Question

我试图将它从朋友表中获取朋友ID，然后从用户表中获取朋友信息。我尝试过使用foreach但没有运气。

这就是我现在所拥有的，它只回应一位朋友，而不是我在桌子上的三位朋友。关于如何解决这个问题的任何想法？也许我没有正确地使用它们？先感谢您！

<?php 
//Gets users information from users. 
$stmt = $DB_con->prepare('SELECT  friendsid FROM friends WHERE userid='.$_SESSION['user']['id']);
$stmt->execute();

if($stmt->rowCount() > 0) {
    $row=$stmt->fetch(PDO::FETCH_ASSOC);
    extract($row);
}   else {
    $friendsid = $row['friendsid'];
} 

$stmt = $DB_con->prepare('SELECT username,userprofile,status FROM users WHERE id='.$friendsid);
$stmt->execute();

if($stmt->rowCount() > 0) {
    while($row=$stmt->fetch(PDO::FETCH_ASSOC))  {
        extract($row);
        ?>  
            <div class="row">
                <div class="col-sm-3">
                    <div class="well">
                        <h4><strong><?php echo $row['username'];?></strong></h4>
                        <img src="images/profile/<?php echo $userprofile;?>" class="img-circle" height="70" width="70" alt="Avatar">
                    </div>
                </div>
                <div class="col-sm-9">
                    <div class="well">
                        <h3 class="text-left"><?php echo $row['status'];?></h3>
                    </div>
                </div>
            </div>

<?php
    }
} else {
?>

<?php
}
?>

Answer 1

预备语句使用参数，您应该使用JOIN。

$stmt = $DB_con->prepare('SELECT u.* FROM users AS u JOIN friends AS f ON u.id = f.friendsid WHERE f.userid = :user_id');
$stmt->bindParam(':user_id', $_SESSION['user']['id']);
$stmt->execute();
while ($row = $stmt->fetch(PDO::FETCH_ASSOC))
{
    // Do whatever you want with $row['id'], $row['username'], $row['userprofile']
}

通过id显示多个用户信息的问题

1 个答案: