newbie problem

[gdfhghjdfghgfhf]

yeah well i'm stuck here... i know my problem is stupid but i'm still a newbie ><

 

$query78 = "SELECT * from pp_files WHERE category = '$id'"; 
$res78 = mysql_query($query78) or die(mysql_error());   
while($row = mysql_fetch_array($res78)) { 
$picture = $row["picture"];
$vidid = $row["id"];
$rows = mysql_num_rows($res78);
}



if ($rows > 0) {
echo "<br><table width=\"85%\" align=\"center\" cellpadding=\"8\"><tr><td style=\"background: url(1.png); border: 1px dashed #FFFFFF; cellpadding: 5px;\" colspan=\"5\">";
echo "<center><b><font size=\"3\">$vbphrase[ilya]il y a <font size=\"5\">$rows</font> $vbphrase[vids]</font></b><br>";

$query78 = "SELECT * from pp_files WHERE category = '$id'"; 
$res78 = mysql_query($query78) or die(mysql_error());   
while($row = mysql_fetch_array($res78)) { 



echo "<a href=\"videos/video.php?id=$vidid\"><img src=\"";
echo $picture;
echo "\" border=\"0\" alt=\"$vbphrase[vids_click]\"></a>  ";
}

echo "<br><font size=\"2\"><a href=\"videos/submit.php\">$vbphrase[vids_add]</a></font></center></td></tr></table>";
}

 

i have multiple items in tables of my database and i am trying to display all of the pictures, with a link to the corresponding picture

 

my problem is that only the last item is displayed.... If there is 3 entry to display, 3 items will be displayed, but with the same picture (none) and the same link (last id of the array)

[DarkWater]

Do NOT run mysql_num_rows inside of the while loop. Take it out and put it before the loop.

[BlueSkyIS]

the problem is you loop over the records loading variables, but you don't do anything with those variables until after the loop. so all you get is the last ones. i suggest that you use the variables as you load them or that you load them all into arrays and use them after the loop

 

while($row = mysql_fetch_array($res78)) { // here, you loop over every record setting values, but never using them.
    $picture = $row["picture"];
    $vidid = $row["id"];
   $rows = mysql_num_rows($res78);
}

[gdfhghjdfghgfhf]

ehhh i don't understand

 

i should remove all mysql_fetch_array parts?

[DarkWater]

No....You're rewriting the results every time you use the while loop with your command though.  Think about it.

[gdfhghjdfghgfhf]

could you just fix the code for me... i'd understand better cuz right now i'm really lost..

[DarkWater]

Where are you getting $vbphrase?

 

Also:

Ton site est en francais? o-O

[gdfhghjdfghgfhf]

i use $vbphrase because i am putting this script inside a vbulletin template... so all $vbphrase variables will be replaced with its corresponding language signification from the vbulletin's language and phrases system.... so basically it's just words and text, but i use it to have both a french and an english version of my site

 

oui mon site est en francais Tu parles francais? D'ou viens tu?

[DarkWater]

Oh, okay.  I'll try to fix up the code.

 

Oui, je parle francais...mais, je suis venu des Etats-Unis. >_>

[DarkWater]

$query78 = "SELECT * from pp_files WHERE category = '$id'"; 
$res78 = mysql_query($query78) or die(mysql_error());   
$rows = mysql_num_rows($res78);
while($row = mysql_fetch_array($res78)) { 
$picture = $row["picture"];
$vidid = $row["id"];

}



if ($rows > 0) {
while($row = mysql_fetch_array($res78)) { 
$picture = $row["picture"];
$vidid = $row["id"];

echo "<br><table width=\"85%\" align=\"center\" cellpadding=\"8\"><tr><td style=\"background: url(1.png); border: 1px dashed #FFFFFF; cellpadding: 5px;\" colspan=\"5\">";
echo "<center><b><font size=\"3\">$vbphrase[ilya]il y a <font size=\"5\">$rows</font> $vbphrase[vids]</font></b><br>";





echo "<a href=\"videos/video.php?id=$vidid\"><img src=\"";
echo $picture;
echo "\" border=\"0\" alt=\"$vbphrase[vids_click]\"></a>  ";

echo "<br><font size=\"2\"><a href=\"videos/submit.php\">$vbphrase[vids_add]</a></font></center></td></tr></table>";
}

 

Je pense qu'il va fonctionner...

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