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[SOLVED] adding username to image


runnerjp

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would this work>>

 

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

 

tried it but getting syntax error, unexpected T_VARIABLE one 1st line

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would this work>>

 

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

 

tried it but getting syntax error, unexpected T_VARIABLE one 1st line

 

Use this, instead. It has no syntax errors:

 

<?php 

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

 

Sam

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Sorry, that's because we had an extra ' in the code

 

<?php 

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

 

That ought to work now

 

Sam

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<?php 

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

?>

 

Try it with the 'mysql_error' part on the query. Also with the first if statement it's just if($q)... if $q is what? set, empty etc..?

 

The second if statement is if(!empty($row)), which is like a double negative in English, if $row is not not full.. :P So you might want to try

 

if(isset($row)){ }

 

If that doesn't work you might want a while loop.. Something along the lines of

 

<?php 

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

 

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$sql = "SELECT `ext` FROM `user_image` WHERE `user_id`=".$_SESSION['user_id'];

$q = mysql_query($sql) or die("Error running query:".mysql_error());

if($q){

  $row = mysql_fetch_array($q);

  if(!empty($row)){

    echo "<img src='http://www.yourdomain.com/members/images/".

        $_SESSION['user_id'] . "." . $row['ext']

        . "' />";

  }

}else{

  echo 'Error: Image not found.';

 

gimmies nexpected $end

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Unexpected end means that you forgot to close a statement.  In this case, you didn't close the if statement. 

 

$sql = "SELECT `ext` FROM `user_image` WHERE `user_id`=".$_SESSION['user_id'];

$q = mysql_query($sql) or die("Error running query:".mysql_error());

if($q){

  $row = mysql_fetch_array($q);

  if(!empty($row)){

    echo "<img src='http://www.yourdomain.com/members/images/".

        $_SESSION['user_id'] . "." . $row['ext']

        . "' />";

  }

}else{

  echo 'Error: Image not found.';

}

 

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I don't understand why your querying the database at all, seems a waste of resources.

 

<?php

  if (isset($_SESSION['user_id'])) {
    foreach(array('jpg','gif','png') as $ext) { // add valid file extensions here.
      if (file_exists('members/images/' . $_SESSION['user_id'] . '.' . $ext)) {
        echo '<img src="members/images/{$_SESSION['user_id']}.$ext" />';
      }
    }
  }

?>

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if the url was www.runningprofiles.com/members/user/index.php?username=helraizer (website doesn't actually exist)

 

$usrname = $_GET['username'] - the username would be helraizer. For instance.

 

Then query the database SELECT * FROM `table` WHERE `username`= $username  (or similar)

then use a while loop, like in the last example I gave.

 

Use that to find the id associated with that username.

 

Sam

 

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question ...  can i get the username from whats typed in the url 

 

Yes, and I have previousy provided you with code (for your profile system) that does just that. As I said when I gave you that example, learn from it. Its no good simply having code handed to you unless you understand it. If you understood the code I provided you with over a week ago, you wouldn't be asking this same question.

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i am currently using

$query = "SELECT * FROM users WHERE Username = '$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();
  }

 

could i not just array the id???

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OK BEEN MESISNG AROUND WITH IT AND I HAVE NEARLY GOT IT

 

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

 

BUT THE CODE DOESNT WORK DO TO unexpected T_VARIABLES

 

$pID gets the users id!

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