我有这句话:
SELECT COUNT(ID) AS numberVote, AWARD_ID, NOmineE_ID, VoteR_ID, MULTI_CODE
FROM b_awards_Vote
WHERE AWARD_ID = 8
GROUP BY MULTI_CODE
我的桌子看起来像这样:
ID | AWARD_ID | NOmineE_ID | VoteR_ID | MUTLI_CODE
1 | 8 | 3 | 1 | 4837
2 | 8 | 4 | 1 | 4837
3 | 8 | 5 | 1 | 4837
但是在PHP中返回:
$numberVote = $row['numberVote'];
该语句返回3.这是我想要的详细信息:
>所有具有相同MULTI_CODE的条目实际上都是1票.
>因此,上表中的这个示例我想算作1票.在下表中:
ID | AWARD_ID | NOmineE_ID | VoteR_ID | MUTLI_CODE
1 | 8 | 3 | 1 | 4837
2 | 8 | 4 | 1 | 4837
3 | 8 | 5 | 1 | 4837
4 | 8 | 4 | 3 | 7480
5 | 8 | 5 | 3 | 7480
>我希望此计数返回:2,因为有两个不同的MULTI_CODE,因为有2个不同的投票组.
解决方法:
SELECT COUNT( distinct MULTI_CODE) AS NumOfGroups ,ID AS numberVote, AWARD_ID, NOmineE_ID, VoteR_ID, MULTI_CODE
FROM b_awards_Vote
WHERE AWARD_ID = 8
参见SQLFiddle作为参考