php - PDO/MySQL 使用 if 语句获取多列

Question

我目前正在尝试从我的数据库中获取两个图像位置，我如何返回两个列，如果它们都是空的，则回显另一个图像。这是我到目前为止所得到的。如何返回 photo 和 photo_small 以便我可以在 php 文件中回应它们。

PUBLIC FUNCTION Profile_Pic($uiD) {
    $sth = $this->db->prepare("SELECT photo,photo_small FROM users WHERE uiD = :id");
    $sth->execute(array(':id' => $uiD));

        if ($sth->rowCount() > 0) {
                $data = $row['photo'];
            return $data; 
        } else {
            $data = './icons/users.png';
            return $data;
        } 
    }

Answer 1

PUBLIC FUNCTION Profile_Pic($uiD) {
    $sql = "SELECT photo,photo_small FROM users WHERE uiD = ?";
    $sth = $this->db->prepare($sql);
    $sth->execute(array($uiD));
    $data = $sth->fetch();
    if (empty($data['photo'])) {
        $data['photo'] = './icons/users.png';
    }
    if (empty($data['photo_small'])) {
        $data['photo_small'] = './icons/users.png';
    }
    return $data;
}

如果你想替换两个图像，即使一个图像都不存在

PUBLIC FUNCTION Profile_Pic($uiD) {
    $sql = "SELECT photo,photo_small FROM users WHERE uiD = ?";
    $sth = $this->db->prepare($sql);
    $sth->execute(array($uiD));
    $data = $sth->fetch();
    if (empty($data['photo']) || empty($data['photo_small'])) {
        $data['photo'] = './icons/users.png';
        $data['photo_small'] = './icons/users.png';
    }
    return $data;
}