2013-04-26 39 views
0

對不起,標題,很難得到標題。但請嘗試閱讀下面的聲明。PHP顯示合併數據表

我想合併具有相同名稱項但具有不同信息的表中的數據。 表舉例:

form_no | name  | status | date_added 
1  | keyboard | OK  | 19-APR-2013 (today) 
2  | keyboard | NG  | 18-APR-2013 (yesterday) 

現在我想只有1信息顯示它在我的頁面合併爲:
表視圖示例:

No | Item Name | Yesterday Status | Today Status 
1 | keyboard | NG    | OK 

只要將它,我想表明它只有1個信息合併的表視圖。它具有相同的名稱,但根據date_added具有不同的狀態。

<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> 

有什麼建議嗎?

+0

[不使用mysql_ * extension](http://www.php.net/en/mysql_query)。改爲使用[PDO](http://php.net/manual/en/book.pdo.php)或[MSQLi](http://php.net/manual/en/book.mysqli.php)。 – 2013-04-26 08:18:45

回答

0

您是否嘗試過做一個自我加入你的狀態表..

SELECT status1, status2 FROM t_production_status status1 JOIN t_production_status status1 
WHERE status1.name = status2.name 
AND status1.date = (today) 
AND status2.data = (yesterday) 

(未測試,但它應該是類似的規定)

http://dev.mysql.com/doc/refman/5.0/en/join.html