MySQL使用逗号分隔值连接两个表

Question

我有2个表格如下

备注表

??????????????????????????????
? nid      ?    forDepts     ?
??????????????????????????????
? 1        ? 1,2,4           ?
? 2        ? 4,5             ?
??????????????????????????????

职位表

??????????????????????????????
? id       ?    name         ?
??????????????????????????????
? 1        ? Executive       ?
? 2        ? Corp Admin      ?
? 3        ? Sales           ?
? 4        ? Art             ?
? 5        ? Marketing       ?
??????????????????????????????

我希望查询我的Notes表并将'forDepts'列与Positions表中的值相关联.

输出应该是:

    ?????????????????????????????????????????
    ? 1        ? Executive, Corp Admin, Art ?
    ? 2        ? Art, Marketing             ?
    ?????????????????????????????????????????

我知道数据库应该规范化,但我无法更改此项目的数据库结构.

这将用于导出带有以下代码的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;

  ?>

此代码仅输出'forDepts'的第一个值

Exa:执行(代替执行,公司行政,艺术)

这可以通过CONCAT或FIND_IN_SET完成吗？

请帮忙!提前致谢!

Answer 1

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

• SQLFiddle演示

Answer 2

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             ?
 ?????????????????????????????????????????