this mysql error is kiling me :(

ovisopa

Hello ppl,
I'm working now at a greetings script, it's all done and now one hour it was everything ok but now when I uploaded on my host I recive this error:

You have an error in your SQL syntax near 'Resource id #12' at line 1

The pege with the erro is this:

	<table width="400" border="0" bgcolor="#FF9F00" class="buttoncolor">
		<tr>
			<td class="linksmall" align="center" valign="top">Ai primit o felicitare de la <b><?echo "$numele";?></b></td></tr>

	<tr>
		<td class="linksmall" align="center" valign="top">
		<?
		require "db.inc.php";
		$query_out = mysql_query("SELECT imagine FROM felicitari WHERE random='$random' and numele='$numele'");
		mysql_query($query_out) or
			die (mysql_error());
		$datele = mysql_fetch_array($query_out);
		$img_n = $datele[imagine];
		$image_name = "felicitari/" . $datele[imagine] . ".jpg";
		$image_name = "felicitari/$img_n.jpg";
		$image_size =  GetImageSize ("$image_name"); 
		echo "<img src=\"felicitari/" . $datele[imagine] . ".jpg\" border=\"1\" $image_size[3]> </td></tr>
		<tr><td class=\"linksmall\"><div style=\"text-align:justify;\"><b>Mesaj :</b>" . $datele[mesaj] . "</div></td></tr>
		<tr><td class=\"linksmall\"><div style=\"text-align:justify;\"><b>Data  : </b>" . $datele[data] . "</div></td></tr>
		<tr><td class=\"linksmall\"><div style=\"text-align:justify;\"><b>Email  : </b><a href=\"contact.php?dest=" . $datele[email] . "\" class=\"linksmall\" title=\"Trimite-i lui " . $datele[numele] . " un email.\">" . $datele[email] . "</a></div></td></tr>
		<tr><td class=\"linksmall\" align=\"center\"><a href=\"felicitari.php?action=alege\" class=\"linkbold\" title=\"Trimite si tu o felicitare.\">Trimite si tu o felicitare.</a></td></tr>

		";
		?>

</table>

Pls help me 'cause I'm gonna get crazy 🙁

10x and see ya all

ovisopa

I forgot to take out some lines, so the real code is this :

	<table width="400" border="0" bgcolor="#FF9F00" class="buttoncolor">
		<tr>
			<td class="linksmall" align="center" valign="top">Ai primit o felicitare de la <b><?echo "$numele";?></b></td></tr>
		<tr>
			<td class="linksmall" align="center" valign="top">
			<?
			require "db.inc.php";
			$query_out = mysql_query("SELECT imagine FROM felicitari WHERE random='$random' and numele='$numele'");
			mysql_query($query_out) or
				die (mysql_error());
			$datele = mysql_fetch_array($query_out);
			$image_name = "felicitari/" . $datele[imagine] . ".jpg";
			$image_size =  GetImageSize ("$image_name"); 
			echo "<img src=\"felicitari/" . $datele[imagine] . ".jpg\" border=\"1\" $image_size[3]> </td></tr>
			<tr><td class=\"linksmall\"><div style=\"text-align:justify;\"><b>Mesaj :</b>" . $datele[mesaj] . "</div></td></tr>
			<tr><td class=\"linksmall\"><div style=\"text-align:justify;\"><b>Data  : </b>" . $datele[data] . "</div></td></tr>
			<tr><td class=\"linksmall\"><div style=\"text-align:justify;\"><b>Email  : </b><a href=\"contact.php?dest=" . $datele[email] . "\" class=\"linksmall\" title=\"Trimite-i lui " . $datele[numele] . " un email.\">" . $datele[email] . "</a></div></td></tr>
			<tr><td class=\"linksmall\" align=\"center\"><a href=\"felicitari.php?action=alege\" class=\"linkbold\" title=\"Trimite si tu o felicitare.\">Trimite si tu o felicitare.</a></td></tr>";
			?>
	</table>

now can someone tell me what is wrong?

siwis

You have an error in your SQL syntax near 'Resource id #12' at line 1

So where in your code is 'Resource id #12'?

Is 'Resource id #12' being fed into the sql as a query value? only (and don't shoot me if I'm mistaken on this cos its late and I'm tired) the hash is a special character and would need to be escaped.

The code you've posted doesn't really help us shed any light on your true problem.

Mind you, you do have a mysql_query() call inside another mysql_query() call which could be the real root of the problem.

$query_out = mysql_query("SELECT imagine FROM felicitari WHERE random='$random' and numele='$numele'");
            mysql_query($query_out) or
                die (mysql_error());

$query_out actually performs the query first time round, so adding the "mysql_query($query_out) or" is asking for trouble.

ovisopa

Hy, and 10x for help.

I though that code is for showing me if there's any error, but I was wrong. ANyway now I don't see that error anymore but I recove other error:

Warning: getimagesize: Unable to open 'felicitari/.jpg' for reading. in /hsphere/local/home/sibiul/sibiul.ro/ecard-display.php on line 13

the problem is that this error apears only online, if I try my script on localhost is everithing fine.

I will tri to see what I've done wrong.

10x again for help.

Bye bye and good night if you're going to sleep

siwis

Once again, the error message holds the key!

Warning: getimagesize: Unable to open 'felicitari/.jpg' for reading. in /hsphere/local/home/sibiul/sibiul.ro/ecard-display.php on line 13

does the file felicitari/.jpg exist in the same directory as your script?
Does the filename really contain a forward slash?
What are the permissions on that file if it does exist?

Most errors like the one you've posted occur because the file isn't where you tell the script that it is, OR because the filename/permissions are wrong.

ovisopa

The problem is that no data is coming from the db, the image file is named 14.jpg and is in the directory felicitari, the file permisions are 644.

You can see here how the image is writen

$image_name = "felicitari/" . $datele[imagine] . ".jpg";

so with this $datele[imagine] I should obtain the name of the pic from the db, actualy I will get only the value 14 or other value for evry pic.

I'm sure is a stupid mistake which I can't see it now maibe because it's late and tomorrow I have some school 🙂

10x again.