[RESOLVED] Display MySQL results

Daniel_Exe

<?php
    $avatars=array();

mysql_select_db('*****', mysql_pconnect('localhost','***********','*************')) or die (mysql_error());
$query = "SELECT * FROM avatars WHERE size='80x80' ORDER BY `id` DESC";
$result = mysql_db_query("*****",$query);
$num_rows = mysql_num_rows($result);

for ($i=1; $i<=$num_rows; $i++) {
	$avatars[] = mysql_fetch_array($result);
	echo "<td><img border=\"0\" src=\"".$avatars[0]."\" /></td>\n";
	if (($i % 4) == 0) {
		echo "</tr>\n<tr>\n";
	}
}
?>

Hello all,
I'm back once again with another problem of mine. I've been spending many hours coding various pages for a CMS that I'm working on, and my brain is starting to hurt. I wanted to code a simple script that would grab all the image urls of images that have a size of 80x80 and then echo the urls inside of an img tag. The table that stores that info looks like the following:

CREATE TABLE `avatars` (
  `id` int(11) NOT NULL auto_increment,
  `url` varchar(255) NOT NULL,
  `size` varchar(50) NOT NULL,
  PRIMARY KEY  (`id`)
)

Because I couldn't figure out a way to display the images in a table, and add a new tr every so many results, I decided to use some code someone else had posted. I'm trying to customise it to work with my site, although my brain just won't do it anymore. So, I was hoping a fellow member/webmaster could lend a hand. Well, my problem is as follows:
$avatars simply echoes "Array" and I'm not sure how to get it to echo the url of the image instead.
Any help would be greatly appreciated as always. =)
Thanks,
Daniel

P.S. I'm sure this is simple enough, but I'm just I'm too tired at the moment to figure this out.

devinemke

$query = "SELECT * FROM avatars WHERE size = '80x80' ORDER BY id DESC";
$result = mysql_query($query) or exit(mysql_error());
if (mysql_num_rows($result))
{
	while ($row = mysql_fetch_assoc($result))
	{
		echo '<img src="' . $row['url'] . '" border="0"><br>';
	}
}
else
{
	echo 'no rows found';
}

Daniel_Exe

@: I had written a code very similar to that, but unfortunately, it does not create a new tr every set results. The code I had posted would make a td for every result, and after every 4, would close the tr and make a new one. How would I go about implementing that into the code you posted?
Thanks,
Daniel

devinemke

$query = "SELECT * FROM avatars WHERE size = '80x80' ORDER BY id DESC"; 
$result = mysql_query($query) or exit(mysql_error()); 
if (mysql_num_rows($result)) 
{ 
	while ($row = mysql_fetch_assoc($result)) {$avatars[] = $row;} 
	$chunks = array_chunk($avatars, 4);
	echo '<table>';
	foreach ($chunks as $value)
	{ 
		echo '<tr>';
		foreach ($value as $subvalue)
		{
			echo '<td><img src="' . $subvalue['url'] . '"></td>';
		}
		echo '</tr>';
	}
	echo '</table>';
} 
else 
{ 
	echo 'no rows found';
}

Daniel_Exe

Thanks, it worked perfectly.

Unfortunately, while coding another page, I've encountered yet another error. I receive the following warning when I view one of my pages:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\AppServ\www\tutorials.php on line 37

Here's the code for that page:

<?php

mysql_select_db('phpBB', mysql_pconnect('localhost','****','*****')) or die (mysql_error());

// Authorise
$allowed_categories = array(
'photoshop',
'flash'
);

$category = $_GET['category'];

//Sets title for page
if(isset($_GET['category'])){

if (in_array($category,$allowed_categories))
{
	if ($category = "photoshop") { $page_title="Photoshop Tutorials"; }
	if ($category = "flash") { $page_title="Flash Tutorials"; }
}

else {
	$page_title = "Tutorial Category does not exist!";
	$warning = "Not a valid category!<br /><br />";
}

