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?
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