所以我想从表 4 中获取特定数据,该表与其他 3 个表互连,用户将选择 $faid 并打印所需的数据
表 1 (dbo.FAID)
FAID(pk)
PCID(fk)
UserID(fk)
表 2 (dbo.users)
UserID(PK)
EmployeeName
表 3(dbo.SubDeptTransfer)
TransferID(pk)
UserID(fk)
SubDeptID(fk)
表 4 (SubDept)
SubDeptID(PK)
DeptID(fk)
表 5(部门)
DeptID(PK)
部门
<?php
$faidf=$_POST['faidf'];
ini_set("display_errors","on");
$conn = new COM("ADODB.Connection");
try {
$myServer = "WTCPHFILESRV\WTCPHINV";
$myUser = "sa";
$myPass = "P@ssw0rd";
$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();
?>
我所能得到的只是“无效的列名'SubDeptID”,尽管我认为我的选择语句搞砸了,但我确定列名是正确的
FAID->用户->子部门转移->子部门->部门
进行了多少内部连接或它不能执行超过 3 个表是否存在冲突?如果是,有什么方法可以连接 5 个表吗?