通过合并三个名为 luckydraw_weekone、luckydraw_weektwo 的表，在用户表中显示现有电子邮件，参与者的字段均为 Email

Posted 2023-02-24

【中文标题】通过合并三个名为 luckydraw_weekone、luckydraw_weektwo 的表，在用户表中显示现有电子邮件，参与者的字段均为 Email

【英文标题】：to show existing email in user table by merging three tables named luckydraw_weekone,luckydraw_weektwo, participants all thre have the fild as Email

【发布时间】：2014-01-27 21:30:27

【问题描述】：

$query = "SELECT GROUP_CONCAT(  luckydraw_weekone.Email),
         GROUP_CONCAT(luckydraw_weektwo.Email),
         GROUP_CONCAT(participants.Email) 
         FROM luckydraw_weekone 
         LEFT JOIN luckydraw_weektwo 
         ON luckydraw_weekone.id = luckydraw_weektwo.id 
         LEFT JOIN participants ON luckydraw_weekone.id = participants.id";

$qry_result = mysql_query($query) or die(mysql_error()); 
$display_string = "<table border=\"1\">"; 
$display_string = "<table border=\"1\">";
$display_string .= "<tr>";
$display_string .= "</tr>";

while($row = mysql_fetch_array($qry_result))
    $display_string .= "<tr>";
    $display_string .= "<td>$row[0]</td>"; 
    $display_string .= "<td>$row[1]</td>"; 
    $display_string .= "<td>$row[2]</td>"; 
    $display_string .= "</tr>"; 
    $query2 = "SELECT * FROM users WHERE mail='.$row[0].' OR mail='.$row[1].' OR mail='.$row[2].'";
    $qry_result2 = mysql_query($query2) or die(mysql_error()); 
    $display_string_user = "<table border=\"1\">"; 
    while($row2 = mysql_fetch_array($qry_result2)) 
        $display_string_user .= "<tr>"; 
        $display_string_user .= "<td>$row[0]</td>"; 
        $display_string_user .= "<td>$row[1]</td>"; 
        // and all details you want to add
        $display_string_user .= "</tr>"; 
    
  
 $display_string .= "</table>"; 
 $display_string_user .= "</table>"; 
 echo $display_string;
 echo $display_string_user;

【参考方案1】：

试试这个：

SELECT DISTINCT u.mail
FROM users u 
INNER JOIN (SELECT email FROM luckydraw_weekone
            UNION 
            SELECT email FROM luckydraw_weektwo
            UNION 
            SELECT email FROM participants
          ) AS A ON u.mail = A.email