Question

我正在使用登录创建一个小帐户，现在我必须显示登录的ID名称。假设我现在有两个帐户进入我的数据库，因为我打印他的名字，它显示ID no 。我使用ID号登录时的名字。你能告诉我发生了什么吗？在哪里我认为是错的？，

这是我的change_setting_db.php：

<?php
$con=mysqli_connect("localhost","root","Bhawanku","members");
// Check connection

if (mysqli_connect_errno())
{
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM admin");
?>

这是我的general_setting.php：

<div id="change_name">
    <label><strong>Name: </strong></label>
        <?php
        include('change_setting_db.php');

        while($row = mysqli_fetch_array($result))
        {
            echo $row['first_name']." ".$row['last_name'];
        }
        ?>
        <a id="display_float" href="change_name.php">Edit</a>
    </div><hr>

EDITED

我试过但它不起作用..

<?php
$con=mysqli_connect("localhost","root","Bhawanku","members");
// Check connection

if (mysqli_connect_errno())
{
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$result = mysqli_query($con,"SELECT * FROM admin");
if ($row = mysqli_fetch_array($result)) {
    $id=$row['id'];
mysqli_query($con,"SELECT * FROM admin WHERE id='$id' ");
}
?>

Answer 1

当您在$result = mysqli_query($con,"SELECT * FROM admin");中对数据库进行查询时，您需要传递存储在会话变量中的用户ID。

看看这个：

$uid = $_SESSION['uid'];
$result = mysqli_query($con, "SELECT * FROM admin WHERE uid = '$uid'");

Answer 2

你的SQL查询应该是

SELECT * FROM admin WHERE id='1';

Answer 3

为什么您希望此代码显示登录用户，因为您正在执行：

SELECT * FROM admin

这将检索表中没有条件的所有行。您需要添加WHERE子句，例如：

SELECT * FROM admin WHERE user_id = ?

通常，登录用户的user_id将存储在会话中。还要记住使用预准备语句，而不是将user_id直接连接到查询中。

Answer 4

这里传递＆＃34; ID＆＃34;用户在查询条件的位置，如果您想要特定用户名。

$con=mysqli_connect("localhost","root","Bhawanku","members");
//Check connection

if (mysqli_connect_errno())
{
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM admin where id=(pass your id's value here)");
?>

或

如果登录凭据正确，您可以将ID保留在会话中。
当你手中有id时，只需点击
之类的条件选择查询
$ result = mysqli_query（$ con，＆＃34; select * from admin其中id =＆＃39;。＆＃34; $ id＆＃34;。＆＃39;＆＃34;）; 如果（计数（$结果大于0） { //现在显示该用户的名称 echo＆＃34; Id =＆＃34;。$ result [0] [＆＃39; id＆＃39;]; echo＆＃34; Name =＆＃34;。$ result [0] [＆＃39; first_name＆＃39;]。＆＃34; ＆＃34; $结果[0] [＆＃39;如first_name＆＃39;]; } 其他 { //在这里处理错误情况 echo＆＃34;未找到结果＆＃34 ;; }

Answer 5

SELECT * FROM admin WHERE username = $username AND password = $password);

Row ID:<?=  $row['id'] ?>

Answer 6

我认为这应该会帮助你，

<!DOCTYPE html>
<html>
<head>
</head>
<body id=""body>
    <div id="body-container">
        <form action="index.php" method="POST">
            <input type="text" id="username" name="username" value="" autocomplete="false" spellcheck="false" placeholder="Username" />
            <input type="password" id="password" name="password" value="" autocomplete="false" spellcheck="false" placeholder="Password" />
            <input type="submit" id="submit" name="submit" value="submit" />
        </form>
    </div>

    <?php
        $username = $_POST['username'];
        $password = $_POST['password'];

        $database = mysqli_connect("database-ip", "database-username", "database-password", "database-name", "database-port"); //fill in the examples correctly!

        $fetch_username = mysqli_query($database, "SELECT * FROM login WHERE username='$username'");

        $fetch_password = mysqli_query($database, "SELECT * FROM login WHERE password='$password'");

        $check_username = mysqli_fetch_array($fetch_username);
        $check_password = mysqli_fetch_array($fetch_password);

        if($username = $check_username && $password = $check_password){
            echo "loged in";
        }

        if($username != $check_username || $password != $check_password){
            echo "incorrect username or password"; 
        }
    ?>


</body>
</html>

我真的建议你查看官方的php文档：

http://www.php.net/manual/en/

我希望这能帮到你（=

如何使用id显示数据库中的名称？

6 个答案: