)
'riqi' => string '2018-09-25' (length=10)
'count' => int 0
5 =>
array (size=2)
'riqi' => string '2018-09-26' (length=10)
'count' => int 2
6 =>
array (size=2)
'riqi' => string '2018-09-27' (length=10)
'count' => int 3
問題二：現有sql如下：
select FROM_UNIXTIME(addtime,'%Y-%m-%d') as riqi, count(1) as count from order where FROM_UNIXTIME(addtime,'%Y-%m-%d') >= now() - interval 7 day group by FROM_UNIXTIME(addtime,'%Y-%m-%d')
如何能通過一句sql實現以下效果
array (size=3)
0 =>
array (size=2)
'riqi' => string '2018-09-21' (length=10)
'count' => int 2
1 =>
array (size=2)
'riqi' => string '2018-09-22' (length=10)
'count' => int 0
2 =>
array (size=2)
'riqi' => string '2018-09-23' (length=10)
'count' => int 0
3 =>
array (size=2)
'riqi' => string '2018-09-24' (length=10)
'count' => int 0
4 =>
array (size=2)
'riqi' => string '2018-09-25' (length=10)
'count' => int 0
5 =>
array (size=2)
'riqi' => string '2018-09-26' (length=10)
'count' => int 2
6 =>
array (size=2)
'riqi' => string '2018-09-27' (length=10)
'count' => int 3
4 回答
鳳凰求蠱
TA貢獻1825條經驗 獲得超4個贊
SELECT FROM_UNIXTIME(addtime,'%Y-%m-%d') as riqi,count(*) from order GROUP BY riqi ORDER BY riqi LIMIT 7
應該是你要的結果
12345678_0001
TA貢獻1802條經驗 獲得超5個贊
用to_days(now()) - to_days(addtime)<= 7來查取近7天的數據，然后用date方法將時間只保留到天數。再對查到的數據，根據時間進行分組，再count(訂單)，得到就是每天對應的訂單數了。
如果要把訂單數為0 的天數也展示出來，就寫個日期的循環，判斷每天的日期是否在查詢結果里有對應的記錄，有就直接取，沒有就是0.
效果圖：
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