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pocobueno1388

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Everything posted by pocobueno1388

  1. pocobueno1388
    Try to change your query to: $queryi = "SELECT DISTINCT(album) FROM `photoalbum` WHERE `userid` = '$idpic'";
  2. pocobueno1388
    Are you calling session_start() at the top of your script?
  3. pocobueno1388
    If what Guardian-Mage suggested doesn't fix it, try to use mysql_real_escape_string() on all your variables.
  4. pocobueno1388
    Take a look at MySQL's DATE_FORMAT() function.
  5. pocobueno1388
    <table border=0 cellpadding=8> <tr> <td>Name:</td> <td>--</td> </tr> <tr> <td>Date:</td> <td>--</td> </tr> <tr> <td>Description:</td> <td>--</td> </tr> <tr> <td>Image:</td> <td>--</td> </tr> </table>
  6. pocobueno1388
    You're missing a colon at the end of this line: $extra = new Extra($output)
  7. pocobueno1388
    You never stated your exact problem, but looking at your code you need to change this line: $team = $_GET['team']; To $team = $_POST['team'];
  8. pocobueno1388
    Look at the function str_replace().
  9. pocobueno1388
    You never close your left() function. function left() { So put the closing bracket wherever that function ends.
  10. pocobueno1388
    We are not here to program scripts for you. We are here to help you along with your struggles as you attempt to write the script yourself. If you don't want to try yourself, you should be posting this in the freelance section and paying someone for it. There may be someone here generous enough to take time away from their life and program this for you, but that's really not what this forum is for. Would you like me to point you to some tutorials so you can try to help yourself?
  11. pocobueno1388
    It's mysql_real_escape_string(), not mysql_escape_string(). Try $delete = mysql_query("DELETE FROM mytable WHERE user='".mysql_real_escape_string($_REQUEST['delete'])."'",$c); Also, what is the point of the line below the query? I don't understand why you just have $delete there.
  12. pocobueno1388
    When they visit a page they have to be logged in to view, store the URL of the page they were trying to visit into a session. Then when they log in, just redirect them to whatever page the session has in it.
  13. pocobueno1388
    You want these variables set BEFORE the form is submitted? So in other words, you want them to automatically set as soon as they type text into the fields? If that's what you want, I'm sure you can get it done using AJAX. You can't do that with just PHP. If what your trying to say is that they have already submitted one form and you want to fill the second one with the values of the first, then you would do this: <html> <body> <?php $imageURL = $_POST['imageURL']; $imageHeight = $_POST['imageHeight']; $imageWidth = $_POST['imageWidth']; $URL = "http://localhost/thumbnails/thumbs.php?image=".$imageURL."&height=".$imageHeight."&width=".$imageWidth; ?> <form action="<?php print $URL; ?>" method="GET"> <input type="text" name="imageURL" value="<?php echo $imageURL; ?>"> If I completely missed the question, please explain in more detail.
  14. pocobueno1388
    What is ":cid"? I don't know why your putting a colon in there, I think you mean for it to be "$cid".
  15. pocobueno1388
    You can use a switch statement. It's really not going to shorten your code much, but it was made for situations like this.
  16. pocobueno1388
    You still haven't made it clear to what you want.
  17. pocobueno1388
    Why do you need it as a separate loop? You need to explain exactly what your trying to do.
  18. pocobueno1388
    I only see one loop...? Exactly how do you want them separated?
  19. pocobueno1388
    Why are you echoing $i++? The problem your having now is just an HTML issue. You need to put <td>'s around the $i. while($member = $query->fetchrow()) { echo "<tr>\n"; echo "<td>$i</td>; echo "<td><a href=\"profile.php?id=" . $member['username'] . "\">"; echo ($member['username'] == $player->username)?"<b>":""; echo $member['username']; echo ($member['username'] == $player->username)?"</b>":""; echo "<td>" . $member['money'] . "</td>\n"; echo "<td>" . $member['level'] . "</td>\n"; echo "</tr>\n"; $i++; } Also, you never close any of your <th>'s.
  20. pocobueno1388
    $i = 1; while($member = $query->fetchrow()) { echo "<tr>\n"; echo "<td>$i</td>"; echo "<td><a href=\"profile.php?id=" . $member['username'] . "\">"; echo ($member['username'] == $player->username)?"<b>":""; echo $member['username']; echo ($member['username'] == $player->username)?"</b>":""; echo "<td>" . $member['money'] . "</td>\n"; echo "<td>" . $member['level'] . "</td>\n"; echo "</tr>\n"; $i++; } EDIT: Beat to it.
  21. pocobueno1388
    When you update the database, are you also updating the session? If you are, show the related code.
  22. pocobueno1388
    user $_SERVER['PHP_SELF'] instead fo $PHP_SELF.
  23. pocobueno1388
    Replace the question mark in your query $query = "SELECT user_id, username, pwd, user_email FROM users WHERE user_id = '$user_id' LIMIT 1"; Make sure the query doesn't return an error. It's hard to tell you how to find errors in your code, because you have your own database class. If it comes down to it, echo out the query and try manually putting it in phpmyadmin to see if it works.
  24. pocobueno1388
    Try to echo it out $user_id = $_GET['user_id']; echo $user_id; If that doesn't display anything, double check that "user_id" is in the URL, and is spelled correctly.
  25. pocobueno1388
    Maybe there is an error in your query for the view records script. Try changing $r = @mysqli_query ($dbc, $q); To $r = mysqli_query ($dbc, $q)or die(mysql_error()); I'm not familiar with using "mysqli", so hopefully the error function is the same. Also, is there a reason your suppressing errors with the @?
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