Having issuie printing each row. in array

Mon, 05/17/2010 - 05:00 — russellharrower

<?
foreach($photos as $photo => $html) 
	{
		print $html . "";
	}
?>

What I am wanting to do, is show the html part of the array. $photos

Tue, 05/18/2010 - 13:48 — dmorison

Hi Russell, Your loop should

Hi Russell,

Your loop should be:

<?
foreach($photos as $photo)
{
print $photo["html"] . "";
}
?>

Cheers,
David.

Tue, 05/25/2010 - 06:00 — russellharrower

So I changed the code to

So I changed the code to this

<?php
  $galleries = array();
  $sql = "SELECT * FROM gallery ORDER BY id";
  $result = mysql_query($sql);
  while($row = mysql_fetch_assoc($result))
  {
    $galleries[] = array("id" => $row["id"], "name" => $row["name"]); // I am assuming that name is a field in the `gallery` table
  }
  foreach($galleries as $gallery)
  { 
    $sql = "SELECT * FROM photos WHERE galleryid = ".$gallery["id"]." ORDER BY RAND() LIMIT 1";
    $result = mysql_query($sql);
    $row = mysql_fetch_assoc($sql);
    print "<p>";
    print "<img src='http://ghhotel.com.au/images/photos/".$row["address"]."' />";  // I am assuming that url is a field in the `images` table
    print "<br />"; 
    print "<a href='http://ghhotel.com.au/gallery/".$gallery["name"]."'>".$gallery["name"]."  ".$gallery["id"]."</a>"; // this is the link to your gallery page
    print "</p>";
  } 
}
?>

however I get the following error
Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/ghhotel/public_html/photos.php on line 151

Tue, 05/25/2010 - 13:22 — dmorison

Hello Russell, My apologies

Hello Russell,

My apologies - what I presume is your line 151:

$row = mysql_fetch_assoc($sql);

...should have been:

$row = mysql_fetch_assoc($result);

Cheers,
David.