1

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?,

here is my 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");
?>

and here is my 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

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' ");
}
?>
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6 Answers 6

Reset to default
1

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.

Look at this:

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

i tried your code but it's showing this error: Undefined index: id in C:\Users\Raj\PhpstormProjects\new linkvessel\change_setting_db.php on line 10
0

Why would you expect this code to show the logged in user, given that you are doing:

SELECT * FROM admin

This retrieves all rows from the table with no condition. You need to add a WHERE clause such as:

SELECT * FROM admin WHERE user_id = ?

Typically the logged in user's user_id would be stored in the session. Also remember to use a prepared statement instead of concatenating the user_id directly into the query.

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1 Comment

How to display id using row??
0

your sql query shoulb be 

SELECT * FROM admin WHERE id='1';
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Comments

0 
SELECT * FROM admin WHERE username = $username AND password = $password);

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

and how to pass $username and $password please help :(
In your login form, username field has name='username' and password field has name='password' and on the page that receives the form submission has $username = $_POST['username']; and the same for password
0

Here pass the "ID" of user in where condition for query if you want particular users name.

$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)");
?>

or

  1. you can just keep the id in session if the login credentials are correct.

  2. and when you have id in your hand just fire a select query with where condition like

    $result = mysqli_query($con,"select * from admin where id='."$id".'"); if(count($result>0) { // now display the name of that user echo "Id = ".$result[0]['id']; echo "Name = ".$result[0]['first_name']." ".$result[0]['first_name']; } else { // handle the error condition here echo "no result found"; }

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2 Comments

You are accessing all columns from database table with particular id so you will get all data for that like(ID,FIRST_NAME,LAST_NAME)as per your code.
see my edited it's not working and as you said i did all but it's not working
0

I think this should help you out,

<!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>

i really recommend you check out the official php documentation at:

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

i hoped this helped you out (=

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