Question

我使这里的内容更易于理解，因此：我想在此处选择在Mysql表的“专业”字段中具有“例如”编码器的用户。因为正如您在此代码中所看到的，它将选择一个随机用户，但已登录的用户除外。而且我需要在AND之后再做一个条件（请参见代码行2），因此它会查找并仅推出带有“编码器”的用户写在行业领域。您还需要什么信息来帮助解决问题？

ps。它应该从列中选择-专业

public function whoToFollow1($user_id, $profileID){
        $stmt = $this->pdo->prepare("SELECT * FROM `users` WHERE `user_id` != :user_id AND (what stetement need to be here)  ORDER BY rand() LIMIT 1");
        $stmt->execute(array("user_id" => $user_id));       
        $users = $stmt->fetchAll(PDO::FETCH_OBJ);
        echo '<div class="wrap"><div class="inner"><div class="title"></div>';
        foreach ($users as $user) {

Answer 1

您只需要将您描述的条件添加到查询中，即

$profession = "coder";
$stmt = $this->pdo->prepare("SELECT * FROM `users` WHERE `user_id` != :user_id AND profession = :profession  ORDER BY rand() LIMIT 1");
$stmt->execute(array("user_id" => $user_id, "profession" => $profession));

Answer 2

 $stmt = $this->pdo->prepare("SELECT * FROM `users` WHERE `user_id` != :user_id AND `profession`=:job ORDER BY rand() LIMIT 1");
 $stmt->execute(array("user_id" => $user_id, "job" => $job));

Note: $job will be used to represent the current job you wish to target.

Answer 3

// I am assuming you have profession table with id and profession name/type
SELECT * 
FROM `users` 
INNER JOIN profession on users.profession_id = profession.id
WHERE `user_id` = :user_id 
AND profession.id =  :profession_id

PHP $ stmt获得合适的用户

3 个答案: