查詢多個表格，顯示錯誤

Question

<?php 

DEFINE ('DB_USER', 'username'); 
DEFINE ('DB_PASSWORD', 'password'); 
DEFINE ('DB_HOST', 'xxxxxxxxxxxxxxxx'); 
DEFINE ('DB_NAME', 'bigbobsveterinarysurgery'); 


$dbc = @mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME) OR die ('Could not connect, check the details ' . mysqli_connect_error()); 
mysqli_set_charset($dbc, 'utf8'); 
$q = "SELECT dog_name, owner_name, vet_date, Breed, cost, address, contact_number FROM owner, dogs ORDER BY dog_name ASC";  
$r = @mysqli_query ($dbc, $q); 


    echo '<table> 
    <tr> 
    <th scope="col">Dog</th> 
    <th scope="col">Owner</th> 
    <th scope="col">Entry Date</th> 
    <th scope="col">Breed</th> 
    <th scope="col">Cost</th> 
    <th scope="col">Address</th> 
    <th scope="col">Contact Number</th> 
    </tr>' 
    ; 


    while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC)) { 
     echo '<tr><td>' . $row['Dog_name'] . '</td><td>' . $row['owner_name'] . '</td><td>' . $row['vet_Date'] . 
     '</td></tr>' . $row['Breed'] . '</td></tr> ' . $row['cost'] . '</td></tr> '. $row['address'] . '</td></tr> ' . $row['contact_number'] . '</td></tr> '; 
    } 

echo '</table>'; 
?> 

它將所有數據的品種，成本和聯繫號碼顯示在一個大塊中。我想要的表格的格式可以在下面找到，但是每條狗都列在每個其他所有者列表中，而其他所有列都是空白的。發佈日期心不是顯示的，但我認爲是有問題的，當我從一個Excel工作表導入數據和日期是錯誤的格式

編輯

我把它更新爲：

while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC)) { 
     echo '<tr><td>' . $row['Dog_name'] . '</td><td>' . $row['owner_name'] . '</td><td>' . $row['vet_Date'] . 
    '</td><td>' . $row['Breed'] . '</td><td> ' . $row['cost'] . '</td><td> '. $row['address'] . '</td><td> ' . $row['contact_number'] . '</td></tr> '; 
} 

這已經停止了開始時的整個混亂，所有數據現在都格式化在一張表中。然而，每隻狗仍然與每個單獨的主人匹配。這是在這篇文章或數據庫中的代碼中的問題？

Answer 1

我認爲你需要連接兩個表

$q = "SELECT dog_name, owner_name, vet_date, Breed, cost, address, contact_number 
     FROM owner o 
     LEFT JOIN dogs d ON o.id = d.owner_id 
     ORDER BY d.dog_name ASC";  
// make sure you join on valid columns. 
// also, if there is a column that appears in both tables, 
// to avoid ambiguity, prefix it with the corresponding table alias 
$r = @mysqli_query ($dbc, $q); 

echo "<table> 
    <tr> 
    // skipping this so it doesn't eat up too much space 
    </tr>"; 

// also, every `$row[something]` should be placed inside a `<td></td>`, like so: 
while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC)) { 
    echo '<tr> 
      <td>' . $row['Dog_name'] . '</td> 
      <td>' . $row['owner_name'] . '</td> 
      <td>' . $row['vet_Date'] . '</td> 
      <td>' . $row['Breed'] . '</td> 
      <td>' . $row['cost'] . '</td> 
      <td>' . $row['address'] . '</td> 
      <td>' . $row['contact_number'] . '</td> 
     </tr> '; 
} 

兩個表中的選擇使他們之間的笛卡爾乘積。
使用join，您只能得到相應的行（根據ON子句）。

在你的特殊情況下，使用join將保持狗主的關係，而使用FROM owner, dogs，你會得到每個所有者的所有狗，這可能不是你所追求的。