Php 获取关联错误 你好

我正试图用这个查询创建一个表，但我无法使它工作。问题是它不能一次选择两个查询，但我不知道另一种方法

$active_ids = '1, 3, 4';
$query = "SELECT * FROM users WHERE id IN ({$active_ids})";
$result = $mysqli->query($query);   

$query = "SELECT dj, count(*) AS n FROM timetable WHERE dj IN ({$active_ids}) GROUP BY dj";
$result = $mysqli->query($query);

while($row = $result->fetch_assoc()){
    echo $row['username'], "<br/>";
    echo $row['n'], "<br/>";
}

试试这个：

$active_ids = '1, 3, 4';
$result = $mysqli->query("
    (SELECT * FROM users WHERE id IN ({$active_ids}))
    UNION 
    (SELECT dj, count(*) AS n FROM timetable WHERE dj IN ({$active_ids}) GROUP BY dj)
") ;

if (!$rec = mysqli_fetch_array($result)) {
    echo ("Sorry, no records found");
}
else {
    do {
        echo ($rec["username"]);
        echo "<br />";
        echo ($rec["n"]);
        echo "<br />";
    }
    while ($rec = mysqli_fetch_array($result));
}

---

我不知道这肯定是真的，但你能试试这个代码吗

这是用户表中的djin列吗？您已经发布了此列-与此列有什么不同？您也得到了答案。我需要将查询添加到1中query@Imran否dj在时间表中，用户名在用户中SSO需要传递给时间表的表用户的外键是什么？致命错误：在C:\xampp\htdocs\Admin\u res\core\home.php第75行的非对象上调用成员函数fetch_assoc，您可以写入你的桌子？这会更容易。

     $active_ids = '1, 3, 4'; 

    $query = "SELECT users.* from users 
              LEFT JOIN timetable ON users.id=timetable.dj
                  WHERE users.id IN ({$active_ids})
              UNION
              SELECT timetable.dj,timetable.count(*) as n from timetable
              RIGHT JOIN users ON timetable.dj=users.id
                  WHERE timetable.dj IN ({$active_ids}) GROUP BY timetable.dj
    ";
    $result = $mysqli->query($query);    

    while($row = $result->fetch_assoc()){
        echo $row['username'], "<br/>";
        echo $row['n'], "<br/>";
    }