Question

我一直在讨论这个PHP代码很长一段时间。我正在尝试实现像

这样的东西

- ＆gt;获取每个状态
- ＆gt;让每个用户进入用户的好友列表
- ＆gt;显示用户朋友列表中每个用户的状态

并重复直到没有更多。我一直在寻找一个解决方案，这几天真的很烦我。这是我试过的代码：

编辑：根据请求发布架构 https://kjf-tech.net/files/schema.png

<?php
$connect = new MySQLi($DBhost,$DBuser,$DBpass,$DBname);
$querya = "SELECT * FROM statuses ORDER BY `id` DESC";
$result = mysqli_query($connect, $querya);
$ALLDATA = array();

$DBcon2 = new MySQLi($DBhost,$DBuser,$DBpass,$DBname);
if ($DBcon2->connect_errno) {
    die("ERROR : -> ".$DBcon2->connect_error);
}


while ($record = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
    array_push($ALLDATA, $record);
        $queryb = "SELECT * FROM friendslist WHERE idOfPerson1='".$record['idOfUser']."' OR idOfPerson2='".$record['idOfUser']."' OR idOfPerson2='".$userRow['user_id']."' OR idOfPerson1='".$userRow['user_id']."' ORDER BY `id` DESC";
        $result2 = mysqli_query($connect, $queryb);
        $ALLDATA2 = array();

    while ($record2 = mysqli_fetch_array($result2, MYSQLI_ASSOC)) {
        array_push($ALLDATA2, $record2);

            if($record['idOfUser'] == $userRow['user_id']) {
            echo '<div>You Posted on '.$record['whenPosted'].'<br />'.$record['content'].'</div>';
            }
            elseif($record2['idOfPerson1'] == $userRow['user_id']) {
            $query2 = $DBcon2->query("SELECT * FROM tbl_users WHERE user_id='".$record2['idOfPerson2']."'");
            $userRow2=$query2->fetch_array();
            echo '<div>'.$userRow2['username'].' Posted on '.$record['whenPosted'].'<br />'.$record['content'].'</div>';
            }
            elseif($record2['idOfPerson2'] == $userRow['user_id']) {
            $query2 = $DBcon2->query("SELECT * FROM tbl_users WHERE user_id='".$record2['idOfPerson1']."'");
            $userRow2=$query2->fetch_array();
            echo '<div>'.$userRow2['username'].' Posted on '.$record['whenPosted'].'<br />'.$record['content'].'</div>';
            }
        }
mysqli_free_result($result2);
}


$DBcon2->close();
mysqli_free_result($result);

?>

Answer 1

您的架构看起来不错，但让我们再看一下这些关系。为方便起见，我将重命名一些列。

   Users:
+- user_id
|  user_name
|
|
|  Friendships:
|  fid
+- user_id
|  friend_id --------+
|  friendship_start  |
|                    |
|                    |
|  Statuses:         |
|  sid               |
+- user_id ----------+
   post_date
   content

如果您只是想查找朋友的状态，那么查询将如此：

SELECT statuses.content 
FROM friendships, statuses 
WHERE friendship.user_id=? AND 
    friendships.friend_id = statuses.user_id

当您bind声明时，您当然会user_id ->prepare()值。

（请参阅http://php.net/manual/en/mysqli.prepare.php了解正确的sql方法。过去想要做类似mysql_query的事情（＆＃34; select * from table where id = ＆＃34;。$ _ POST [＆＃39; id＆＃39;]）因为它对SQL注入开放了）

不幸的是，这不包括您在查询结果中的状态。我们将不得不在查询上做更多的工作......

SELECT statuses.content
FROM friendships, statuses
WHERE
  ( friends.user_id = ? AND
    friends.friend_id = stuatuses.user_id ) 
  OR
  statuses.user_id = ?
ORDER BY statuses.post_date DESC

到目前为止，这么好......但我们没有姓名或发布日期。发布日期很简单，只需将其添加到选择中： SELECT statuses.content, statuses.post_date

要添加名称，我们还必须从用户处获取数据。

SELECT users.user_name, statuses.content, statuses.post_date
FROM users, friendships, statuses
WHERE 
  users.user_id = ? AND
  (
    ( users.user_id = friendships.user_id AND 
      friendships.friend_id = statuses.user_id )
    OR
    statuses.user_id = users.user_id
  )
ORDER BY statuses.post_date DESC

你有它;数据库为您完成所有工作。不需要嵌套查询等。这只会为您提供在页面上打印的简单列表。请记住，这不在我的头脑中，所以如果我忽略了某些东西，你可能不得不调整它。

循环遍历每个结果会产生多个不正确的结果

1 个答案: