User not found in script :/

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dalle1996
Posts: 8
Joined: Tue Dec 11, 2012 1:25 pm

User not found in script :/

Post by dalle1996 »

Well, i am doing an avatar upload for profile but somehow, it always says that the user is not found and i have no idea why..
Here is my code:

I splitted it up so it will echo out the image with the first script and echo out the users image in the second script

[syntax=php]<?php
$username = isset($_SESSION['username']);
$query = mysql_query("SELECT * FROM `users` WHERE `username`='$username'");

if (mysql_num_rows($query)==0)
die("User not found!");
else
{

$row = mysql_fetch_array($query);
$location = $row['profile'];

echo "<img src='$location'>";
}
?>
<?php
isset($_SESSION['username']);
$username = $_SESSION['username'];

if (isset($_POST['submit']))
{
$name = $_FILES['myfile']['name'];
$tmp_name = $_FILES['myfile']['tmp_name'];

if ($name) {

$location = "images/$name";
move_uploaded_file($tmp_name,"images/profile/".$name);

$query = mysql_query("UPDATE `users` SET profile='$location' WHERE `username`='$username'");

die("Your images have been has been uploaded!");
} else die("Please select a file!");
}

echo "Upload your image:

<div class='profile'>
<form action='' method='post' enctype='multipart/form-data'>
File: <input type='file' name='myfile'> <input type='submit' name='submit' value='Upload!'>
</form>
</div>
";
?>[/syntax]
Image
ExtremeGaming
Posts: 205
Joined: Mon Jul 09, 2012 11:13 pm

Re: User not found in script :/

Post by ExtremeGaming »

$username will return the value of true because you are using isset(); Remove that and give it a try.
<?php while(!$succeed = try()); ?>
dalle1996
Posts: 8
Joined: Tue Dec 11, 2012 1:25 pm

Re: User not found in script :/

Post by dalle1996 »

Yes thanks, it worked (Y)
Image
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