BetterPHP

Post your two inc.php files.

[syntax=php]<?php foreach ($albums as $album) { foreach ($images as $image) { echo'<a href="view_album.php?album_id=', $album['id'],'&image_id=uploads/', $image['album'], '/', $image['id'], '.', $image['ext'], '"><img src="uploads/thumbs/', $image['album'], '/', $image['id'], '.',...

Sounds like you aren't even starting the session. Add session_start(); to the very top of your script right after your first <?php tag. If that does nothing post your two inc.php files I believe it is init.inc.php and user.inc.php I'm not sure as I've never followed this tutorial.

You have parentheses around $username I don't think it matters but might be worth changing.

I believe Temor is on to something as well.

I believe your query should be like this. I have never needed to do such a query before so I don't know. [syntax=sql] function get_posts() { $sql = "SELECT `posts`.`post_id` AS `id`, `posts`.`post_title` AS `title`, LEFT(`posts.post_body`, 512) AS `preview`, `posts`.`post_user` AS `user`, DATE_...

Change all the apostrophes around your column names to backticks `

Change all the apostrophes around your column names to backticks `

"Parse error: syntax error, unexpected $end in /usr/local/www/project/A._____/website + blog/blog/blog_post.php on line 35" Lines 6-10: [syntax=php]if (isset($_POST['user'], $_POST['title'], $_POST['body'])){ add_post($_POST['user'], $_POST['title'], $_POST['body']); header('Location: blo...

I don't see anywhere that you are generating $uid from. Add echo mysql_error() after your update query like this [syntax=php]mysql_query("UPDATE `users` SET `grad` = '{$grad}', `drzava` = '{$drzava}',`fan` = '{$fan}', `website` = '{$website}' WHERE `uid`='{$uid}'"); echo mysql_error();[/sy...

or you can simply [syntax=php]$query = mysql_query("UPDATE `table` SET `field1` = '{$field1}', `field2` = '{$field2}' WHERE `id`='{$id}'"); print_r = $query;[/syntax] print_r is a php function and will give a syntax error the way you are using. As far as I see there is no reason why they ...

Has uid been set in the url? Profile.php?uid=1234

$_SERVER['PHP_SELF'] is a huge security risk, but that's not the point I guess. PHP_SELF looks for the file itself so it is not possible.

I'm pretty sure if he minded, he wouldn't use it in a tutorial.

[syntax=php]`drzava` AS `drzava',[/syntax]

You have an apostrophe after drzava instead of a backtick

There is most likely a problem with your query. Add [syntax=php]echo mysql_error();[/syntax]

You can not have any output before a header. This includes any html or echo/printing. I suspect this is what is causing your error.

And in case the user has javascript disabled you can always put it in a <noscript>

Question to ExtremeGaming; would I be able to use that, but with value? Because IE simply refuses to use PlaceHolder. Yes just change this.placeholder to this.value However, you will need to remove the onblur because it will submit whatever you are setting the default value to, regardless of what y...

If you mean when clicked, try: [syntax=xhtml]<input type="text" placeholder="Enter Message..." onfocus="this.placeholder = ''" onblur="this.placeholder = 'Enter Message...'" />[/syntax] If you really mean hover, try: [syntax=xhtml]<input type="text" ...

-- Solved (ty Helx ) --

Make sure there are no spaces before the <?php tag and after the ?> tag