我对标题感到抱歉,获得标题很复杂。但请尝试阅读我在下面的声明。
我想合并表中具有相同名称但具有不同信息的数据。表示例:
form_no | name | status | date_added
1 | keyboard | OK | 19-APR-2013 (today)
2 | keyboard | NG | 18-APR-2013 (yesterday)
现在我想在我的页面中显示它,只有 1 个信息合并为:
表格视图示例:
No | Item Name | Yesterday Status | Today Status
1 | keyboard | NG | OK
只是它,我想在表格视图中显示它,只需 1 个信息合并。它具有相同的名称,但根据添加的日期具有不同的状态。
<table>
<td>Item Name</td><td>Yesterday Status</td><td>Today Status</td>
</table>
include ("includes/_db_.php");
$today = date("j-F-Y");
$yesterday = date("j-F-Y", time() - 60 * 60 * 24);
$query = "SELECT * from t_production_status";
$result = mysql_query($query);
while ($data = mysql_fetch_array($result))
{
$item_name = $data['name'];
$status = $data['status'];
$date_added = $data['date_added'];
<tr>
<td><?php echo $item_name; ?></td>
<td><?php echo $status; ?></td> <- this what I want show for yesterday status
<td><?php echo $status; ?></td> <- this what I want show for today status
<td><?php echo $date_added; ?></td>
有什么建议么 ?