Question

如果显示消息，如果没有显示从db消息名称“你没有任何朋友”这里是PHP代码

$mid = mysql_real_escape_string($_COOKIE['member_id']);
$res = mysql_query("
SELECT profile_friends.friends_friend_id, members.name, profile_friends.friends_member_id,members.member_id,pp_main_photo,pp_photo_type
FROM profile_friends
INNER JOIN members
ON profile_friends.friends_friend_id=members.member_id
INNER JOIN profile_portal on pp_member_id = member_id
WHERE friends_member_id  = '$mid'");
while($row = mysql_fetch_array($res)){  
?>



<?php echo $row['name'];?>
<img src="../forum/uploads/<?php echo $row['pp_main_photo'];?>
" width="50" height="50" alt=""/><br>


<?php
}
?>

Answer 1

第一条评论是正确的。

if (mysql_num_rows($res) < 1) {

   echo "no friends found";
} else {
    echo $row['name'];?>
    <img src="../forum/uploads/<?php echo $row['pp_main_photo'];
     ?>
   " width="50" height="50" alt=""/><br>

此外，您发布的错误通常只是意味着您错过了;某处...

Answer 2

如我的评论中所述，使用mysql_num_rows()检查> 0是否循环显示结果。

$mid = mysql_real_escape_string($_COOKIE['member_id']);
$res = mysql_query("
       SELECT profile_friends.friends_friend_id, members.name, profile_friends.friends_member_id,members.member_id,pp_main_photo,pp_photo_type
       FROM profile_friends
       INNER JOIN members
       ON profile_friends.friends_friend_id=members.member_id
       INNER JOIN profile_portal on pp_member_id = member_id
       WHERE friends_member_id  = '$mid'");

if(mysql_num_rows($res) == 0) 
     echo "you dont have any friend"; 
else {
    while($row = mysql_fetch_array($res)){  
?>
        <?php echo $row['name'];?>
        <img src="../forum/uploads/<?php echo $row['pp_main_photo'];?>" width="50" height="50" alt=""/><br>  
<?php
    }
}
?>

如果没有从db显示任何内容的错误消息

2 个答案: