,所以我想从互连到其他3台 用户会选择$ FAID,将打印数据所需选择内连接4台
表1(DBO表4号特定数据.FAID)
FAID(PK)
PCID(FK)
用户ID(FK)
表2(dbo.users)
用户ID(PK)
EmployeeName
表3(dbo.SubDeptTransfer)
TransferID(PK)
用户ID(FK)
SubDeptID(FK)
表4(SubDept)
SubDeptID(PK)
DEPTID(FK)
表5(部门)
DEPTID(PK)
部
<?php
$faidf=$_POST['faidf'];
ini_set("display_errors","on");
$conn = new COM("ADODB.Connection");
try {
$myServer = "WTCPHFILESRV\WTCPHINV";
$myUser = "sa";
$myPass = "[email protected]";
$myDB = "wtcphitinventory";
$connStr = "PROVIDER=SQLOLEDB;SERVER=".$myServer.";UID=".$myUser.";PWD=".$myPass.";DATABASE=".$myDB;
$conn->open($connStr);
if (! $conn) {
throw new Exception("Could not connect!");
}
}
catch (Exception $e) {
echo "Error (File:): ".$e->getMessage()."<br>";
}
if (!$conn)
{exit("Connection Failed: " . $conn);}
echo "<center>";
echo "<table border='0' width ='100%' style='margin-left:90px'><tr><th></th><th></th></tr>";
$sql_exp = "SELECT e.Department
FROM dbo.FA_PC a
INNER JOIN dbo.users b
on a.UserID = b.UserID
INNER JOIN dbo.SubDeptTransfer c
ON a.UserID = c.UserID
INNER JOIN dbo.SubDept d
ON a.SubDeptID = d.SubDeptID
INNER JOIN dbo.department e
ON a.DeptID = e.DeptID
WHERE a.FAID = $faidf";
$rs = $conn->Execute($sql_exp);
echo "<tr><td>".$rs->Fields("Department")."</tr></td>";
$rs->Close();
?>
所有我能得到的是“无效的列名称SubDeptID”,这是IM肯定的是列名是正确的,虽然我认为我乱用我的select语句
FAID->用户 - > subdepttransfer-> subdept - >部门
有多少内部连接已经取得或不能执行超过3个表什么冲突?如果是的话,有没有办法连接5个表格?
有妳数据库中创建你的表之间的关系 – freaky 2013-02-11 05:29:18
是与他们的外键 – Yinks 2013-02-11 05:30:32
沿着互连,因为你使用的是SQL服务器,我建议你尝试设计查询编辑器你的sql服务器。这可以帮助您查询 – freaky 2013-02-11 05:34:31