将2个表与来自不同查询的数据合并?

时间:2013-03-21 14:13:07

标签: php merge html-table

我目前有2个表格,如下所示:

enter image description here

这两个表的代码如下所示:

<table width="600" border="1">  
<tr>   
   <th width="91"> <div align="center">Mail_ID  </div></th> 
   <th width="91"> <div align="center">Status  </div></th> 
   <th width="91"> <div align="center">Datum  </div></th> 
</tr>
<?php 
$j = 0;  
while($objResult2 = mysql_fetch_array($objQuery2))  
{  
$j++;  
?> 
<tr>   
   <td><div align="center"><?=$objResult2["Mail_ID"];?> </div></td> 
   <td><div align="center"><?=$objResult2["Status"];?> </div></td> 
   <td><div align="center"><?=$objResult2["Datum"];?> </div></td>  
</tr>
<?php  
}  
?>
</table>
<br/>
<table width="600" border="1">  
<tr> 
   <th width="91"> <div align="center">ID  </div></th>  
   <th width="91"> <div align="center">Titel  </div></th> 
   <th width="91"> <div align="center">Subscribe  </div></th> 
   <th width="91"> <div align="center">Unsubscribe  </div></th> 
</tr>

<?php for ($i = 0; $objResult1 = mysql_fetch_array($objQuery1); $i++) : ?>
<tr>  
    <td><div align="center"><?=$objResult1["ID"];?><input type="hidden" name="mailid[]" value="<?=$objResult1["ID"];?>"> </div></td>
    <td><div align="center"><?=$objResult1["Titel"];?> </div></td>  
    <td><div align="center"><input type="radio" name="sub[<?php echo $i; ?>]" value="10"> </div></td>  
    <td><div align="center"><input type="radio" name="sub[<?php echo $i; ?>]" value="90"> </div></td> 
</tr>
<?php endfor; ?>

</table>  

查询和变量如下所示:

$strSQL1    = "SELECT ID, Titel FROM Mail"; 
$strSQL2    = "SELECT * FROM Subscriptions,
(SELECT MAX(ID) as ids, Mail_ID FROM Subscriptions
    WHERE Klant_ID = '".$_GET["ID"]."' GROUP BY Mail_ID) table2
WHERE ID=table2.ids";  
$objQuery1  = mysql_query($strSQL1);
$objQuery2     = mysql_query($strSQL2); 

现在我想将这两个表合并为:

enter image description here

我无法让这个工作,因为我尝试的所有内容由于其中的2个循环而混合了一切。我希望你们理解我的问题,我添加了尽可能多的信息和图片,以使一切都清楚^^如果你还有问题,只要问他们评论。欢迎任何帮助!

3 个答案:

答案 0 :(得分:0)

我不确定这是否适合您,但它应该是一个开始:

SELECT Mail.ID Mail_ID, Mail.Titel, Subscriptions.status, Subscriptions.Datum, Subscriptions.Subscribe, Subscriptions.Unsubscribe
FROM Mail left join Subscriptions on Mail.ID = Subscriptions.Mail_ID

答案 1 :(得分:0)

使用SQL Join形成一个返回所有数据的SQL查询。然后只使用一个循环。

SQL可能类似于:

SELECT ID, Titel, Subscriptions.* FROM Mail INNER JOIN Subscriptions ON Subscriptions.ID = Mail.ID ...

答案 2 :(得分:0)

你应该像其他人已经说过的那样使用联接。通过连接两个或多个表您不需要使用多个while循环,代码将更加清洁

SELECT 
  Mail.ID Mail_ID,
  Mail.Titel,
  Subscriptions.status,
  Subscriptions.Datum,
  Subscriptions.Subscribe,
  Subscriptions.Unsubscribe 
FROM
  Mail 
LEFT JOIN Subscriptions 
ON (Subscriptions.Mail_ID = Mail.ID)

其中一种连接类型适用于您:LEFT JOIN或INNER JOIN。

HTML应如下所示:

<table width="600" border="1">  
<tr>   
   <th width="91"> <div align="center">Mail_ID  </div></th> 
   <th width="91"> <div align="center">Status  </div></th> 
   <th width="91"> <div align="center">Datum  </div></th> 
   <th width="91"> <div align="center">Titel  </div></th> 
   <th width="91"> <div align="center">Subscribe  </div></th> 
   <th width="91"> <div align="center">Unsubscribe  </div></th> 
</tr>
<?php 
while ($objResult2 = mysql_fetch_array($objQuery2))  
{
?> 
<tr>   
   <td><div align="center"><?=$objResult2["Mail_ID"];?> </div></td> 
   <td><div align="center"><?=$objResult2["Status"];?> </div></td> 
   <td><div align="center"><?=$objResult2["Datum"];?> </div></td>  
   <td><div align="center"><?=$objResult1["Mail_ID"];?><input type="hidden" name="mailid[]" value="<?=$objResult1["Mail_ID"];?>"> </div></td>
   <td><div align="center"><?=$objResult1["Titel"];?> </div></td>  
   <td><div align="center"><input type="radio" name="sub[<?=$objResult1["Subscriptions_ID"];?>]" value="10"> </div></td>  
   <td><div align="center"><input type="radio" name="sub[<?=$objResult1["Subscriptions_ID"];?>]" value="90"> </div></td> 
</tr>
<?php
}
?>
</table>  

如果您使用LEFT JOIN,则所有邮件行都将可见但如果您使用INNER JOIN,那么只有那些具有与其关联的订阅行的邮件行

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