58

我有 2 个表如下

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       ║
╚══════════╩═════════════════╝

我正在查询我的 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，但我无法更改该项目的数据库结构。

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：Executive（而不是Executive、Corp Admin、Art）

这可以通过 CONCAT 或 FIND_IN_SET 来完成吗？

请帮忙！提前致谢！

4

2 回答 2

119 
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
于 2013-09-30T18:53:17.553 回答
14 
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             ║
 ╚══════════╩════════════════════════════╝
于 2015-04-29T09:47:57.467 回答