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運行到使用PHP選擇和分組結果

2013-05-13 25 views
0

我試圖從我的主表（父母詳細信息）中顯示一行數據，然後在這些行的每一行下顯示另一個表中的子女姓名。我不得不使用下面的代碼一展身手：運行到使用PHP選擇和分組結果

$sql="select distinct members_main.first_name as main_firstname, 
    members_main.first_name as main_firstname, 
    members_main.last_name as main_lastname, 
    members_main.address_1 as main_address_1, 
    members_main.address_2 as main_address_2, 
    members_main.address_3 as main_address_3, 
    members_main.address_4 as main_address_4, 
    members_main.post_code as main_post_code, 
    members_main.home_tel as main_home_tel, 
    members_main.mobile as main_mobile, 
    members_main.home_email as main_home_email 
    from members_main, members_family where members_main.contact_id=members_family.contact_id"; 

$sql2="select members_family.first_name as fam_firstname, 
    members_family.last_name as fam_lastname 
    from members_family, members_main 
    where members_main.contact_id=members_family.contact_id"; 

$result=mysql_query($sql); 

while($rows=mysql_fetch_array($result)){ 

echo '<table width="100%" cellpadding="3" cellspacing="0"> 
<tr> 
<th align="left" valign="top">First Name</th> 
<th align="left" valign="top">Last Name</th> 
<th align="left" valign="top">Address 1</th> 
<th align="left" valign="top">Address 2</th> 
<th align="left" valign="top">Address 3</th> 
<th align="left" valign="top">Address 4</th> 
<th align="left" valign="top">Post Code</th> 
<th align="left" valign="top">Home Tel</th> 
<th align="left" valign="top">Mobile</th> 
<th align="left" valign="top">Email</th> 
</tr>'; 

echo '<tr> 
<td valign="top">'.$rows['main_firstname'].'</td> 
<td valign="top">'.$rows['main_lastname'].'</td> 
<td valign="top">'.$rows['main_address_1'].'</td> 
<td valign="top">'.$rows['main_address_2'].'</td> 
<td valign="top">'.$rows['main_address_3'].'</td> 
<td valign="top">'.$rows['main_address_4'].'</td> 
<td valign="top">'.$rows['main_post_code'].'</td> 
<td valign="top">'.$rows['main_home_tel'].'</td> 
<td valign="top">'.$rows['main_mobile'].'</td> 
<td valign="top">'.$rows['main_home_email'].'</td> 
</tr>'; 

$result2=mysql_query($sql2); 

echo '<table width="100%" cellpadding="3" cellspacing="0"> 
<tr> 
<th align="left" valign="top">First Name</th> 
<th align="left" valign="top">Last Name</th> 
</tr>'; 

while($rows=mysql_fetch_array($result2)){ 

echo '<tr> 
<td valign="top">'.$rows['fam_firstname'].'</td> 
<td valign="top">'.$rows['fam_lastname'].'</td> 
</tr>'; 

} 

} 

echo "</table>";

我得到的所有行從表members_family每每條記錄從members_main。 members_family中可以有多個記錄，這些記錄將與members_main中的一條記錄相關聯。基本上，members_main（parent）可以在members_main中擁有多個子項。

來源

2013-05-13 user2281899

+0

您$ SQL2必須是您的主，而（）{}內，它會從你也先需要輸入選擇只有孩子，該父 – Dave 2013-05-13 11:52:27

+0

感謝戴夫。我怎麼做？我是一個PHP新手！ – user2281899 2013-05-13 11:53:43

+0

'$ sql2 =「select members_family.first_name as fam_firstname， members_family.last_name as fam_lastname from members_family where contact_id =」。$ rows ['contact_id'];'這樣的東西就在'$ result2 ='位之上，您需要確保將主要SQL查詢中的contact_id拉出。 – Dave 2013-05-13 11:55:22

回答

2

我想你只需要從第一張表中檢索你的members_main.contact_id以便在第二個while循環中使用。然後，你將不得不搬進去查詢while loop裏面你的第二個查詢，以便代碼看起來像：

$sql="select distinct 
members_main.contact_id as ID, 
members_main.first_name as main_firstname, 
members_main.first_name as main_firstname, 
members_main.last_name as main_lastname, 
members_main.address_1 as main_address_1, 
members_main.address_2 as main_address_2, 
members_main.address_3 as main_address_3, 
members_main.address_4 as main_address_4, 
members_main.post_code as main_post_code, 
members_main.home_tel as main_home_tel, 
members_main.mobile as main_mobile, 
members_main.home_email as main_home_email 
from members_main, members_family where members_main.contact_id=members_family.contact_id"; 