}

// header
include("./header.php");

if(isset($category)){
	echo "$warning";
	if (in_array($category,$allowed_categories)){
		$query="SELECT * FROM tutorials WHERE category=$category ORDER BY id DESC";
		$row=mysql_query($query);
		while ($result=mysql_fetch_array($row)){
				echo "<img class=\"difficulty\" src=\"***/".$result['difficulty'].".gif\" alt=\"".$result['difficulty']."\" />
						<table border=\"0\" width=\"501px\" cellspacing=\"0\" cellpadding=\"0\">
						<tr>
							<td valign=\"top\" colspan=\"2\"><img src=\"***/tuts_top.gif\" border=\"0\" alt=\"\" /></td>
						</tr>
						<tr>
							<td valign=\"middle\" class=\"t_bg\">
								<div class=\"fl\">&nbsp;&nbsp;&nbsp;&nbsp;<a href=\"\"><img height=\"90\" with=\"90\" src=\"".$result['thumbnail']."\" alt=\"\" border=\"\" /></a></div>
								<b>&nbsp;&nbsp;Title:&nbsp;".$result['title']."</b><br />
								&nbsp;&nbsp;Author:&nbsp;".$result['author']."<br />
								&nbsp;&nbsp;Date Added:&nbsp;".$result['date']."<br />
								&nbsp;&nbsp;Hits:&nbsp;".$result['hits']."
							</td>
						</tr>
						<tr>
							<td valign=\"top\" class=\"t_bg\" colspan=\"2\">Description:&nbsp;".$result['description']."</td>
						</tr>
						<tr>
							<td valign=\"top\" colspan=\"2\"><img src=\"***/tuts_bottom.gif\" border=\"0\" alt=\"\" /></td>
						</tr>
					</table>";
		}
	}
}

if(!isset($_GET['category'])){
	echo "You must select a category!<br /><br />";
}

include("./footer.php");

?>
<br /><br />

A bit messy, I know, but I kinda slacked off while trying to filter and validate the results. Luckily, just validating is necessary for this script. Anyway, I have no idea what's wrong. This whole problem appeared when I added some code to set the title of the page depending on $_GET. Unfortunately, it seems like that isn't even working properly as when I type in "tutorials.php?category=photoshop" the title is "Flash Tutorials". I had this working fine before, but I changed something to cause this problem and I don't remember what the change was.

So, I'm hoping someone can help me fix this code and possibly point out a way to make it shorter, cleaner, etc.
Thanks again,
Daniel =)

P.S. Ignore all those tables and stuff. I'll make it valid XHTML using Divs later. =P

NogDog

That "not a valid resource id" error means that MySQL was not able to process your query, due to some sort of syntax error or invalid column/table name. To debug, check the result of your query and output some debug info if it fails:

$row=mysql_query($query) or die("Query failed: $query - " . mysql_error());

Daniel_Exe

Heh, oops =P I suppose there was a syntax problem with the SQL. I just changed it to category='".$category."' and it lists the data perfectly. Unfortunately, one other problem just won't go away.. The

        if ($category = "photoshop") { $page_title="Photoshop Tutorials"; }
        if ($category = "flash") { $page_title="Flash Tutorials"; }

Doesn't seem to work. I tried changing it to:

		if ($_GET['category'] = "photoshop") { $page_title="Photoshop Tutorials"; }
		if ($_GET['category'] = "flash") { $page_title="Flash Tutorials"; }

But that didn't fix it either. No matter what $_GET is, the title is always Flash Tutorials. For some reason the Flash IF statement always overwrites the previous one, although I'm not sure why. Could anyone shed some light on this?
Thanks,
Daniel

bradgrafelman

That's because you're using the wrong operator. What you want here is a comparison operator (man page: [man]operators.comparison[/man]), e.g. "==".

Daniel_Exe

Wow, thanks for the help. =) That fixed it. Also, I'm not very familiar with Comparison Operators, so I was wondering what the difference between == and === is? The manual says that both the value and type must be the same, but I'm not sure what they mean by type.
Thanks again for all the help,
Daniel

bradgrafelman

Type, meaning type of the variable. For example:

echo var_dump("1" == 1); // bool(TRUE)
echo var_dump("1" === 1); // bool(FALSE)

"1" is a string, 1 is an integer. The type of the two things being compared is different, so a strict comparison w/ 3 equal signs returns false.

NogDog

Also: 0 (zero), FALSE, and "" all are treated as the same with a "==" comparison, but are not the same with a "===" comparison. This can be very important with functions that may correctly return a numeric value including zero, while returning a boolean FALSE if the function fails.