我的情况是这样的:
情况是这样的:
public function get_opportunity($customer_id, $group_id, $action)
{
$sql = "SELECT COUNT(transaction_id) AS total_transaction
FROM `transaction`
WHERE
CASE
WHEN '$action' = 'view_all' THEN transaction_id IS NOT NULL
WHEN '$action' = 'view_group' THEN group_id IN($group_id)
ELSE transaction_created_by = ?
END
";
$result = $this->db->query($sql, array($customer_id))->result_array();
return ($result[0]['total_transaction']) ? $result[0]['total_transaction'] : 0;
}
如果其他是这样的:
public function get_opportunity($customer_id, $group_id, $action)
{
if($action == "view_all")
$condition = " transaction_id IS NOT NULL";
else if($action == 'view_group')
$condition = " group_id IN($group_id)";
else
$condition = " customer_id = ?";
$sql = "SELECT COUNT(transaction_id) AS total_transaction
FROM `transaction`
WHERE $condition";
$result = $this->db->query($sql, array($customer_id))->result_array();
return ($result[0]['total_transaction']) ? $result[0]['total_transaction'] : 0;
}
我已经尝试过了。上述工作的所有类型。但这里我想问一下,哪一个更好?
最佳答案
这是不对的,第一个代码不应该工作!在 SQL 中,CASE
是一个表达式,而不是一个语句。您不能在 THEN
部分添加条件,只能添加单个值。 MySQL 将 bool 表达式计算为 1\0 ,所以你的情况基本上是这样的:
WHERE CASE... THEN (Cond) -- equals to 1/0
THEN (Cond) -- equals to 1/0
ELSE (Cond) -- -- equals to 1/0
大多数 RDBMS 都会抛出错误。
您可以像这样格式化它:
SELECT COUNT(transaction_id) AS total_transaction
FROM `transaction`
WHERE ('$action' = 'view_all' AND transaction_id IS NOT NULL) OR
('$action' = 'view_group' THEN group_id IN($group_id)) OR
transaction_created_by = ?
只要没有性能问题,就不要考虑这个事情, 它只会浪费你的时间。选择一个更容易维护的。
关于php - 哪一个更好?何时或如果其他情况?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38582874/