$result = mysql_query($sql); 
while($rows=mysql_fetch_array($result)) { 

//script goes on with table printing untill 
<td valign="top">'.$rows['main_home_email'].'</td> 
</tr>';

現在是時候了查詢

$sql2="select members_family.first_name as fam_firstname, 
members_family.last_name as fam_lastname 
from members_family, members_main 
where members_main.contact_id=members_family.contact_id and members_main.contact_id = '".$rows['ID']."'";

正如你看到的我加入WHERE條件限制結果爲指定ID至於我從第一個查詢檢索。現在您可以打印表格的其餘部分。哦，順便說一句，你關閉表應該看起來像這樣

echo "</table></td></tr></table>";

這是因爲你打開了兩個表。

那麼我想你要記住，mysql_*功能已被棄用，所以我會建議你切換到mysqli或PDO

來源

2013-05-13 11:58:24 Fabio

+0

我剛剛刪除了我的帖子，因爲它與您的帖子非常相似。 – 2013-05-13 12:07:58

+0

@ KarlM.W。是的，我看到，很好。您仍然可以上傳我的文件：P – Fabio 2013-05-13 12:08:55

+0

現在您已經添加了mysqli遷移信息，您可以使用Upvote：D – 2013-05-13 12:22:26

0

我感到有些困惑與你的第二個查詢。我在想，你必須用來自第一個查詢（members_main）的contact_id構造第二個查詢。我以這種方式假設我已經編碼如下。可能它會幫助你

 <?php 
     $sql = "select distinct members_main.first_name as main_firstname, 
      members_main.first_name as main_firstname, 
      members_main.last_name as main_lastname, 
      members_main.address_1 as main_address_1, 
      members_main.address_2 as main_address_2, 
      members_main.address_3 as main_address_3, 
      members_main.address_4 as main_address_4, 
      members_main.post_code as main_post_code, 
      members_main.home_tel as main_home_tel, 
      members_main.mobile as main_mobile, 
      members_main.home_email as main_home_email 
      from members_main, members_family where members_main.contact_id=members_family.contact_id"; 
     $result = mysql_query($sql); 
     ?> 
     <table width="100%" cellpadding="3" cellspacing="0"> 
      <tr> 
       <th align="left" valign="top">First Name</th> 
       <th align="left" valign="top">Last Name</th> 
       <th align="left" valign="top">Address 1</th> 
       <th align="left" valign="top">Address 2</th> 
       <th align="left" valign="top">Address 3</th> 
       <th align="left" valign="top">Address 4</th> 
       <th align="left" valign="top">Post Code</th> 
       <th align="left" valign="top">Home Tel</th> 
       <th align="left" valign="top">Mobile</th> 
       <th align="left" valign="top">Email</th> 
      </tr> 
      <?php 
      while ($rows = mysql_fetch_array($result)) 
      { 
       $conId = $rows['contact_id']; //Which is comming from first Query 
       $sql2 = "select members_family.first_name as fam_firstname, 
      members_family.last_name as fam_lastname 
      from members_family, members_main 
      where members_family.contact_id=$conId"; 

       $result2 = mysql_query($sql2); 
       ?> 
       <tr> 
        <td valign="top"><?= $rows['main_firstname']; ?></td> 
        <td valign="top"><?= $rows['main_lastname']; ?></td> 
        <td valign="top"><?= $rows['main_address_1']; ?></td> 
        <td valign="top"><?= $rows['main_address_2']; ?></td> 
        <td valign="top"><?= $rows['main_address_3']; ?></td> 
        <td valign="top"><?= $rows['main_address_4']; ?></td> 
        <td valign="top"><?= $rows['main_post_code']; ?></td> 
        <td valign="top"><?= $rows['main_home_tel']; ?></td> 
        <td valign="top"><?= $rows['main_mobile']; ?></td> 
        <td valign="top"><?= $rows['main_home_email']; ?></td> 
       </tr>  

       <tr> 
        <td colspan="10"> 
         <table width="100%" cellpadding="3" cellspacing="0"> 
          <tr> 
           <th align="left" valign="top">First Name</th> 
           <th align="left" valign="top">Last Name</th> 
          </tr> 
          <?php 
          while ($rowsFamily = mysql_fetch_array($result2)) 
          { 
           ?> 
           <tr> 
            <td valign="top"><?= $rowsFamily['fam_firstname']; ?></td> 
            <td valign="top"><?= $rowsFamily['fam_lastname']; ?></td> 
           </tr> 
           <?php 
          } 
          ?> 
         </table> 
        </td> 
       </tr> 
       <?php 
      } 
      ?> 
     </table>

來源

2013-05-13 12:07:46 Hearaman

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