PHP，僅從數據庫中選擇登錄的用戶名

Question

我基本上希望它僅選擇已登錄的當前用戶，然后在此處顯示用戶名和點，而不是顯示數據庫中的所有用戶

這是顯示它的代碼

    <?php
include("connect.php"); // Includes the file connect.php to connect to database
session_start(); // Starting session cookies
if($_SESSION['LOGGEDIN'] == 1)  //Checking if they have the session cookie
{


$result = mysql_query("SELECT * FROM userdata");

echo "<table border='1'>
<tr>
<th>Username</th>
<th>Points</th>
</tr>";

while($row = mysql_fetch_array($result))
  {
  echo "<tr>";
  echo "<td>" . $row['username'] . "</td>";
  echo "<td>" . $row['points'] . "</td>";
  echo "</tr>";
  }
echo "</table>";

mysql_close($con);
}

else
{
    echo "<title>Error!</title>";
    //Doesn't have session cookie
    echo "YOU ARE NOT LOGGED IN!";
}
?>

Answer 1

好吧，您正在選擇整個表格。

只需將其更改為

$result = mysql_query("SELECT * FROM `userdata` WHERE `username`='".$_SESSION["LOGGEDIN"]."' LIMIT 1");

請注意，您應該更改會話變量LOGGEDIN以包含已登錄用戶的用戶名，或者使用另一個會話變量，並替換我在上面的代碼行中對LOGGEDIN引用。

例如，在您的登錄腳本中，不要執行以下操作：

if($_POST["user"] == $user and $_POST["password"] == $pass)
$_SESSION["LOGGEDIN"] = 1;

做這個：

if($_POST["user"] == $user and $_POST["password"] == $pass)
$_SESSION["LOGGEDIN"] = $user;

如果確實使用LOGGEDIN ，則將需要更新初始的if子句，以便它不會檢查它是否等於1，而是檢查它是否已設置：

if(isset($_SESSION["LOGGEDIN"]))

因此，您的文件應如下所示：

<?php
include("connect.php"); // Includes the file connect.php to connect to database
session_start(); // Starting session cookies
if(isset($_SESSION['LOGGEDIN']))  //Checking if they have the session cookie
{


$result = mysql_query("SELECT * FROM `userdata` WHERE `username`='".$_SESSION["LOGGEDIN"]."' LIMIT 1");

echo "<table border='1'>
<tr>
<th>Username</th>
<th>Points</th>
</tr>";

while($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['username'] . "</td>";
echo "<td>" . $row['points'] . "</td>";
echo "</tr>";
}
echo "</table>";

mysql_close($con);
}

else
{
    echo "<title>Error!</title>";
    //Doesn't have session cookie
    echo "YOU ARE NOT LOGGED IN!";
}
?>

Answer 2

<?php 

        include("connect.php");
        $email= $_POST['userid'];
        $password= $_POST['password1']; 
        $papas=base64_encode($password);
        $check = $_POST['rememberme'];      
            $tablename="userdata";
        $select_qry = $jeob->SqlQuery("SELECT * FROM ".$jeob->dbprefix.$tablename." WHERE email ='$email' AND password ='$papas' AND active_link='1' ");
            if($jeob->SqlRows($select_qry) == "0"){                 
             echo "Invalid Username and Password";
            } else {

            $getuser = $jeob->SqlFetch($select_qry);
            $_SESSION['userid'] = $getuser['user_id'];  
            $_SESSION['oauth_provider'] = "normal";
            $_SESSION['email'] = $getuser['email']; 
            }
                if($_SESSION['userid'] == ""
                {
                echo "You are not logged in";
                }
                else
                {
                Welcome "Fetch username using the session id or emial"
                }
    ?>

希望這對您有用