我有2张表格如下
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。
输出应为:
╠══════════╬════════════════════════════╣
║ 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
Exa:执行(代替执行,公司管理,艺术)
这可以通过CONCAT或FIND_IN_SET完成吗?
请帮忙!提前谢谢!
答案 0 :(得分:72)
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
答案 1 :(得分:9)
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 ║
╚══════════╩════════════════════════════╝