Php MySQL使用逗号分隔的值连接两个表

我有如下两张桌子

注释表

╔══════════╦═════════════════╗
║ nid      ║    forDepts     ║
╠══════════╬═════════════════╣
║ 1        ║ 1,2,4           ║
║ 2        ║ 4,5             ║
╚══════════╩═════════════════╝

位置表

╔══════════╦═════════════════╗
║ id       ║    name         ║
╠══════════╬═════════════════╣
║ 1        ║ Executive       ║
║ 2        ║ Corp Admin      ║
║ 3        ║ Sales           ║
║ 4        ║ Art             ║
║ 5        ║ Marketing       ║
╚══════════╩═════════════════╝

我希望查询Notes表，并将“forDepts”列与Positions表中的值相关联

输出应为：

    ╠══════════╬════════════════════════════╣
    ║ 1        ║ Executive, Corp Admin, Art ║
    ║ 2        ║ Art, Marketing             ║
    ╚══════════╩════════════════════════════╝

我知道数据库应该被规范化，但是我不能改变这个项目的数据库结构

这将用于导出包含以下代码的excel文件

<?PHP

    $dbh1 = mysql_connect($hostname, $username, $password); 
    mysql_select_db('exAdmin', $dbh1);


    function cleanData(&$str)
  {
    $str = preg_replace("/\t/", "\\t", $str);
    $str = preg_replace("/\r?\n/", "\\n", $str);
    if(strstr($str, '"')) $str = '"' . str_replace('"', '""', $str) . '"';
  }

    $filename = "eXteres_summary_" . date('m/d/y') . ".xls";

    header("Content-Disposition: attachment; filename=\"$filename\"");
    header("Content-Type: application/vnd.ms-excel");
    //header("Content-Type: text/plain");

    $flag = false;


    $result = mysql_query(
    "SELECT p.name, c.company, n.nid, n.createdOn, CONCAT_WS(' ',c2.fname,c2.lname), n.description 
     FROM notes n 
     LEFT JOIN Positions p ON p.id = n.forDepts
     LEFT JOIN companies c ON c.userid = n.clientId
     LEFT JOIN companies c2 ON c2.userid = n.createdBy"
     , $dbh1);



    while(false !== ($row = mysql_fetch_assoc($result))) {
        if(!$flag) {
            $colnames = array(
                'Created For' => "Created For",
                'Company' => "Company",
                'Case ID' => "Case ID",
                'Created On' => "Created On",
                'Created By' => "Created By",
                'Description' => "Description"
            );
            // display field/column names as first row
            echo implode("\t", array_keys($colnames)) . "\r\n";
            $flag = true;
    }


        $row['createdOn'] = date('m-d-Y | g:i a', strtotime($row['createdOn']));

    array_walk($row, 'cleanData');
    echo implode("\t", array_values($row)) . "\r\n";
  }
  exit;

  ?>

如果您无法更改糟糕的数据库布局，您可能应该在数据库之外进行修复，即在将ID推送到Excel文件之前，查询您获得的数据并用PHP中的文本值交换ID。遗憾的是，恶劣的数据库布局迫使您创建恶劣的布局修复代码。您应该在答案中添加一个解释。这种方法的问题是，当存在大量数据时，查询需要很长时间才能完成。@JohnWoo可以告诉我此查询的性能吗？？。我需要在最低1M记录的非常高的DB上实现这一点？我有一个几乎类似的问题。请你看一下好吗？这真是一个很棒的答案。如果我在几列中需要它怎么办？如果
Notes
表中的一行在
forDepts
列中有一个空值，那么结果将丢失该行。
SELECT  a.nid,
        GROUP_CONCAT(b.name ORDER BY b.id) DepartmentName
FROM    Notes a
        INNER JOIN Positions b
            ON FIND_IN_SET(b.id, a.forDepts) > 0
GROUP   BY a.nid

Table 1
╔══════════╦═════════════════╗
║ nid      ║    forDepts     ║
╠══════════╬═════════════════╣
║ 1        ║ 1,2,4           ║
║ 2        ║ 4,5             ║
╚══════════╩═════════════════╝
Table 2
╔══════════╦═════════════════╗
║ id       ║    name         ║
╠══════════╬═════════════════╣
║ 1        ║ Executive       ║
║ 2        ║ Corp Admin      ║
║ 3        ║ Sales           ║
║ 4        ║ Art             ║
║ 5        ║ Marketing       ║
╚══════════╩═════════════════╝

SELECT * FROM table1 as t1 LEFT JOIN table2 as t2 ON find_in_set(t2.id, 
t1.forDepts)

Output 

 ╠══════════╬════════════════════════════╣
 ║ 1        ║ Executive, Corp Admin, Art ║
 ║ 2        ║ Art, Marketing             ║
 ╚══════════╩════════════════════════════╝