顯示2個表中的數據

Question

我有兩個表，需要以以下方式顯示一些數據：

我要從表1中選擇具有相同id_ref的所有用戶。 選擇它們之后，我想顯示它們的編號，然后在列表中顯示它們中的每一個，還要顯示它們在Table2中有多少行，但不是僅顯示某些符合我的條件的行。 另外，我想顯示從Table2中選擇的行的ID。 我的代碼是：

    $query = 'SELECT * FROM Table1  WHERE id_ref = '.$id_user.' AND active = 1 AND approved = 1 ORDER BY `id_aff` DESC';
$users_brought = mysql_query($query, $conn) or die(mysql_error());
$num_users_brought  = mysql_num_rows($users_brought);  


    while($user_bro = mysql_fetch_assoc($users_brought)):


    $query = 'SELECT * FROM Table2 WHERE id_aff = '.$user_bro['id_aff'].' AND ref = '.$income['id_user'].
$user_bro_brought = mysql_query($query, $conn) or die(mysql_error());
$num_user_bro_brought = mysql_num_rows($user_bro_brought); 

但是有些東西不起作用...對此有任何想法嗎？

Answer 1

您在這行中有語法錯誤

$query = 'SELECT * FROM Table2 WHERE id_aff = '.$user_bro['id_aff'].' AND ref = '.$income['id_user'].

您應該將其格式化為

$query = 'SELECT * FROM Table2 WHERE id_aff = '.$user_bro['id_aff'].' AND ref = '.$income['id_user'];

順便說一句，您不應該使用mysql_ *函數，因為它們已被棄用

Answer 2

您可以像這樣連接兩個表

$query = 'SELECT * FROM Table1
         JOIN table Table2 
         ON Table2.id_aff = Table1.id_aff AND Table2.ref = Table1.id_user  
         WHERE id_ref = '.$id_user.' AND active = 1 AND approved = 1 
         ORDER BY `id_aff` DESC';

以及更多您在第二次選擇查詢中遇到語法錯誤，請嘗試將其替換為

$query = 'SELECT * FROM Table2 
          WHERE id_aff = '.$user_bro['id_aff'].' 
          AND ref = '.$income['id_user'];