如何在php中分组两个相似的字段?
我尝试了GROUP BY DATE(bill.date),bill.agent_id,但它不适用于我
表结构1:http://i.stack.imgur.com/yvBF0.jpg(表名为bill_agents) 表结构2:http://i.stack.imgur.com/38tKh.jpg(表名是账单)
当前结果
+---------+----+----+----+-----+----+
| | 1 | 2 | 3 | 4 | 5 |
+---------+----+----+----+-----+----+
| Agent 1 | 35 | 0 | 0 | 0 | 0 |
| Agent 2 | 0 | 10 | 0 | 0 | 0 |
| Agent 1 | 0 | 0 | 12 | 0 | 0 |
| Agent 3 | 0 | 0 | 0 | 100 | 0 |
| Agent 6 | 0 | 0 | 0 | 9 | 0 |
| Agent 2 | 0 | 0 | 0 | 9 | 14 |
+---------+----+----+----+-----+----+
But I want To get Like The Following
+---------+----+----+----+-----+----+
| | 1 | 2 | 3 | 4 | 5 |
+---------+----+----+----+-----+----+
| Agent 1 | 35 | 0 | 12 | 0 | 0 |
| Agent 2 | 0 | 10 | 0 | 0 | 14 |
| Agent 3 | 0 | 0 | 0 | 100 | 0 |
| Agent 6 | 0 | 0 | 0 | 9 | 0 |
+---------+----+----+----+-----+----+
Php Code粘贴在我正在使用的下方。
<table width="100%" border="1" cellspacing="4" cellpadding="1">
<tr>
<td> </td>
<?php
for ($i=01; $i<=31; $i++)
{?>
<td><?php echo $i; ?></td>
<?php
}
?>
<td>Total</td>
</tr>
<?php
$query4 = "SELECT bill.agent_id, bill.date, SUM(bill.amount + bill.cheque) AS total, bill_agents.id,bill_agents.name ".
"FROM bill, bill_agents ".
"WHERE bill.agent_id = bill_agents.id AND YEAR(date) = YEAR(CURDATE()) AND MONTH(date) = MONTH(CURDATE()) ".
"GROUP BY bill.agent_id , DATE(bill.date) ".
// "GROUP BY bill.agent_id , DATE(bill.date) ".
"ORDER BY bill.date ASC";
$result4 = mysql_query($query4) or die('Error, query failed1');
if (mysql_num_rows($result4)>0){
mysql_data_seek($result4, 0);
?>
<?php $total_1 = 0; while($row4 = mysql_fetch_array($result4, MYSQL_ASSOC)){?>
<?php $date = $row4['date'];
$var = $date;
$date = date("d-m-Y", strtotime($var) );
$date=substr($date, 0, -8);
echo $date;
?>
<tr>
<td><?php echo $row4['name']; ?></td>
<?php
for ($i=01; $i<=31; $i++)
{?>
<td><?php if ($date == $i) echo $row4['total']; ?></td>
<?php
}
?>
<td></td>
</tr>
<?php } } ?>
<tr>
<td colspan="31"></td>
<td>Total</td>
<td></td>
</tr>
</table>
答案 0 :(得分:0)
我的理解是,您需要一个代理商处理的总金额以及一年中的特定月份。 然后按如下所示更改您的查询,
<?php
$query = "SELECT b.date,b.agent_id,(SUM(b.amount) + SUM(b.cheque)) AS total,ba.id,ba.name
FROM bill_agent ba LEFT JOIN bill b ON b.agent_id = ba.id
WHERE YEAR(b.date) = YEAR(CURDATE()) AND MONTH(b.date) = MONTH(CURDATE())
GROUP BY b.date";
?>