[SOLVED] mysql call not working

cornwall57 · January 6, 2008

I'm not a php programmer, in fact I am only here because I cannot for the life of me get this script to work, although it works just fine on a different server.

ok

file 1 holds :

connectdb();
$sid = $_GET["sid"];
$uid = getuid_sid($sid);

  echo "<p align=\"center\">";
    $sql = "select * from fun_users where val='0'";

    $result = mysql_query($sql);
    echo mysql_error();
    while ($item = mysql_fetch_array($result))
    {
      $lnk = "<a title=\"Enter\" href=\"xxxx.php?sid=$sid&user2=$item[1]\">$item[1]</a>";
      echo "$lnk <br/>";


}

This works just fine as far as I can see : clicking the link gives :

http://xxx.com/xxxx.php?sid=5f75cf73631769998acb3ff7047ac45e&user2=test

Now, entering xxx.php the code is:

<?php


include("core.php");
include("config.php");

connectdb();
............ headers all correct.......................

echo "<head>";

echo "<title>$stitle</title>";
echo "<link rel=\"StyleSheet\" type=\"text/css\" href=\"style/style.css\" />";
echo "</head>";

echo "<body>";

connectdb();
$sid = $_GET["sid"];
$uid = getuid_sid($sid);

  echo "<p align=\"center\">";
    $sql = "select * from fun_users where name='user2'";
    $items = mysql_query($sql);
    echo mysql_error();
    while ($item = mysql_fetch_array($items))
{
echo "<br/>";
echo "Name: ";
echo $item[1] ;
echo "<br/>D.O.B: ";
printf($item[3]);
echo "<br/>Sex: ";
printf($item[4]);
echo "<br/>Location: ";
printf($item[5]);
echo "<br/>Registered With: ";
printf($item[14]);
     $jdt = date("d m y-H:i:s",$item[17]);
echo "<br/>Registered Date: $jdt";
echo "<br/>Ip: $item[15]";

}

?>
<br/>
<img src="images/line.jpg" alt="----"/>
<br/>
<?php
echo "<a title=\"Enter\" href=\"zzzzz.php?sid=$sid&user2=$user2&action=1\">Validate Users</a>";
echo "<br/>";
echo "<a title=\"Enter\" href=\"zzzzz.php?sid=$sid&user2=$user2&action=3\">Dont Validate</a>";
echo "<br/>";
echo "<a title=\"Enter\" href=\"yyyyy.php?sid=$sid\">Back</a>";
echo "<br/>";
echo "<a title=\"Enter\" href=\"index.php?action=main&sid=$sid\">Main Menu</a>";

echo "</p>";
  



echo "</body>";
echo "</html>";
?>

It appears to me that 'user2' is not being picked up, as the resulting web page does not print any of the data between the "while" and the html divider.

Can anyone help with this? Is there something fundamentally wrong or am I being even thicker than a newbie?

LOL

Thanks

Steve

AndyB · January 6, 2008

...
$sid = $_GET['sid'];
$user2 = $_GET['user2']; // retrieve BOTH passed values
...

eRott · January 6, 2008

Ok, so far from your code, i've come up with these conclusions:

[#1] I have no idea what on earth $uid = getuid_sid($sid); does, is supposed to do, or for that matter, i cannot see it being used for anything, so get rid of it. Not to mention, unless i am unaware of some PHP built in function, getuid_sid doesn't exist.

[#2] It does not grab the value of user2 because you didn't include a get statement.

$user2 = $_GET["user2"];

[#3] You may want to clean up your excessive amount of echo's in your script and store it in a nice neat variable or something to call later.

$jdt = date("d m y-H:i:s",$item[17]);

$show = '<br/>Name: ' . $item[1] . '<br/>D.O.B: ' . $item[3] . '<br/>Sex: ' . $item[4] . '<br/>Location: ' . $item[5] . '<br/>Registered With: ' . $item[14] . '<br/>Registered Date: ' . $jdt . '<br/>Ip: ' . $item[15];

echo $show;

So change those couple of things and it should work.

Take care.

cornwall57 · January 6, 2008

Thanks! That got it!

Many thanks for the help without any hint of sarcasm to a newbie like myself. I salute you!

Cheers

Steve

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[SOLVED] mysql call not working

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