numbered list in php

[kurtos123]

I am displaying the top 10 games on my website but i want this list numbered from 1 to 10.

How do i inplement this in my code ? Is it posible also the ad numbered images, and how do i do that?

 

Here is my code so far:

 

<?php
include "config.php";
$rQuery = mysql_query("SELECT * FROM " . $pre . "$sqlgame WHERE cijfer / stemmen >= 10 ORDER BY views DESC LIMIT 10");
while($aGame = mysql_fetch_assoc($rQuery))
{
  if($aGame['cijfer'] != 0 AND $aGame['stemmen'] != 0)
  {
    $cijfer = floor($aGame['cijfer'] / $aGame['stemmen']);
  }
  else
  {
    $cijfer = 0;
  }
  $sSpelNaam = str_replace(" ", "-",$aGame['name']);
  echo "<font color='#333333'>• <a href=\"http://www.crocgame.com/games.php/" . $sSpelNaam . ".html\" target='blank' onmouseover=\"ddrivetip('<img src=\'" . $aGame['screenshoturl'] . "\' width=\'170\' height=\'140\' alt=\'" . $aGame['name'] . "\'><br /><b>" . $aGame['name']  . "</b>');\" onmouseout='hideddrivetip();'>" . $aGame['name'] . " " . $aGame['extra'] . "</a></font><br />" . PHP_EOL;
}
?>
 

[Ch0cu3r]

On way would be to use counter

// initiate counter
$i = 0;

while($aGame = mysql_fetch_assoc($rQuery)
{
    if($aGame['cijfer'] != 0 AND $aGame['stemmen'] != 0)
    {
        $cijfer = floor($aGame['cijfer'] / $aGame['stemmen']);
    }
    else
    {
        $cijfer = 0;
    }
    $sSpelNaam = str_replace(" ", "-",$aGame['name']);

    // increment counter by one
    $i++;

    // output counter
    echo "$i. <font color='#333333'>• <a href=\"http://www.crocgame.com/games.php/" . $sSpelNaam . ".html\" target='blank' onmouseover=\"ddrivetip('<img src=\'" . $aGame['screenshoturl'] . "\' width=\'170\' height=\'140\' alt=\'" . $aGame['name'] . "\'><br /><b>" . $aGame['name']  . "</b>');\" onmouseout='hideddrivetip();'>" . $aGame['name'] . " " . $aGame['extra'] . "</a></font><br />" . PHP_EOL;

}

The code will increment $i by one each time a row is outputted in the while loop

 

Or you could output the results in a HTML ordered list

 

 

Is it posible also the ad numbered images, and how do i do that?

Yes, depending on how have named your images, I assume you have named like 1.png, 2.png etc, you can change  echo "$i.   in my example code above to to be  echo "<img src=\"$i.png\" />

[kurtos123]

Updated my browser and got error message

Parse error: syntax error, unexpected '{' in /home/crocgame/domains/crocgame.com/public_html/beste_games1.php on line 7

I am bad in php so please help me further...

Many thanks for help!

 

<?php
include "config.php";
// initiate counter
$i = 0;

while($aGame = mysql_fetch_assoc($rQuery)
{
    if($aGame['cijfer'] != 0 AND $aGame['stemmen'] != 0)
    {
        $cijfer = floor($aGame['cijfer'] / $aGame['stemmen']);
    }
    else
    {
        $cijfer = 0;
    }
    $sSpelNaam = str_replace(" ", "-",$aGame['name']);

    // increment counter by one
    $i++;

    // output counter
    echo "$i. <font color='#333333'>• <a href=\"http://www.crocgame.com/games.php/" . $sSpelNaam . ".html\" target='blank' onmouseover=\"ddrivetip('<img src=\'" . $aGame['screenshoturl'] . "\' width=\'170\' height=\'140\' alt=\'" . $aGame['name'] . "\'><br /><b>" . $aGame['name']  . "</b>');\" onmouseout='hideddrivetip();'>" . $aGame['name'] . " " . $aGame['extra'] . "</a></font><br />" . PHP_EOL;

}

?>

[Barand]

Missing a closing ) at the end of line 6

[kurtos123]

where is line 6 exactly and with closing you mean this "}" ?

[Barand]

You will find line 6 immediately before line 7, and by "closing )" i mean )

while($aGame = mysql_fetch_assoc($rQuery) )
                                          ^
                                          |
                                    this is missing

[kurtos123]

Get a mysql error on line 6 with: while($aGame = mysql_fetch_assoc($rQuery) ) now:  Mysql_fetch _assoc():supplied argument is not a valid MySQL result resource in /home/crocgame/domains/crocgame.com/public_html/beste_games1.php on line 6

[kurtos123]

i agree in code of moderator Ch0cu3r " )"  on line 6 was missing but something more but be wrong ....

[kurtos123]

Code so far:

 

<?php
include "config.php";
// initiate counter
$i = 0;
$rQuery = mysql_query("SELECT * FROM " . $pre . "$sqlgame WHERE cijfer / stemmen >= 10 ORDER BY views DESC LIMIT 10");
while($aGame = mysql_fetch_assoc($rQuery))
    if($aGame['cijfer'] != 0 AND $aGame['stemmen'] != 0)
    {
        $cijfer = floor($aGame['cijfer'] / $aGame['stemmen']);
    }
    else
    {
        $cijfer = 0;
    }
    $sSpelNaam = str_replace(" ", "-",$aGame['name']);

    // increment counter by one
    $i++;

    // output counter
    echo "$i.<font color='#333333'>• <a href=\"http://www.crocgame.com/games.php/" . $sSpelNaam . ".html\" target='blank' onmouseover=\"ddrivetip('<img src=\'" . $aGame['screenshoturl'] . "\' width=\'170\' height=\'140\' alt=\'" . $aGame['name'] . "\'><br /><b>" . $aGame['name']  . "</b>');\" onmouseout='hideddrivetip();'>" . $aGame['name'] . " " . $aGame['extra'] . "</a></font><br />" . PHP_EOL;

}
?>
 

 

STill not working with above code get mysql error: Parse error: syntax error, unexpected '}' in/home/crocgame/domains/crocgame.com/public_html/beste_games1.phpon line 23                     Please help !

[Ch0cu3r]

There should be a { after the while loop statement on line 6

while($aGame = mysql_fetch_assoc($rQuery))
{ // <--- this is missing

[kurtos123]

Everybody thanks for great help, it works now !!!!

[kurtos123]

small detail don't now it can be solved the list is printed like this (watch position number 10 under 9):

1

2

..

9

10

 

Possible to print like this:

  1

  ..

  9

10

[Ch0cu3r]

Could use either

str_pad, sprintf or use a condition if $i less than 10 to prefix $i with a space

 

See if you can implement either of the 3 options above into your code.

 

NOTE: You may need to use      rather than the actual space character. As web browsers tend to ignore (white)space characters.

[kurtos123]

topic can be closed thanks for great help again !