PHP SQL select只使用in语句中数组的第一个结果

Question

我试图根据从我放入数组的表中选择的id来求和列。由于某种原因，Where子句中只使用第一个ID。当我回显变量时，所有的id都在那里。我做错了什么?

$counttheid = array();
$stmt3 = $mysqli->prepare("SELECT 
id 
FROM account
WHERE  level <= '5' AND door = ? AND `group_name` = ? AND betaald = 'Yes'");
$stmt3->bind_param("ss",$usernamesession,$groupname);
$stmt3->execute();
$result3 = $stmt3->get_result(); //only works when nd_mysli is set on the server!

while ($rowid = $result3->fetch_assoc())
{
$counttheid[] = $rowid['id'];
$countid = implode(',', $counttheid); // contains all the ids !!

}
$sql = "SELECT SUM(mobcash) AS totalcash FROM account WHERE id IN (?)  
";
$stmt = $mysqli->prepare($sql);
$stmt->bind_param("i",$countid);
$stmt->execute();
$stmt->bind_result($row['totalcash']);
while($stmt->fetch()) $sumcash = $row['totalcash'];
echo $sumcash; // Somhow only the sum of the first ID of the array !!

echo $countid;// all the ids from the array !!

Answer 1

不仅是in，而且绑定参数的数量也需要匹配。

试着用这个例子从while到execute的代码:

while ($rowid = $result3->fetch_assoc())
{
$counttheid[] = $rowid['id'];
// $countid = implode(',', $counttheid); // contains all the ids !!
}
$in = str_repeat('?,', count($counttheid) - 1) . '?';
$types = str_repeat('i', count($counttheid));
$sql = "SELECT SUM(mobcash) AS totalcash FROM account WHERE id IN ($in)";
$stmt = $mysqli->prepare($sql);
$stmt->bind_param($types, ...$counttheid);
$stmt->execute();

在bind_param,...$counttheid部分，...部分是参数解包操作符。