本文介绍了加入3个表以显示某些数据PHP-MSSQL的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
所以我有这个表,我想获得某些数据供用户查看,并能够POST到其他页
so i have this tables and i want to get certain datas for user to view and be able to POST to other page
我不能张贴图像,所以我有打破这一点,请耐心等待。
i cant post images so i have to break this down so please bear with me
第一页
- dbo.users
- pkey(UserID)
- EmployeeName
第二表
- dbo.PC
- pkey(PCID)
- PC_Number
第三表
- dbo.FA_PC
- pkey(FAID)
- fkey(UserID)
- fkey / li>
- dbo.FA_PC
- pkey(FAID)
- fkey(UserID)
- fkey(PCID)
如何以相同的形式显示当前选择的$ rs-> Fields('UserID')的PC_Number,发布到printd.php
how could i display the PC_Number of the currently selected $rs->Fields('UserID') in the same form and still be able to post it on printd.php
我不知道如何连接dbo.users-> dbo.FA_PC-> dbo.PC
i dont know how to connect the dbo.users->dbo.FA_PC->dbo.PC
<form action="printd.php" method="post" target="_blank">
<?php
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);}
$sql_exp = "select * from dbo.users";
$rs = $conn->Execute($sql_exp);
echo "<select name='empt'>";
while (!$rs->EOF) {
set_time_limit(0);
echo "<option value=".$rs->Fields('UserID')." >".$rs->Fields('EmployeeName')."</option>";
$rs->MoveNext();
}
$rs->Close();
?>
推荐答案
the join will look like this,
SELECT a.*, c.PC_Number
FROM users a
INNER JOIN FA_PC b
ON a.UserID = b.UserID
INNER JOIN PC c
ON b.PCID = c.PCID
$ b b
要完全获得关于联接的知识,请访问以下链接:
To fully gain knowledge about joins, kindly visit the link below:
- Visual Representation of SQL Joins
这篇关于加入3个表以显示某些数据PHP-MSSQL的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!