Php MySql如何在1个查询中选择具有2个不同where条件的sum列
我有上面的示例表,我正在使用CI,我想选择“列余额有值dr的列金额”的总额减去“列余额有值cr的列金额”的总额。现在我如何将其写入一个查询 我目前正在做的是使用如下2个查询Php MySql如何在1个查询中选择具有2个不同where条件的sum列,php,mysql,codeigniter,Php,Mysql,Codeigniter,我有上面的示例表,我正在使用CI,我想选择“列余额有值dr的列金额”的总额减去“列余额有值cr的列金额”的总额。现在我如何将其写入一个查询 我目前正在做的是使用如下2个查询 +-----+---------+--------+---------+ | PID | account | amount | balance | +-----+---------+--------+---------+ | 1 | 1 | 100 | dr | | 2 | 5
+-----+---------+--------+---------+
| PID | account | amount | balance |
+-----+---------+--------+---------+
| 1 | 1 | 100 | dr |
| 2 | 5 | 100 | cr |
| 3 | 2 | 30 | dr |
| 4 | 1 | 30 | cr |
| 5 | 1 | 50 | cr |
| 6 | 4 | 50 | dr |
+-----+---------+--------+---------+
我想一定有办法不必查询两次就可以得到$total,但我找不到任何线索。像这样的方法应该可以做到:
// Get total amount that has balance dr of account 1
$this->db->select_sum('amount');
$query = $this->db->get_where('table', array('account' => '1', 'balance' => 'dr');
$result = $query->result();
$total_dr = $result[0] -> amount;
// Get total amount that has balance cr of account 1
$this->db->select_sum('amount');
$query = $this->db->get_where('table', array('account' => '1', 'balance' => 'cr');
$result = $query->result();
$total_cr = $result[0] -> amount;
// Minus total_dr to total_cr
$total = $total_dr - $total_cr;
您可以使用SQL查询执行此操作:
SELECT SUM(IF(balance = 'dr', amount, (-1) * amount))
FROM table
WHERE account = 1
AND balance IN ('dr', 'cr')
将上面的查询放到CI query()方法中,并从结果数组中获取列:“sum\u dr”和“sum\u cr”。您真是太棒了。非常感谢您的简单高效的查询。非常感谢您的回答。非常感谢您的回答。:)
SELECT
SUM(CASE WHEN balance = 'dr' THEN amount ELSE NULL END) AS sum_dr,
SUM(CASE WHEN balance = 'cr' THEN amount ELSE NULL END) AS sum_cr
FROM table
WHERE account = 1
AND balance IN ('dr', 'cr')
SELECT
account,
SUM(CASE WHEN balance = 'dr' THEN amount
WHEN balance = 'cr' THEN -amount
ELSE 0
END
) amount
FROM
table
GROUP BY
account