Question

+-----+---------+--------+---------+
| 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 |
+-----+---------+--------+---------+

我有上面的示例表，我正在使用CI，我想选择总额'列数量WHERE列余额有值dr'减去'列数WHERE列余额有值cr'的总量。现在我如何将其写入一个查询？

我目前正在做的是使用2个查询，如下面的

// 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;

我认为必须有一种方法可以在不查询两次的情况下获得$ total，但我找不到SO中的任何线索。

Answer 1

SELECT
   account,
   SUM(CASE WHEN balance = 'dr' THEN amount
      WHEN balance = 'cr' THEN -amount
      ELSE 0
      END
   ) amount
FROM
   table
GROUP BY
   account

Answer 2

您可以使用SQL查询执行此操作：

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')

将上面的查询放入CI query（）方法并从结果数组中获取列：“sum_dr”和“sum_cr”。

Answer 3

这样的事情可以解决问题：

SELECT SUM(IF(balance = 'dr', amount, (-1) * amount))
FROM table
WHERE account = 1
AND balance IN ('dr', 'cr')

MySql如何在1个查询中选择2个不同where条件的sum列

3 个答案: