Php Help

[leafypiggy]

I need help with a script that displays my mysql table. I keep getting this error.

Parse error: syntax error, unexpected $end in /home/leafy/public_html/roster.php on line 31

Here is the code:

<?php
$username="leafy_admin";
$password="removed";
$database="leafy_roster";
mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
$query="SELECT * FROM leafy_clanemail";
$result=mysql_query($query);
$num=mysql_numrows($result);
mysql_close();
echo "<b><center>Night Assassins Clan Roster</center></b><br><br>";
$i=0;
while ($i < $num) {
$name=mysql_result($result,$i,"field1-name");
$email=mysql_result($result,$i,"field2-name");
$rank=mysql_result($result,$i,"field3-name");

echo "<b>$field1-name 
$field2-name2</b><br>$field3-name<br>;
$i++;
}
?>

Can you find out anything wrong with it?

 

[knightcon]

Your missing a closing quotation marks. Try this code instead...

 <?php 
$username="leafy_admin"; 
$password="nightassassins"; 
$database="leafy_roster"; 
mysql_connect(localhost,$username,$password); 
@mysql_select_db($database) or die( "Unable to select database"); 
$query="SELECT * FROM leafy_clanemail"; 
$result=mysql_query($query); 
$num=mysql_numrows($result); 
mysql_close(); 
echo "<b><center>Night Assassins Clan Roster</center></b><br><br>"; 
$i=0; 
while ($i < $num) { 
$name=mysql_result($result,$i,"field1-name"); 
$email=mysql_result($result,$i,"field2-name"); 
$rank=mysql_result($result,$i,"field3-name"); 

echo "<b>$field1-name  
$field2-name2</b><br>$field3-name<br>"; 
$i++; 
}
?>

 

[Mohron]

Here's a great way to find php syntax errors.
http://www.meandeviation.com/tutorials/learnphp/php-syntax-check/

 

[leafypiggy]

thanks, ill check it out, and can you remove my pass from the script, lol, I forgot to take it out.
Edit:
It just gave me another error. Meanwhile, I tried this script from w3schools:

<?php
$con = mysql_connect("localhost","leafy_admin","*REMOVED*");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("leafy_roster", $con);

$result = mysql_query("SELECT * FROM leafy_clanemail");

echo "<table border='1'>
<tr>
<th>Name</th>
<th>Email</th>
<th>Rank</th>
</tr>";
while($row=mysql_fetch_array($result));
  {
  echo "<tr>";
  echo "<td>" . $row['name'] . "</td>";
  echo "<td>" . $row['email'] . "</td>";
  echo "<td>" . $row['rank'] . "</td>";
  echo "</tr>";
  }
echo "</table>";mysql_close($con);
?>

I get this error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/leafy/public_html/rosterw3schools.php on line 18

 

[xbeto78]

The while loop does not need a semicolon at the end of the line. It only needs an opening and closing brace.


while($row=mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . $row['email'] . "</td>";
echo "<td>" . $row['rank'] . "</td>";
echo "</tr>";
}
echo "</table>";mysql_close($con);
?>

 

[leafypiggy]

changed that, now getting this:

Warning: mysql_connect() [function.mysql-connect]: Can't connect to local MySQL server through socket '/var/lib/mysql/mysql.sock' (2) in /home/leafy/public_html/rosterw3schools.php on line 2
Could not connect: Can't connect to local MySQL server through socket '/var/lib/mysql/mysql.sock' (2)

Temporery or do I need to link it to the new MYSQL server.

 

[Slothie]

Are you by any chance, on stoli?

If you are, link yourself to the new mysql server

 

[leafypiggy]

yeah Im on Stoli. What is the new IP?
Edit:
I changed the server for MYSQL but I am still getting the

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/leafy/public_html/rosterw3schools.php on line 18

error