在PHP中获取mysql数据库中特定用户的ID

Question

当用户登录时，我想从 there 帐户回显 ID（由于 phpMyAdmin 中的 auto_increment 而创建），这是我的 login.PHP：

<?php

$conn = mysqli_connect("xxxxxx", "xxxxx", "xxxx", "BuyerAccounts");

$Email = $_POST["Email"];
$Password = $_POST["Password"];


$sql_query = "select Buyer_Email from user_info where Buyer_Email like '$Email' and Buyer_Password like '$Password';";

$result = mysqli_query($conn, $sql_query);

if(mysqli_num_rows($result) > 0 ){

$row = mysqli_fetch_assoc($result);
$name = $row["Buyer_Email"]; 
echo "Welcome: Buyer";

}else{
$int = 1;
//echo "Buyer login failed...";
}
}else{
echo "Login failed...";
}
}


mysqli_stmt_close($statement);

mysqli_close($conn);


?>

Answer 1

在您的 sql 查询中添加列名 id。假设您的 id 列名是 ID

$sql_query = "select ID,Buyer_Email from user_info where Buyer_Email like '$Email' and Buyer_Password like '$Password';";
$result = mysqli_query($conn, $sql_query);

if(mysqli_num_rows($result) > 0 ){

$row = mysqli_fetch_assoc($result);
$name = $row["Buyer_Email"]; 
$user_id =  $row['ID'];
echo $user_id;
echo "Welcome: Buyer";

}

由于您在 php 中进行登录，因此使用 $_SESSION 是一个不错的选择。 您需要做的就是在需要使用会话的任何 php 脚本的顶部添加 session_start();。

<?php
session_start();
$conn = mysqli_connect("xxxxxx", "xxxxx", "xxxx", "BuyerAccounts");

$Email = $_POST["Email"];
$Password = $_POST["Password"];
 $sql_query = "select ID,Buyer_Email from user_info where Buyer_Email like '$Email' and Buyer_Password like '$Password';";

    $result = mysqli_query($conn, $sql_query);

    if(mysqli_num_rows($result) > 0 ){

    $row = mysqli_fetch_assoc($result);
    $name = $row["Buyer_Email"]; 
    $user_id =  $row['ID'];

    //using session
    $_SESSION["user_id"] = $user_id;

    echo $user_id;
    echo "Welcome: Buyer";

    }

现在您可以使用 $_SESSION 变量访问 php 脚本中的任何位置。

echo $_SESSION["user_id"] ;

Answer 2

让我们从头开始。您创建一个登录表单，根据值存储会话：

login.php

session_start();
$_SESSION["username"] = $username;

主页面

$username = $_SESSION["username"];
echo "Hi $username";

编辑 2

好的，所以你想检查用户名是否存在，然后回显他们的 ID。无论如何，几乎所有的登录系统都有会话。

登录后，假设您有一个 id 的$_SESSION。

php

session_start();
$id = $_SESSION["id"];

$db = mysqli_connect("xxxxxx", "xxxxx", "xxxx", "BuyerAccounts");
$check = $db->query("SELECT * FROM users WHERE id='$id'"); 
$num_check = $check->num_rows; 
$fetch_check = $check->fetch_object();
$id2 = $fetch_check->id;

if($num_check) {
// User Exists
echo $id2; 
} else {
echo "You don't exist."
} 

请注意，通常情况下，我只会回显$id。但是，OP 要求从数据库中回显 id，所以我回显了 $id2。