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 |
+---------+----+----+----+-----+----+
1,2,3,4,5 ....是从日期开始的日子
但我希望得到以下
+---------+----+----+----+-----+----+----+
| | 1 | 2 | 3 | 4 | 5 |total
+---------+----+----+----+-----+----+----+
| Agent 1 | 35 | 0 | 12 | 0 | 0 |47 |
| Agent 2 | 0 | 10 | 0 | 0 | 14 |28 |
| Agent 3 | 0 | 0 | 0 | 100 | 0 |100 |
| Agent 6 | 0 | 0 | 0 | 9 | 0 | 9 |
+---------+----+----+----+-----+----+----+
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)
因为我不知道我是否理解你的问题,这是一个简短的样本,可以指出你正确的方向:
$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) "
. "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);
$agents = array();
while (true == ($row4 = mysql_fetch_assoc($result4))) {
$agents[$row4['agent_id']][$row4['date']] = $row4;
}
var_dump($agents);
}
查看输出,看看是否可以使用$agents
变量继续工作。