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Question

我正在开发一个小型会计应用。以下函数用于获取供应商的交易。我希望在此循环中的每个事务之后获得平衡。有人可以为我提供最好的方法。

public function transaction_by_id_paginated($supplier_id,$start_from,$limit) {
$sql="
SELECT *
  FROM transactions  
 WHERE tr_sup_id = $supplier_id 
 ORDER 
    BY tr_date DESC 
 LIMIT $start_from, $limit
";
$suppliers = DB::fetch($sql);
return $suppliers;          
}

使用此功能，我想得到一个计算的余额值。我不是在DB中存储余额。

transactions表中的字段

tr_id (Primary) int(11) 
tr_sup_id   int(11) 
bill_id int(11)     
tr_date date        
tr_type varchar(250)        
tr_bill int(250)        
tr_payment  int(250)        
tr_payment_note varchar(250)    

我希望余额等于balance = balance + tr_payment - tr_bill

Answer 1

请考虑以下事项：

SELECT * FROM ints;
+---+
| i |
+---+
| 0 |
| 1 |
| 2 |
| 3 |
| 4 |
| 5 |
| 6 |
| 7 |
| 8 |
| 9 |
+---+

SELECT i, @j:=@j+i FROM ints, (SELECT @j:=0) vars ORDER BY i;
+---+----------+
| i | @j:=@j+i |
+---+----------+
| 0 |        0 |
| 1 |        1 |
| 2 |        3 |
| 3 |        6 |
| 4 |       10 |
| 5 |       15 |
| 6 |       21 |
| 7 |       28 |
| 8 |       36 |
| 9 |       45 |
+---+----------+
10 rows in set (0.00 sec)

SELECT i, @j:=@j+i FROM ints, (SELECT @j:=0) vars ORDER BY i LIMIT 3,6;
+---+----------+
| i | @j:=@j+i |
+---+----------+
| 3 |        3 |
| 4 |        7 |
| 5 |       12 |
| 6 |       18 |
| 7 |       25 |
| 8 |       33 |
+---+----------+
6 rows in set (0.00 sec)