Question

我有一个消息页面，我只需要从每个用户获取最后一条消息，我的数据库如下：msg_id，message，userid，toid，datecreated 它用于php聊天，我只需要每个用户的最后一条消息

我以这种方式获得结果

 <?php
 $sqlmessages = mysql_query("SELECT * FROM messages WHERE toid=$logid");

 while($runrows = mysql_fetch_assoc($sqlmessages))
 {

 $aid = $_POST ['m_id'];
 $message = $runrows ['message'];
 $userid = $runrows ['uid'];
 $toid = $runrows ['toid'];
 $date = $runrows ['datecreated'];


 ?>            

 <?php echo $message;?>


 <?php } ?>

我需要更改sql以获取来自数据库的最后一条消息，任何想法？

Answer 1

试试这个：

$sqlmessages = mysql_query("SELECT * FROM messages WHERE toid=$logid ORDER BY msg_id DESC LIMIT 1");

修改

然后这样做：

$sqlmessages = mysql_query("SELECT * FROM messages WHERE toid=$logid GROUP BY(userid) ORDER BY msg_id DESC");

Answer 2

试试这个：我假设$userid是您用户的字段

$query = "SELECT * FROM messages WHERE toid=$logid GROUP BY $userid ORDER BY $date DESC";
$result = mysql_query($query) or die(mysql_error());

while($row = mysql_fetch_array($result)){
    echo $row['$userid']. " - $". $row['$message'];
    echo "<br />";
}

Answer 3

集思考：

SELECT
    u.user_name,
    u.uid,
    m.m_id,
    m.toid,
    m.datecreated
FROM
    users u
LEFT JOIN
    /* auxilary table to create set of uid with max m_id*/
    (SELECT
        uid,
        MAX(m_id) max_id
    FROM
        messages
    GROUP BY uid
    ) a
    ON u.uid = a.m_id
LEFT JOIN
    /* now join the messages again to get the latest message */
    messages m
    ON a.uid = m.uid AND a.m_id = m.m_id

如果只需要有消息的用户在INNER join

中更改LEFT

如何仅从表中获取最后一个条目

3 个答案: