MySQL SELECT列COUNT(name)在哪里条件GROUP BY列?
作者:互联网
嘿,这是我正在使用的SQL语句
$sql = "SELECT number, COUNT(name) FROM people WHERE id='$id' GROUP BY number";
$result = mysql_query($sql);
while($row = mysql_fetch_array($result)){
$statarray['stat'.$i] = $row['COUNT(name)'];
$i++; }
数字列针对每个名称存储一个0、1、2或3,因此该数组返回总计每个数字对应的名称数.
唯一的问题是,如果所有名称都存储有一个数字1,则我的数组条目’stat0’不会告诉我有多少个0,因为它被1的数量代替了(这是因为在其中没有0.数组,因此第一个输出是1s).
我如何找出多少个数字列中包含0s 1s 2s和3s的名称,包括什么时候都没有?
解决方法:
SELECT numbers.number ,
COUNT(name)
FROM ( SELECT 0 AS number
UNION ALL
SELECT 1
UNION ALL
SELECT 2
UNION ALL
SELECT 3
) numbers
LEFT JOIN people ON people.number = numbers.number AND id='$id'
GROUP BY numbers.number
或者,如果您有根据注释的永久数字表
SELECT numbers.number ,
COUNT(name)
FROM numbers
LEFT JOIN people ON people.number = numbers.number
AND id = '$id'
WHERE numbers.number BETWEEN 0 AND 3
GROUP BY numbers.number
标签:count,mysql,group-by 来源: https://codeday.me/bug/20191208/2088172.html