[SOLVED] adding username to image

[roopurt18]

Yes, but just to be sure you didn't leave out the other call to mysql_query(), here it is in completeness.

          $sql = "DELETE FROM `user_images` WHERE `user_id`={$Clean['user_id']}";
          $q = mysql_query($sql);
          if($q === FALSE){
            dbg_out('ERROR: ' . mysql_error());
          }else{
            dbg_out('Database delete successful');
          }
          $sql = "
            INSERT INTO `user_images` 
              (`user_id`, `ext`, `created`, `modified`)
            VALUES (
              {$Clean['user_id']}, {$Clean['ext']}, NOW(), NOW()
            )
          ";
          $q = mysql_query($sql);
          if($q === FALSE){
            dbg_out('ERROR: ' . mysql_error());
          }else{
            dbg_out('Database insert successful');
          }

[runnerjp]

hey i did it and it doesnt delete the image

Array
(
    [user_id] => 1
    [logged_in] => 1
)

Array
(
    [image] => Array
        (
            [name] => logoo.bmp
            [type] => image/bmp
            [tmp_name] => /tmp/phpkbw3Ja
            [error] => 0
            [size] => 135054
        )

)

image/bmp

/home/runningp/public_html/members/images/1.bmp

SUCCESS!

Database delete successful

Database insert successful 

[roopurt18]

Well, if it deletes the old image and then inserts a new one, the record will still be there.  To test it, upload a .gif followed by a .jpg and make sure in the DB the record changes from gif to jpg.  Or just check that the created field has changed.

 

 

[runnerjp]

it does chnage in the db sorry lol

 

what is the next part to follow then

[roopurt18]

That's it as far as uploading is concerned.

 

The next part is to display the avatars where appropriate.  That should be as easy as selecting a user's avatar from the database and creating an img tag.

[runnerjp]

what you mean by creating an img tag.

[roopurt18]

Well these are avatars.  I assume they're meant to be displayed while viewing a member's profile or forum post.  In order to do so you have to SELECT the data from the database and then create an HTML img tag.

[runnerjp]

hey i treid this code to show image but gets notting and no error shown :S

 

$sql = "SELECT * FROM user_image WHERE user_id=".$_GET["id"];
$result = mysql_query ($sql, $conn);
if (mysql_num_rows ($result)>0) {
$row = @mysql_fetch_array ($result);
$image_type = $row["ext"];
Header ("Content-type: $image");
print $image;
}
?>

[runnerjp]

is my code correct?

[roopurt18]

If it were correct it would be working, which you already said it wasn't.  You're using a variable $image in your header() call but I don't see where you're setting it.

 

Anyways, you don't have to make any calls to header().  The file has the extension attached so the browser will know what to do with it.  All you need to do is look up the record in the table for the appropriate extension.

 

<?php
$sql = 'SELECT `ext` FROM `user_image` WHERE `user_id`=' . $_GET['id'];
$q = mysql_query($sql);
if($q){
  $row = mysql_fetch_array($q);
  if(!empty($row)){
    echo '<img src="http://www.yourdomain.com/members/images/'
         . $_GET['id'] . '.' . $row['ext']
         . '" />';
  }
}else{
  echo 'Error: Image not found.';
}
?>

[runnerjp]

humm it not selecting any img... it just echos image not found even though 1 is uploaded

[roopurt18]

Try echo'ing the SQL statement and running it directly in phpMyAdmin.  Try echo'ing mysql_error().

 

At some point you need to learn how to debug your application.  Other than copying and pasting what I put, running it once, and then coming back to say it doesn't work, what have you done to try and troubleshoot it yourself?

[runnerjp]

well i have been echoing it and i have found that

 

$q = mysql_query($sql);

if($q){

  $row = mysql_fetch_array($q);

 

is the error

[runnerjp]

mesingarround i got #1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '<?php

$sql = 'SELECT `ext` FROM `user_image` WHERE `user_id`=' . $_GET['id']' at line

[roopurt18]

It seems that the PHP snippet I gave you is somehow being sent as the query.  How are you calling this code?

[runnerjp]

calling it... surly the code alone will show the image??

