我有2张桌子如下
笔记表
╔══════════╦═════════════════╗ ║ 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 ║ ╚══════════╩═════════════════╝
我想查询我的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 ║ ╚══════════╩════════════════════════════╝
我知道数据库应该规范化,但是我不能更改此项目的数据库结构。
这将用于使用以下代码导出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
考试:执行人员(而不是执行人员,公司行政人员,Art)
可以通过CONCAT或FIND_IN_SET完成吗?
请帮忙!提前致谢!
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