PHP的AJAX聊天问题

Question

这是我的PHP代码...........

$root = "http://localhost/"; 
$sql = mysql_query("SELECT * FROM mychat WHERE (too='$mid' AND froom='$uid') OR (too='$uid' AND froom='$mid') ORDER BY id ASC"); 
    while($ro=mysql_fetch_array($sql)){ 
     $from = $ro['froom']; 
     $to = $ro['too']; 
     $text = $ro['text']; 
     $time = $ro['time']; 
     $last_count = $ro['last_count']; 
     echo '<li class="chat_li"> 
     <font face="MS Sans Serif" size="1">'.$froom.'</font></a> says: &nbsp; '.wordwrap($text, 1000,"<br>\n", true).' 
      <br><font size="1" color="#777777">'.$time.'</font> 
     </li>'; 
} 

但只显示用户1名，所有文本，但并没有显示文字信息相应的用户。 ...........

我怎样才能解决这个问题

我的数据库表.........

id, froom, too, text, time 

，这里是我的输出.. ........ 如果user1类型........ [ssss]和user2类型.....

输出总是。

user1says：SSSS 1min之前

user1says：1244 50秒前

Answer 1

试试这个

$root = "http://localhost/"; 
    $sql = mysql_query("SELECT * FROM mychat WHERE (too='$mid' AND froom='$uid') OR (too='$uid' AND froom='$mid') ORDER BY id ASC"); 

    $ro= mysql_fetch_array($sql); 
    for($i=0;$i< count($ro);$i++) 
    {   
      $to[$i] = $ro[$i]['too']; 
      $last_count[$i] = $ro[$i]['last_count']; 
      echo '<li class="chat_li"> 

      <font face="MS Sans Serif" size="1">'.$ro[$i]['froom'].'</font></a> says:.wordwrap($ro[$i]['text'], 1000,"<br>\n", true).' 
        <br><font size="1" color="#777777">'.$ro[$i]['time'].'</font> 
      </li> 

}