试图从第一个查询中获取结果以搜索第二个查询

Question

我一直在开发一个社交网络，我有一个粉丝页面表，用户可以为他们最喜欢的乐队或名人创建个人资料，还有一个名为朋友的表，他们可以订阅粉丝页面。然后，我希望请求出现在粉丝页面管理员的通知中，以除外。 在第一个 sql 查询之后，它返回具有多个值的粉丝页面，但是如果我回显下一个 $sql 查询，它表明它只选择一个结果来查询而不是全部。

所以基本上我需要查询用户登录的所有粉丝页面（$log_username）的粉丝页面表，然后我需要获取这些粉丝页面并查询朋友表以了解是否有人请求订阅用户的粉丝页面??

这行代码 $fanpage_requests = "$fansubSql";输出 SELECT * FROM friends WHERE user2='fan4' AND accepted='0' ORDER BY datemade ASC 作为唯一结果，但应该有 2 个结果，因为 fan3 和 fan4 都有请求。

感谢您的帮助迈克尔

$fanpage_requests = '';

$fansql = "SELECT created_by, fanpage_name FROM `fanpages`  WHERE created_by = '$log_username' ";
$fanquery = mysqli_query($db_conx, $fansql);
$fannumrows = mysqli_num_rows($fanquery);

if($fannumrows < 1){
    $fanpage_requests = 'No friend requests';
} else {

    while($row = mysqli_fetch_array($fanquery, MYSQLI_ASSOC)) {
        $fanpage_name = $row["fanpage_name"];
        $created_by = $row["created_by"];

        $fansubSql = "SELECT * FROM friends WHERE user2='$fanpage_name' AND accepted='0' ORDER BY datemade ASC";
        $fansubQuery = mysqli_query($db_conx, $fansubSql);
        $fansubNumrows = mysqli_fetch_row($fansubQuery);
        $fanpage_requests = "$fansubSql";

        if($fansubNumrows < 1){
            $fanpage_requests = "blah blah"; 
        }

        while ($fansubRow = mysqli_fetch_array($fansubQuery, MYSQLI_ASSOC)) {
            $fansubreqID = $fansubRow["id"];
            $fansubuser1 = $fansubRow["user1"];
            $fansubdatemade = $fansubRow["datemade"];
            $fansubdatemade = strftime("%B %d", strtotime($datemade));

            $fansubthumbquery = mysqli_query($db_conx, "SELECT avatar FROM users WHERE username='$user1' LIMIT 1");
            $fansubthumbrow = mysqli_fetch_row($thumbquery);
            $fansubuser1avatar = $thumbrow[0];
            $fansubuser1pic = '<img src="user/'.$user1.'/'.$user1avatar.'" alt="'.$user1.'" class="user_pic">';

            if($fansubuser1avatar == NULL){
                $fansubuser1pic = '<img src="images/avatardefault.jpg" alt="'.$user1.'" class="user_pic">';
            }

            $fanpage_requests .= '<div id="friendreq_'.$fansubreqID.'" class="friendrequests">';
            $fanpage_requests .= '<a href="user.php?u='.$fansubuser1.'">'.$fansubuser1pic.'</a>';
            $fanpage_requests .= '<div class="user_info" id="user_info_'.$fansubreqID.'">'.$fansubdatemade.' <a href="user.php?u='.$fansubuser1.'">'.$fansubuser1.'</a> requests friendship<br /><br />';
            $fanpage_requests .= '<button onclick="fanReqHandler(\'accept\',\''.$fansubreqID.'\',\''.$fansubuser1.'\',\'user_info_'.$fansubreqID.'\')">accept</button> or ';
            $fanpage_requests .= '<button onclick="fanReqHandler(\'reject\',\''.$fansubreqID.'\',\''.$fansubuser1.'\',\'user_info_'.$fansubreqID.'\')">reject</button>';
            $fanpage_requests .= '</div>';
            $fanpage_requests .= '</div>';
        }
    }
}

?>

Answer 1

阅读文档。 mysqli_fetch_row 不返回数字。它从查询中返回一条记录。查看文档中的示例：您需要在循环中使用mysqli_fetch_row 来检索所有记录。

要知道查询返回的行数，请使用mysqli_num_rows。