PhP會話傳遞並結合變量和字符串

Question

我的登錄代碼：

<?php
session_start();
$f_usr= $_POST["userid"];
$f_pswd= $_POST["password"];
$_SESSION['user']=$f_usr;
$con=mysql_connect("localhost","root","");
if(! $con)
{
        die('Connection Failed'.mysql_error());
}
mysql_select_db("finaltest",$con);
$result=mysql_query("select * from user");
while($row=mysql_fetch_array($result))
{
    if($row["username"]==$f_usr && $row["password"]==$f_pswd)
        header('Location: selectdata.php');
    else
        echo"Sorry : $f_usr";
}
?>

selectdata.php

<?php
session_start();
$s= $_SESSION['user'];
// Make a MySQL Connection
mysql_connect("localhost", "root", "") or die(mysql_error());
mysql_select_db("finaltest") or die(mysql_error());
$select="temperature".$s;
// Get all the data from the "example" table
$result = mysql_query("SELECT * FROM $select") 
or die(mysql_error());  

echo "<table border='1'>";
echo "<tr> <th>username</th> <th>password</th> </tr>";
// keeps getting the next row until there are no more to get
while($row = mysql_fetch_array( $result )) {
    // Print out the contents of each row into a table
    echo "<tr><td>"; 
    echo $row['username'];
    echo "</td><td>"; 
    echo $row['password'];
    echo "</td></tr>"; 
} 

echo "</table>";
?>

實際上會話varibale沒有得到解析，我得到一個錯誤：

Notice: Undefined index: user in C:\xampp\htdocs\bars\selectdata.php on line 3

而我還有另一個問題，我想選擇一個名為“ temperaturexyz”的數據庫，其中要存儲在字符串中的溫度，xyz是我通過會話獲取的變量，我想將兩者結合起來，這樣我就可以獲得一個變量我可以在查詢中使用

Answer 1

關於您的會話：您嘗試在登錄代碼中設置的會話變量$_SESSION['user']並未正確設置該變量。 嘗試：

<?php
 session_start();
 $f_usr= $_POST["userid"];
$f_pswd= $_POST["password"];
$_SESSION['user']=$f_usr;
echo $_SESSION['user'] //<-------- see if this echo's out the value you are expecting

至於您的變量：我假設您的意思是想要一個變量名$ temperaturexyz，但您正在動態創建該變量名。 所以

$select="temperature".$s;
$$select = /* Insert whatever value you need here*/ //<-- you can then call this variable like this (ie. $temperaturexyz)