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);
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?