在sql中从数据库中选择时如何排除值

2ekbmq32 于 2021-06-25 发布在 Mysql

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如何从已添加到调度程序表中的主题表中排除要显示的主题？我有这个代码，但它仍然显示已添加到调度表的主题。

<?php
    $sem = $_GET['semChooser'];

    $res = mysql_query("SELECT * FROM scheduler WHERE semester = '$sem'") or die(mysql_errno());
   $row=mysql_fetch_array($res);

           $subName = $row['subject_name'];

        $res1 = mysql_query("SELECT * FROM subject WHERE semester IN ('$sem', 3) AND subject_name != '$subName'") or die(mysql_errno());
      while ($row=mysql_fetch_array($res1)) {

      echo "
         <tr class='active'> 
            <td align='center'>
              <button type='submit' data-dismiss='modal' class='btn btn-primary btn-block btn-xs subject'><em class='fa fa-plus'></em> Add</button>
            </td>
            <td>".$row['subject_code']."</td> 
            <td>".$row['subject_name']."</td> 
            <td>".$row['subject_type']."</td>
            <td>".$row['room_type']."</td>
            <td>".$row['subject_category']."</td>
            <td>".$row['year_level']."</td>  
            <td>".$row['units']."</td> 
         </tr> "; }
    ?>

mysql php

来源：https://stackoverflow.com/questions/39899727/how-to-exclude-value-when-selecting-from-database-in-sql

1条答案

按热度按时间

f0brbegy1#

您可以使用subselect并仅通过一个查询获得结果

SELECT * 
      FROM subject 
      WHERE semester IN ('$sem', 3)  
      AND subject_name not in ( SELECT * FROM scheduler 
                                WHERE semester = '$sem' )

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在sql中从数据库中选择时如何排除值

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