我正在使用登录名创建一个小帐户,现在我必须显示登录的ID的姓名.假设,当我打印他的姓名时,我的数据库中现在有两个帐户,显示ID号.我使用ID号登录时为1. 2你能告诉我发生了什么吗?我应该在哪里错了?,
i'm making a small account using login, now i have to display name from ID who is logged in. suppose, i have two accounts into my database now as i print his name, it shows ID no. 1's name while i'm logged in using ID no. 2 can you tell me what's going on? where i'm suppose to be wrong?,
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>已编辑
我尝试过,但是没有用..
i tried but it's not working..
<?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' "); } ?>推荐答案
在$result = mysqli_query($con,"SELECT * FROM admin");中对数据库进行查询时,需要传递存储在会话变量中的用户ID或其他内容.
When you make the query to the database in $result = mysqli_query($con,"SELECT * FROM admin"); you need to pass the user id stored in a session variable or something.
看看这个:
$uid = $_SESSION['uid']; $result = mysqli_query($con, "SELECT * FROM admin WHERE uid = '$uid'");更多推荐
如何使用ID显示数据库中的名称?
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