[roopurt18]

That error message is telling you that this is your query:

'<?php
$sql = 'SELECT `ext` FROM `user_image` WHERE `user_id`=' . $_GET['id']'

 

Whatever you are doing, mysql is running something like this:

mysql_query("<?php\n$sql = 'SELECT `ext` ...");

which is surely wrong.

 

So if this code is in a single file that you are referring to in your browser, yah it should work.  But if you are including() or requiring() that code I gave you, that could explain it.

[runnerjp]

no single file like so...

<?php
  
  session_start();
  require_once '../settings.php';

  // see if a name was passed via the url.
  if (isset($_GET['username'])) {
    $username = mysql_real_escape_string($_GET['username']);
  } elseif (isset($_SESSION['username'])) {
    $username = $_SESSION['user_id'];
  } else {
    die("No profile to search for");
  }
  
  $query = "SELECT * FROM users WHERE Username = '$username' LIMIT 1"; 
  if ($result = mysql_query($query)) {
    if (mysql_num_rows($result)) { 
      $array = mysql_fetch_assoc($result);
      $pemail = $array['Email'];
      $puser = $array['Username'];
    } else {
      echo "No users found with id $id<br />";
    }
  } else {
    echo "Query failed<br />$sql<br />" . mysql_error();
  }

?>
<?php $sql = 'SELECT `ext` FROM `user_image` WHERE `user_id`=' . $_GET['id'];


$q = mysql_query($sql);
if($q){
  $row = mysql_fetch_array($q);
  
	 echo 'emma';
  if(!empty($row)){
    echo '<img src="http://www.runningprofiles.com/members/images/'
         . $_GET['id'] . '.' . $row['ext']
         . '" />';
  }
}else{
  echo 'Error: Image not found.';
}
?>

so it should work

[roopurt18]

What does this print:

<?php
  
  session_start();
  require_once '../settings.php';

  // see if a name was passed via the url.
  if (isset($_GET['username'])) {
    $username = mysql_real_escape_string($_GET['username']);
  } elseif (isset($_SESSION['username'])) {
    $username = $_SESSION['user_id'];
  } else {
    die("No profile to search for");
  }
  
  $query = "SELECT * FROM users WHERE Username = '$username' LIMIT 1"; 
  if ($result = mysql_query($query)) {
    if (mysql_num_rows($result)) { 
      $array = mysql_fetch_assoc($result);
      $pemail = $array['Email'];
      $puser = $array['Username'];
    } else {
      echo "No users found with id $id<br />";
    }
  } else {
    echo "Query failed<br />$sql<br />" . mysql_error();
  }

echo '<pre style="text-align: left;">id: '  . $_GET['id'] . '</pre>';
$sql = 'SELECT `ext` FROM `user_image` WHERE `user_id`=' . $_GET['id'];
$q = mysql_query($sql);
echo '<pre style="text-align: left;">sql: '  . $sql . '</pre>';
echo '<pre style="text-align: left;">q: '  . $q . '</pre>';
echo '<pre style="text-align: left;">error: '  . mysql_error() . '</pre>';
if($q){
  $row = mysql_fetch_array($q);
  
	 echo 'emma';
  if(!empty($row)){
    echo '<img src="http://www.runningprofiles.com/members/images/'
         . $_GET['id'] . '.' . $row['ext']
         . '" />';
  }
}else{
  echo 'Error: Image not found.';
}
?>

[runnerjp]

id:

 

sql: SELECT `ext` FROM `user_image` WHERE `user_id`=

 

q:

 

error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1

 

 

why is it not getting the id??

[roopurt18]

What does the URL look like?

[runnerjp]

well u gotta be signed in lol  www.runningprofiles.com

 

username test

password test

 

 

then go to update profile and will show you the page

[runnerjp]

why will it not work?

[runnerjp]

did u try it out?

[roopurt18]

It won't work because the URL looks like this:

http://www.runningprofiles.com/members/test

 

Where if you're going to use $_GET it needs to look like this:

http://www.runningprofiles.com/members/test?id=<users_number_id>

 

So either you change the URL or you change the script to not use $_GET.  I recommend changing the script to not use $_GET since you are already passing the username into it.