收到错误信息
Unknown column 'code' in 'where clause'
SELECT * FROM `payment` WHERE `code` = 'ORD00023'
尽管我已经使用了join方法来加入支付表和服务表。在这里我找到了前一行值的解决方案，但它没有得到。
这是我的模特
public function order_balance($order_code)
{
$this->db->query("
SELECT p1.*
, p2.balance AS previous_balance
FROM payment p1
JOIN payment p2
ON p1.order_id = p2.order_id + 1
AND p1.customer_id = p2.customer_id
LEFT
JOIN services s
ON p1.customer_id = s.customer_id
ORDER BY p1.id DESC
");
$query = $this->db->get_where('payment p1', array('code' => $order_code));
return $query;
}
这是我的餐桌付款
id order_id customer_id amount actual_amount paid_amount balance type
25 11 16 100.00 50.00 50.00 Cash
26 12 16 200.00 100.00 100.00 Cash
27 13 16 150.00 100.00 50.00 Cash
28 14 16 300.00 250.00 50.00 Cash
29 14 16 170.00 100.00 70.00 Cash
30 15 16 100 170.00 70.00 100.00 Cash
31 16 16 400 500.00 300.00 200.00 Cash
这是桌面服务
id code customer_id particulars
11 ORD00011 16 phone
12 ORD00012 16 gdf
13 ORD00013 16 ghgfh
14 ORD00014 16 tv
15 ORD00015 16 ghfg
16 ORD00016 16 tv
17 ORD00017 16 gdfg
18 ORD00018 16 desk
19 ORD00019 16 gdf
在这里，我加入了付款表和服务表，但仍未获得
见我的表
id order_id customer_id amount actual_amount paid_amount balance type
50 31 16 650 750.00 250.00 500.00 Cash
51 1 16 100 600.00 300.00 300.00 Cash
52 2 16 100 400.00 200.00 200.00 Cash
53 3 16 800 1000.00 600.00 400.00 Cash
54 4 15 400 400.00 300.00 100.00 Cash
55 5 15 500 600.00 575.00 25.00 Cash
56 6 16 350 750.00 600.00 150.00 Cash
在此表中，16的customer_id具有余额为25的上一行值，但我希望最后的customer_id 16余额值为先前的customer_id 16值。
例如，我的结果应该是这样的
id order_id customer_id amount actual_amount paid_amount balance type
56 6 16 350 750.00 600.00 400.00 Cash
答案 0 :(得分：1)
你想要这个吗？
$this->db->query("SELECT p1.*, p2.balance AS previous_balance FROM payment p1 INNER JOIN payment p2 ON p1.order_id = p2.order_id + 1 AND p1.customer_id = p2.customer_id LEFT JOIN services s ON p1.customer_id = s.customer_id AND s.code = ? ORDER BY p1.id DESC", array($order_code));
答案 1 :(得分：0)
因为你正在写两个不同的查询
要查看两者之后添加$this->db->last_query()。您将看到正在执行两个查询。
//1st query
$this->db->query("SELECT p1.*, p2.balance AS previous_balance FROM payment p1
INNER JOIN payment p2 ON p1.order_id = p2.order_id + 1 AND p1.customer_id
= p2.customer_id LEFT JOIN services s ON p1.customer_id = s.customer_id
ORDER BY p1.id DESC");
echo $this->db->last_query();
//2nd query
$query = $this->db->get_where('payment p1', array('code' => $order_code));
echo $this->db->last_query();
彼此独立。
试试这个
$query = $this->db->query("SELECT p1.*, p2.balance AS previous_balance FROM payment p1
INNER JOIN payment p2 ON p1.order_id = p2.order_id + 1 AND p1.customer_id
= p2.customer_id LEFT JOIN services s ON p1.customer_id = s.customer_id
where code='".$order_code."' ORDER BY p1.id DESC");
或
$this->db->select('p1.*, p2.balance AS previous_balance');
$this->db->from('payment p1');
$this->db->join('payment p2','p1.order_id = p2.order_id + 1 AND p1.customer_id
= p2.customer_id','');
$this->db->join('services s','p1.customer_id = s.customer_id','left');
$this->db->where('s.code',$order_code);
$this->db->order_by('p1.id', 'DESC');
$query = $this->db->get();
答案 2 :(得分：0)
这显示错误，因为您缺少表名的别名，其中列代码可用。替换代码
$query = $this->db->get_where('payment p1', array('code' => $order_code));
与
$query = $this->db->get_where('payment p1', array('services.code' => $order_code));
或
$query = $this->db->get_where('payment p1', array('s.code' => $order_code));
答案 3 :(得分：0)
我在模型中所做的这些改变得到了结果
public function order_balance($order_code)
{
$query=$this->db->query("SELECT payment.*, t3.balance AS pre_balance FROM payment
INNER JOIN (SELECT max(id) AS id, customer_id FROM payment GROUP BY customer_id) t ON payment.id = t.id AND payment.customer_id = t.customer_id
LEFT JOIN (SELECT t1.* FROM payment t1
WHERE (
SELECT count(*) FROM payment t2 WHERE t1.customer_id = t2.customer_id AND t1.id <= t2.id
) <= 2
AND t1.id not IN (SELECT max(id) FROM payment GROUP BY customer_id)) t3 ON payment.customer_id = t3.customer_id LEFT JOIN services s ON payment.customer_id = s.customer_id AND s.code = ? ", array($order_code));
return $query;
}