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DTV
Forum Newbie
Posts: 3 Joined: Sat Feb 06, 2010 1:22 am
Post
by DTV » Sat Feb 06, 2010 1:24 am
im trying to show info about the user on the page there id and stuff but for some resson it ent showing
Code: Select all
<?php
mysql_connect("..", "_", "") or die(mysql_error());
mysql_select_db("_") or die(mysql_error());
$sql = ("SELECT username FROM users WHERE username = '$username'")or die(mysql_error());
$result = mysql_query($sql) or die(mysql_error());
$values = mysql_fetch_array($result);
print $values['username'];
?>
- Trainer #
<?php
$sql = ("SELECT ID FROM users WHERE username = '$username'")or die(mysql_error());
$result = mysql_query($sql) or die(mysql_error());
$values = mysql_fetch_array($result);
print $values['ID'];
?>
</p>
<p class="style6"> </p>
<p>
<?php
$sql = ("SELECT pokemon1_pic FROM users WHERE username = '$username'")or die(mysql_error());
$result = mysql_query($sql) or die(mysql_error());
$row = mysql_fetch_array($result);?>
</p>
<p align="center"><img src="<?php echo $row['pokemon1_pic'] ?>" width="94" height="93" /> </p>
<p><span class="style6">Pokemon:</span>
<?php
$sql = ("SELECT pokemon FROM users WHERE username = '$username'")or die(mysql_error());
$result = mysql_query($sql) or die(mysql_error());
$values = mysql_fetch_array($result);
print $values['pokemon'];
?>
</p>
<p><span class="style6">Level:</span>
<?php
$sql = ("SELECT poke1lvl FROM users WHERE username = '$username'")or die(mysql_error());
$result = mysql_query($sql) or die(mysql_error());
$values = mysql_fetch_array($result);
print $values['poke1lvl'];
?>
it does not show anything none of them work im not getting any errors i am starting the session at top of the page two but still ent working
i am making the variable $username on the login page
Code: Select all
// if login is ok then we add a cookie
$_POST['username'] = stripslashes($_POST['username']);
$hour = time() + 3600;
setcookie(ID_my_site, $_POST['username'], $hour);
setcookie(Key_my_site, $_POST['pass'], $hour);
//if there is, it logs you in and directes you to the members page
{
$username = $_COOKIE['ID_my_site'];
$pass = $_COOKIE['Key_my_site'];
$check = mysql_query("SELECT * FROM users WHERE username = '$username'")or die(mysql_error());
while($info = mysql_fetch_array( $check ))
{
if ($pass != $info['password'])
{
}
else
{
header("Location: members.php");
all the login works fine just showing the info dont and i dont know why ?
social_experiment
DevNet Master
Posts: 2793 Joined: Sun Feb 15, 2009 11:08 am
Location: .za
Post
by social_experiment » Sat Feb 06, 2010 3:34 am
You have 3 seperate queries to display information about just one user. Try using the code below.
Code: Select all
<?php $query = "SELECT id, username, pokemon1_pic, poke1lvl FROM table WHERE username = '$username'";
$result = mysql_query($query) or die(mysql_error());
while ($pointer = mysql_fetch_array($result)) {
echo $pointer['id']."\n";
echo $pointer['username']."\n";
echo $pointer['pokemon1_pic']."\n";
echo $pointer['poke1lvl']."\n";
} ?>
I dont think there is something wrong with the other code but i'd assign the username slightly diffrently
Instead of using :
Code: Select all
<?php $_POST['username'] = stripslashes($_POST['username']); ?>
I would :
Code: Select all
<?php $username = stripslashes($_POST['username']); ?>
“Don’t worry if it doesn’t work right. If everything did, you’d be out of a job.” - Mosher’s Law of Software Engineering
DTV
Forum Newbie
Posts: 3 Joined: Sat Feb 06, 2010 1:22 am
Post
by DTV » Wed Feb 10, 2010 7:00 am
social_experiment : I tryed putting your code in but i got an error i think i put it in the wrong place
Could u copy the first code and add
the new code u gave me please?
social_experiment
DevNet Master
Posts: 2793 Joined: Sun Feb 15, 2009 11:08 am
Location: .za
Post
by social_experiment » Wed Feb 10, 2010 9:16 am
Hope this helps
Code: Select all
<?php
$connect = mysql_query('host', 'username', 'password');
mysql_select_db('db') or die(mysql_error());
$sql = "SELECT ID, username, pokemon1_pic, poke1lvl, pokemon FROM users WHERE username = '$username'";
$result = mysql_query($sql) or die(mysql_error());
while ($value = mysql_fetch_array($result)) {
echo "<p class=\"style6\">";
//will echo username
echo $value['username'];
echo "</p>";
echo "<p class=\"style6\">";
//will echo "Trainer #3" or whatever the ID value is
echo "<p class=\"style6\">";
echo "- Trainer #".$value['ID']."";
echo '</p>';
//will display the image
echo "<p class=\"style6\">";
echo "<img src=\"".$value['pokemon1_pic']."\" />";
echo '</p>';
//will echo Pokemon : pokemon or whatever the value is
echo "<p class=\"style6\">";
echo "Pokemon : ".$value['pokemon'];
echo "</p>";
//will echo the value of the poke1lvl
echo "<p class=\"style6\">";
echo "Level : ".$value['poke1lvl'];
echo '</p>';
}
?>
“Don’t worry if it doesn’t work right. If everything did, you’d be out of a job.” - Mosher’s Law of Software Engineering
DTV
Forum Newbie
Posts: 3 Joined: Sat Feb 06, 2010 1:22 am
Post
by DTV » Wed Feb 10, 2010 10:59 pm
Thx (:
but it didnt work :'(
PHP Error Message
Parse error: syntax error, unexpected T_DNUMBER in /home/a8310104/public_html/register.php on line 77
Sorry, im so noobish but im 13 and im trying to prove my teacher wrong he said that it would be a massive challenge to a yr 8 like you it would even challenge a yr 12 student and im like i can do it
social_experiment
DevNet Master
Posts: 2793 Joined: Sun Feb 15, 2009 11:08 am
Location: .za
Post
by social_experiment » Wed Feb 10, 2010 11:32 pm
Could you paste the code giving the error?
“Don’t worry if it doesn’t work right. If everything did, you’d be out of a job.” - Mosher’s Law of Software Engineering
social_experiment
DevNet Master
Posts: 2793 Joined: Sun Feb 15, 2009 11:08 am
Location: .za
Post
by social_experiment » Wed Feb 10, 2010 11:46 pm
There is a error in one of the lines of the script i gave you (this is not relevant to your current problem i think)
Code: Select all
<?php $connect = mysql_query('host', 'username', 'password'); ?>
should be :
Code: Select all
<?php $connect = mysql_connect('host', 'username', 'password'); ?>
“Don’t worry if it doesn’t work right. If everything did, you’d be out of a job.” - Mosher’s Law of Software Engineering