Member Avatar for prawin@123

Hai can any tel me , How i can get username from database where i logged in using userid . here is the code below.

if(isset($_SESSION['u_name'])){
include 'dbcon.php';
$qry = "SELECT * FROM tbl_user where user_id='".$_SESSION['u_name']."'";
$result = mysql_query($qry,$con);
if($result)
{
$row=mysql_fetch_array($result);

}
while($row=mysql_fetch_array($result))
{ 
   echo ' Welcome  ' .$row['user_name'];
} 
}

i want to display only the username of the person who logged in using his userid. The above code displays nothing.

  • 4 Contributors
  • 7 Replies
  • 98 Views
  • 1 Day Discussion Span
  • Latest Post Latest Post by ahmksssv

Dani AI

Generated

The behaviour seen in this thread (no username shown or the first row missing from a result set) is usually a symptom of reading from the query result before you iterate it, and of using the old ext/mysql API. and pointed toward fixes that make the immediate problem go away. Below are practical, modern steps to fix the bug and avoid common security/compatibility problems going forward.

  • Only retrieve the column you need (for example user_name) and add LIMIT 1 when looking up the logged-in user. This is cheaper and clearer than SELECT *.
  • Do not consume a row before your loop. If you need a single username, fetch once and stop; if you need many rows, iterate directly.
  • Move away from the deprecated mysql_* functions. Use prepared statements (mysqli or PDO) to avoid SQL injection instead of manual escaping.
  • Always validate the session value (type and presence) before using it in a query, and escape output with htmlspecialchars() when echoing names into HTML to avoid XSS.

Example using PDO (prepared statement + safe output):

<?php
// assume session_start() has already run and $_SESSION['user_id'] is set
$pdo = new PDO('mysql:host=localhost;dbname=your_db;charset=utf8mb4','dbuser','dbpass',[
    PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION,
]);
$stmt = $pdo->prepare('SELECT user_name FROM tbl_user WHERE user_id = :id LIMIT 1');
$stmt->execute([':id' => $_SESSION['user_id']]);
$user = $stmt->fetch(PDO::FETCH_ASSOC);
if ($user) {
    echo 'Welcome ' . htmlspecialchars($user['user_name'], ENT_QUOTES, 'UTF-8');
}

For reference on prepared statements and safe output: see the PHP docs for PDO prepared statements and htmlspecialchars(). Also confirm session_start() runs before checking $_SESSION so the stored id is available. Thanks to and for the troubleshooting direction in the thread.

Recommended Answers

All 7 Replies

Member Avatar for Shanti C

Don't use this statement $row=mysql_fetch_array($result); twice in your code...
Just remove first one , then you will find your expected result...

Member Avatar for cwarn23

Try the following:

if(isset($_SESSION['u_name'])){
include 'dbcon.php';
$qry = "SELECT * FROM `tbl_user` where `user_id`='".mysql_real_escape_string($_SESSION['u_name'])."'";
$result = mysql_query($qry,$con) or die(mysql_error());
if(mysql_num_rows($result)>0) {
    while($row=mysql_fetch_array($result))
        { 
        echo ' Welcome ' .$row['user_name'];
        }
    }
}

Note: Please use code tags.

Member Avatar for prawin@123

Hi cwarn23, I am really glad bcoz ur solution to me solved the problem.
YOU r the man . Please keep doing this job for begineers like me.
Again Thanks a lot to u.

Member Avatar for prawin@123

Hi any one please solve this,
I am new to php, i displayed the datas from db , via select ,
whats the problem i got is , i didnt got the first entry of the database , But it displays Rest of the elements in db.
Here is the code

<?php
/*$link = mysql_connect('localhost', 'root', ''); 
if (!$link) {
  die('Could not connect: ' . mysql_error());
}
mysql_select_db("b2bassociates");*/
include 'dbcon.php';

$qry = "select * from tbl_location order by loc_name";
$result = mysql_query($qry,$con);
if($result)
{
$row=mysql_fetch_array($result);

}
?>
<select name="sel_loc" title="Select Location">
<option value="">----Select Location----</option>
<?php 
while($row=mysql_fetch_array($result))
{ 
?>
   <option value="<?php echo $row['loc_code']; ?>" 
   >
   <?php echo $row['loc_name']; ?>
   </option>
<?php 
} 
?>

ie if i have 5 names in the above table named tbl_location , it displays only 4. Please help me.

Member Avatar for ahmksssv

Hi any one please solve this,
I am new to php, i displayed the datas from db , via select ,
whats the problem i got is , i didnt got the first entry of the database , But it displays Rest of the elements in db.
Here is the code

  1. <?php

  3. /*$link = mysql_connect('localhost', 'root', '');

  5. if (!$link) {

  7. die('Could not connect: ' . mysql_error());

  9. }

  11. mysql_select_db("b2bassociates");*/

  13. include 'dbcon.php';

$qry = "select * from tbl_location order by loc_name";
$result = mysql_query($qry,$con);
if($result)
{
$row=mysql_fetch_array($result);

}
?>
<select name="sel_loc" title="Select Location">
<option value="">----Select Location----</option>
<?php
while($row=mysql_fetch_array($result))
{
?>
<option value="<?php echo $row; ?>"
>
<?php echo $row; ?>
</option>
<?php
}
?>

ie if i have 5 names in the above table named tbl_location , it displays only 4. Please help me.

hi...
plz use code tags....here u have to check 2 ways....
1) u wrote order by loc_name, check location is empty or not?
2)u have to use mysql_fetch_array() only once...
try to use this below code..

<?php
/*$link = mysql_connect('localhost', 'root', ''); 
if (!$link) {
  die('Could not connect: ' . mysql_error());
}
mysql_select_db("b2bassociates");*/
include 'dbcon.php';

$qry = "select * from tbl_location order by loc_name";
$result = mysql_query($qry,$con);
?>
<select name="sel_loc" title="Select Location">
<option value="">----Select Location----</option>
<?php 
while($row=mysql_fetch_array($result))
{ 
?>
   <option value="<?php echo $row['loc_code']; ?>" 
   >
   <?php echo $row['loc_name']; ?>
   </option>
<?php 
} 
?></select>
Member Avatar for prawin@123

Thank u ahmksssv,
It works good for me, I am really appreciate ur work, Thanks a lot , Keep it up.

Member Avatar for ahmksssv

Thank u ahmksssv,
It works good for me, I am really appreciate ur work, Thanks a lot , Keep it up.

Welcome....plz mark as solved this thread....

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