嘿,我需要帮助,我想在一个功能
中完成两个功能,但我不知道为什么,但我就是不能像我想的那样做。这是我的两个函数的代码
第一:
$users = User::getAllUsers();
第二个:
$userCheck = Suggestion::getIfUserWrotesSuggestion($user->getId());
if(!is_null($userCheck)){
?>
<option><?php echo $user->getId; ?></option
}
现在这两个函数的工作方式是这样的
第一:
public static function getAllUsers(){
$db = new database();
$mysqli = $db->getConnect();
$statement = "SELECT `id` FROM `users` ORDER BY `name` ASC";
var_dump($statement);
$sql = mysqli_query($mysqli, $statement);
$userArray = array();
while($result = mysqli_fetch_array($sql)){
$id = $result['id'];
$user = new User($id);
$userArray[] = $user;
}
return $userArray;
}
第二:
public static function getIfUserWrotesSuggestion($userId){
$db = new Database();
$mysqli = $db->getConnect();
$statement = "SELECT `user_id` FROM `suggestions` WHERE `user_id`= '$userId'";
$sql = mysqli_query($mysqli, $statement);
$sql = mysqli_fetch_array($sql);
return $sql;
}
我希望你能帮助我。
万事如意,提前感谢:)
我认为您要查找的是在一个查询中加载所有用户和相应的建议(如果有(。
这样的事情会有所帮助
SELECT
u.id,
COUNT(s.*) as suggestionCount
FROM
users AS u
LEFT JOIN suggestions as s ON u.id = s.user_id
GROUP BY
u.id
ORDER BY
u.name` ASC