Question

我有这个功能：

function dulide($uid) {
    global $mysqli, $sm;
    $city = $mysqli->query("SELECT city FROM users WHERE id = '" . $sm['user']['id'] . "'");
    $haha = $city->fetch_assoc();
    $citye = $haha['city'];
    $sql = "SELECT id,age,name,email,city,fixed FROM users WHERE fixed=1 AND city = '" . $citye . "' LIMIT 5";
    $query = $mysqli->query($sql);

    if ($query->num_rows > 0) {
        while ($user = $query->fetch_object()) {
            $return .= '<tr><td> ' . $user->name . ' </td></tr><tr><td>';
            $uid = $user->id;
            $photos = $mysqli->query("SELECT photo FROM users_photos WHERE approved = 1 and profile = 1 and u_id = '" . $uid . "' order by id desc LIMIT 1");
            if ($photos->num_rows > 0) {
                while ($up = $photos->fetch_object()) {

                    $return .= '<img src="' . $up->photo . '" alt="Smiley face" height="50" width="50">';
                }
            }
            $return .= ' </td></tr> ';
        }
    }
    return $return;
}

当我尝试用

调用该函数时

<?=dulide($uid); ?>

什么都没发生。有人有解决方案吗？

发出大量的魔法和祝福！ :)

代表感谢

Answer 1

结果发现有一个数据库错误导致我的功能无法正常工作。谢谢你的帮助，另一个时间会很好做}其他{在num行之后发布如果没有结果:)

函数不会返回结果php

1 个答案: