Question

我有两张表：用户和消息

我想在messages表中加入users表，并从users表中选择用户信息，其中users表中的ID在messages中没有行使用哈希列检查表。

请注意：user_two表中的messages列是users表

中的ID

我试过了，但没有结果。请帮忙：

<?php 

        //Get the friend a user wants to send message
    if(isset($_POST['recipname']) && !empty($_POST['recipname'])){
        $recipname  =   mysqli_real_escape_string($dbc_conn,htmlentities(trim($_POST['recipname'])));

        $message_group_tatable = "messages";

        $sql    =   "

        SELECT      users.id, users.username,users.FirstName ,
                    users.LastName , users.avatar ,
                    users.cell_group

        FROM        users 
        INNER JOIN  $message_group_tatable 
        ON          $table_name.id=$message_group_tatable.user_two

        WHERE       $message_group_tatable.hash = NULL
        AND         users.id    != $message_group_tatable.user_two
        AND         users.username 
        LIKE        '%$recipname%'   
        LIMIT       6


        ";

        $query  =   mysqli_query($dbc_conn,$sql);
        //die(mysqli_error($dbc_conn));
        if(mysqli_num_rows($query) > 0){
        while($row  =   mysqli_fetch_array($query)){
        $name   =   ucfirst($row['FirstName'])." ".ucfirst($row['LastName']);
        $user_id    =   $row['id'];
        $user_name  =   $row['username'];
        $school     =   $row['cell_group'];
        $avatar     =   $row['avatar'];


        ?>
        <div class="selectmeWrapper this">
        <table class="selectme">
          <tr>
            <td><span class="selectmeavtspan"><img class="selectmeavatar" src="uploaded/<?php echo $avatar; ?>" /></span></td>
            <td><span class="univ"><?php echo $name; ?></span></td>
          </tr>
        </table>

    <span class="uiremovable selected" title="pro/<?php echo $user_name;?>">
        <span> <img class="recipavt" src="uploaded/<?php echo $avatar; ?>" /></span>
        <span class="selectedName">
            <?php echo $name; ?>
        <input type="hidden" autocomplete="off" 
            value="<?php echo $user_name ?>" />

    </span>
    <a href="#" id="<?php echo $user_name?>" class="ulCloseSmall <?php echo "Remove ".$name; ?>"><i class="fa fa-times"></i></a>
    </span>
        </div>
        <?php


        }

        }else{
        echo "<p class='noresult'>No Result Found.</p>";    

        }


        }


?>

表messages的表结构

        CREATE TABLE IF NOT EXISTS `messages` (
      `user_one` int(11) NOT NULL,
      `user_two` int(11) NOT NULL,
      `hash` int(11) DEFAULT NULL,
      `id` int(11) NOT NULL AUTO_INCREMENT,
      PRIMARY KEY (`id`)
    ) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=54 ;

和表users

的表结构

    CREATE TABLE IF NOT EXISTS `users` (
      `id` int(11) NOT NULL AUTO_INCREMENT,
      `username` varchar(64) DEFAULT NULL,
      `FirstName` varchar(32) DEFAULT NULL,
      `LastName` varchar(32) DEFAULT NULL,
      `Email` varchar(64) DEFAULT NULL,
      `Password` varchar(32) DEFAULT NULL,
      `Month` varchar(6) DEFAULT NULL,
      `Day` varchar(6) DEFAULT NULL,
      `Year` varchar(11) DEFAULT NULL,
      `Gender` varchar(6) DEFAULT NULL,
      `cell_group` varchar(100) DEFAULT NULL,
      `active` varchar(11) DEFAULT NULL,
      `avatar` text,

      PRIMARY KEY (`id`)
    ) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=21 ;

Answer 1

您可以使用not in子句

   $sql    =   "
    SELECT      users.id, users.username,users.FirstName ,
                users.LastName , users.avatar ,
                users.cell_group

    FROM        users 
    WHERE       users.id  not in (select distinct user_two from " . $message_group_tatable . " )
    AND         users.username 
    LIKE        concat('%', ". $recipname .", '%')   
    LIMIT       6 
  ";

如何连接两个表并在php

1 个答案: