Question

我正在将一个系统放在一起，我希望获得不包含session['id']用户的用户行数，这样我就可以显示有点像facebook的链接。

示例：You and x.. others likes this

显然，只使用标准num行抓取数据库中的所有记录，但我想排除会话。

$likes = mysqli_query($mysqli, "
    SELECT feedback_streamid,feedback_userid,feedback_rating FROM streamdata_feedback 
        WHERE feedback_streamid=".$streamitem_data['streamitem_id']." 
        AND feedback_rating=1 ORDER BY feedback_id LIMIT 10") 

        or die("SELECT Error: ".mysqli_error($mysqli));

$numRowslikes = mysqli_num_rows($likes);
while ($row = mysqli_fetch_array($likes)) {
    $likesmemberid = rawfeeds_user_core::getuser($row['feedback_userid']);
    $user1_id = $_SESSION['id'];
    $user2_id = $row['feedback_userid'];
    if ($user2_id == $_SESSION['id']) {
        echo '<div class="like_name"><b><a href="profile.php?username='.$likesmemberid['username'].'">You and</a> '.$numRowslikes.' Others likes this </b> ';
    } else {
    echo '<a title="'.$likesmemberid['fullname'].' Likes This" href="profile.php?username='.$likesmemberid['username'].'"> <img border=\'0\' src=\'../userimages/    cropped'.$row['feedback_userid'].'.jpg\' onerror="this.src=\'userimages/no_profile_img.jpeg\'" width=\'20\' ></a> ';

    }
}

Answer 1

听起来您可以更改查询以排除当前用户的ID，然后正常使用num_rows（）。

WHERE feedback_userid NOT IN ('whatever_id_here')

你和X的其他人都是这样的

1 个答案: