How do I make it so..

[Zoofu]

I've got an image field in my users table, linking to an image.

 

But i want to know how I will show this image?

 

$sql = "SELECT * FROM `users` WHERE `id`='".$uid."'";

 

^

|

 

I need the `id`='".$uid."'; part, but I also want to define `image` as a $image var. How?

[Garethp]

$sql = "SELECT * FROM `users` WHERE `id`='$uid' AND `image`='$image'";

[Zoofu]

nope, now it's not working.

[Garethp]

Well, echo mysql_query();

 

What's the error?

[trq]

Define 'not working', your original post is not at all clear.

[Zoofu]

Right, since i'm taking everything from it...

 

I tried doing .uid($rowr['image'])

 

but it comes out as a red x.

[Zoofu]

For anyone who wants to know... this is the function

 

function uid($uid, $link = FALSE){

$sql = "SELECT * FROM `users` WHERE `id`='".$uid."' AND `image`='".$image."'";

$res = mysql_query($sql) or die(mysql_error());

if(mysql_num_rows($res) == 0){

return "Invalid User";

}else {

$row = mysql_fetch_assoc($res);

if(!$link){

 

return $row['username'];

}else {

return "<a href=\"/index.php?act=profile&id=".$uid."\">".$row['username']."</a>";

}

}

}

[TeNDoLLA]

Your image data will be in $row['image'] variable if the field name is image in database. Then just assign $imageVar = $row['image'];

[trq]

The variable $image is not defined anywhere within your function. Assuming it exists outside of the function it will need to be passed in as an argument.

[Zoofu]

it's in the users table, so i'm trying to do it as $row['image']

[Zoofu]

Well I can't test any of the suggestions, since Hostzi's FTP is being really irritating, can't wait till I get my proper server back. x.x

 

Filezilla keeps saying there's 2 or more connected with the same user.

[Zoofu]

Still getting errors, tell me what I should do with these codes:

 

function uid($uid, $link = FALSE){
$sql = "SELECT * FROM `users` WHERE `id`='".$uid."'";
$res = mysql_query($sql) or die(mysql_error());
if(mysql_num_rows($res) == 0){
	return "Invalid User";
}else {
	$row = mysql_fetch_assoc($res);
	if(!$link){

	return $row['username'];	
	}else {
		return "<a href=\"/index.php?act=profile&id=".$uid."\">".$row['username']."</a>";
	}
}
}

 

and the part that matters:

 

$select_sql = "SELECT * FROM `forum_replies` WHERE `tid`='".$id."' ORDER BY id ASC LIMIT ".$end.",".$start.""; 
		$select_res = mysql_query($select_sql) or die(mysql_error()); 

		while($rowr = mysql_fetch_assoc($select_res)){
			echo "<tr><td colspan=\"2\" align=\"left\" class=\"forum_header\">Posted On: <em>".$rowr['date']."</em></td></tr>\n";
			echo "<tr><td align=\"left\" width=\"15%\" valign=\"top\" class=\"forum_header\">".uid($rowr['uid'], true)."<br><img src=".uid($image)."></td>\n"; //< --- There is where the image will go.
			echo "<td align=\"left\" valign=\"top\" class=\"forum_header\">\n";
			echo nl2br(topic($rowr['message']));
			echo "</td>\n";
			echo "</tr>\n";
		}
		echo "<form method=\"post\" action=\"./index.php?act=reply&id=".$row['id']."\">\n";
		echo "<tr><td colspan=\"2\" align=\"center\"><textarea style=\"width:300px; height:100px\" name=\"reply\"></textarea><br><input type=\"submit\" name=\"submit\" value=\"Add Reply\" style=\"width:100px\"></td></tr>\n";
		echo "</table>\n";

[Zoofu]

I really must fix this before I get back to making my anime, I've only got 2 weeks to finish a 30 minute animation. Well, it's only supposed to be 30 seconds but who cares? Lmao.

 

Back on topic: So yeah, I need this pretty fast if anyone could?

[trq]

Your code makes little sense.

 

Firstly you call the function passing at what we assume is a users id. uid($rowr['uid'], true)

 

The next time you call it you seem to pass it an undefined (at least not anywhere in your post) $image variable. uid($image)

 

How exactly do you expect your function to lookup a user based on an image? There is your problem. Now, for homework, fix it.