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Php SQL在页面上显示重复结果以查看用户_Php_Mysql - Fatal编程技术网

Php SQL在页面上显示重复结果以查看用户

Php SQL在页面上显示重复结果以查看用户,php,mysql,Php,Mysql,此查询: SELECT DISTINCT `social_posts` . * , `social_friends` . * , `users` . * FROM `social_posts` JOIN `social_friends` ON `friend` = `pUID` JOIN `users` ON `friend` = `uid` WHERE `fUID` = '$UID' OR `friend` = '$UID' ORDER BY `date` DESC LIMIT 0 , 30

此查询:

SELECT DISTINCT `social_posts` . * , `social_friends` . * , `users` . *
FROM `social_posts`
JOIN `social_friends` ON `friend` = `pUID`
JOIN `users` ON `friend` = `uid`
WHERE `fUID` = '$UID' OR `friend` = '$UID'
ORDER BY `date` DESC
LIMIT 0 , 30
应该选择帖子、用户数据,并验证该用户是否与其他用户是朋友。查看此文件的用户由
$UID
定义,
fUID
是好友UID,
friend
是第二个好友UID

问题是当向用户显示时,所有用户的帖子都会显示两次,而其他用户的帖子只显示一次

PHP:

$query=“选择不同的‘社交’帖子、‘社交朋友’、‘用户’*
来自`社会论坛'的帖子`
在'friend'='pUID上加入'social_friends'`
在'friend'='uid上加入'users'`
其中'fUID`='$UID'或'friend`='$UID'
按“日期”说明订购
限制为0,30”;
$data=mysqli_查询($dbc,$query);
while($row=mysqli\u fetch\u数组($data)){
$ID=$row['pid'];
$parsed=$parser->parse($row['body']);
回声';
echo sprintf($layout['socialViewFormat'],$row['picture'],$row['uid'],$row['username'],$date('M j Y g:i A',strotime($row['date']),$parsed);
$cQ=“选择用户。*,评论。*从`user`=`uid`和`module`=`social`上的`comments`加入`users`和`id`='$id';
$cD=mysqli\u查询($dbc,$cQ);
而($cR=mysqli\u fetch\u数组($cD)){
$body=htmlentities($cR['cBody']);
echo'
; 回声“.$body.”; } }

编辑:我在sql中输入了实际值,而不是使用变量,但这似乎没有帮助。

这是如何“在页面上”完成的?@SpencerWieczorek和PHP,检查我的编辑。有人有答案吗?我看不出查询中的sql与您在PHP中所做的完全匹配。@SpencerWieczorek,我使用的SQL查询显示在文章的PHP页面中。。。
$query = "SELECT DISTINCT `social_posts` . * , `social_friends` . * , `users` . *
    FROM `social_posts`
    JOIN `social_friends` ON `friend` = `pUID`
    JOIN `users` ON `friend` = `uid`
    WHERE `fUID` = '$UID' OR `friend` = '$UID'
    ORDER BY `date` DESC
    LIMIT 0 , 30";
        $data = mysqli_query($dbc, $query);
        while ($row = mysqli_fetch_array($data)){
            $ID = $row['pid'];
            $parsed = $parser->parse($row['body']);
            echo '<div class="shadowbar">';
            echo sprintf($layout['socialViewFormat'], $row['picture'], $row['uid'], $row['username'], date('M j Y g:i A', strtotime($row['date'])), $parsed);
            $cQ = "SELECT users.*, comments.* FROM `comments` JOIN `users` ON `user` = `uid` AND `module` = 'social' AND `id` = '$ID'";
            $cD = mysqli_query($dbc, $cQ);
            while($cR = mysqli_fetch_array($cD)){
                $body = htmlentities($cR['cBody']);
                echo '<pre><a href="/ucp/uid/'.$cR['uid'].'">' . $cR['username'] . '</a><hr style="padding:0; margin:0;" />';
                echo ''.$body.'</pre>';
            }
}