Question

我正在PHP mysqli中创建两个用户对话视图脚本。当我作为最后一条消息发送消息时，我的问题是我的用户名显示在URL上，因此，当我讨论URL上的其他用户名时，我想显示它。而且我还想从用户表中显示我讨论过的用户个人资料图片的个人资料图片。

数据库pm表

id  from_id   from_name      to_id   to_name     msg               sent_date
1   2         john           3       master      hi how are you?   2019-12-05 04:14:20
2   3         master         2       john        fine              2019-12-05 05:15:58
3   2         john           3       master      hi                2019-12-05 03:20:34
4   5         previn         2       john        hi                2019-12-05 08:30:40

users表

userid | username | profile_pic

这是我的网址

 <a href="cons.php?to_id=<?php echo $guaranteed_from_id ?>&to_name=<?php echo $row['from_name'];?>">Replay</a>

这是源代码

<?php
 if (isset($_SESSION['userid'])) {
    $session_id = $_SESSION['userid'];
  }

  if ($stmt = $con->prepare("SELECT * FROM pm WHERE from_id = ? OR to_id = ? ORDER BY sent_time DESC")) {
    $stmt->bind_param('ii', $session_id, $session_id);
    $stmt->execute();
  }

  $tempArray = array();

  $result = $stmt->get_result();

  if ($result->num_rows > 0) {
      while ($row = $result->fetch_assoc()) {

        if (!in_array($row['to_id'].$row['from_id'], $tempArray)) {

          echo "<br>";
          echo $row['from_id']." - " . $row['to_id']." ". $row['msg']. " - " .$row['sent_time'];


        $guaranteed_from_id = str_replace($session_id, null, $row['to_id'].$row['from_id']);


        ?>
<img src="images/<?php echo $row['profile_pic'];?>" height="20px" width="20px"/>
        <a href="cons.php?to_id=<?php echo $guaranteed_from_id ?>&to_name=<?php echo $row['from_name'];?>">Replay</a>

        <?php

 }


        array_push($tempArray, $row['from_id'].$row['to_id']);
        array_push($tempArray, $row['to_id'].$row['from_id']);

      }
  } else {
      echo "NO MESSAGES";
  }

?>

Answer 1

因此，首先，您根本不需要在pm表中拥有用户名。因此，请从“ pm”表中删除这些列。这是因为您可以使用用户表中的id来链接“ from_id”和“ to_id”。

PM表：

id  from_id     to_id    msg               sent_date
1   2           3        hi how are you?   2019-12-05 04:14:20
2   3           2        fine              2019-12-05 05:15:58
3   2           3        hi                2019-12-05 03:20:34
4   5           2        hi                2019-12-05 08:30:40

用户表：

id    username   profile_pic         
1     john       <link to profile picture>
2     master     <link to profile picture>          
3     john       <link to profile picture>          
4     previn     <link to profile picture>
4     robin      <link to profile picture>

所以我正在考虑在SQL中使用JOIN。但是当必须同时选择to_id和from_id时，并没有真正使它工作。因此想出了一些非常奇怪的东西。我认为您必须以其他方式执行此操作，因为这真的很奇怪。但是我终于做到了。但这不是一个好的解决方案。

解决方案的代码：

    if (isset($_SESSION['userid'])) {
        $session_id = $_SESSION['userid'];
      }

        $sql = "SELECT *,
  (SELECT username FROM users WHERE userid=from_id) AS from_username,
  (SELECT username FROM users WHERE userid=to_id) AS to_username,
  (SELECT username FROM users WHERE userid=?) AS my_username,
  (SELECT profile_pic FROM users WHERE userid=from_id) AS from_profile_pic,
  (SELECT profile_pic FROM users WHERE userid=to_id) AS to_profile_pic,
  (SELECT profile_pic FROM users WHERE userid=?) AS my_profile_pic
  FROM pm WHERE from_id = ? OR to_id = ? ORDER BY id DESC";


  if ($stmt = $con->prepare($sql)) {
     $stmt->bind_param('iiii', $session_id, $session_id, $session_id, $session_id);
     $stmt->execute();
   }

   $tempArray = array();

   $result = $stmt->get_result();

   if ($result->num_rows > 0) {
       while ($row = $result->fetch_assoc()) {





         if (!in_array($row['to_id'].$row['from_id'], $tempArray)) {

            $friend_id = str_replace($session_id, null, $row['to_id'].$row['from_id']);

            $title = $row['from_username'].$row['to_username'];
            $num = strlen($title) ;

            $num = $num/2;

            $first_half = substr($title,0, $num);

            $second_half = substr($title, $num);


            if ($first_half == $second_half) {
              $friend_username = $row['from_username'];
            } else {
              $friend_username = str_replace($row['my_username'], null, $row['from_username'].$row['to_username']);
            }

            $friend_profile_pic = str_replace($row['my_profile_pic'], null, $row['from_profile_pic'].$row['to_profile_pic']);

            echo "<br>";
            echo $row['from_username'] . " - " . $row['to_username']." ". $row['msg']. " - " .$row['sent_time'];

         ?>

           <img src="images/<?php echo $friend_profile_pic;?>" height="20px" width="20px"/>
           <a href="cons.php?to_name=<?php echo $friend_username ?>&to_id=<?php echo $friend_id ?>">Reply</a>

         <?php

  }


         array_push($tempArray, $row['from_id'].$row['to_id']);
         array_push($tempArray, $row['to_id'].$row['from_id']);

       }
   } else {
       echo "NO MESSAGES";
   }

如何在用户表的URL和个人资料图片显示上显示消息对话用户名

1 个答案: