unitedintruth

I have a page of photos that can be viewed at http://www.littlemi-kis.com/puppy.php I need to make it so when someone clicks on a photo it goes to a page of photos of just that puppy. I have created the extra pages for puppy pictures. I can make the page where each photo has its own query in a table, is there an easier way to do this?

Works perfectly, thank you very much for your help. I would be lost without you guys.

I have this added but it is not showing any data from the database. $sql = mysql_query("SELECT * FROM 'newsletter'");//if you just want to select every entry on the table, leave the WHERE part out. while ($row = mysql_fetch_assoc($sql)){//loop through the entries $month = $row['month'];//should be $row[month column name] $year = $row['year'];//should be $row[year column name] $link = $row['newsletter'];//should be $row[newsletter column name] echo "$month<&nbsp>$year. <a href='admin/newsletter/$link'>Click here to view!</a>"; }

I have a table in a mysql database for newsletters. The fields in the database are month, year, newsletter. The newsletters are in pdf format and in an /admin/newsletter folder on the server but the filename is in the database. How would I get a list showing the month, and year with a "Read" button linking to the newsletter? I am a little lost on how to write the table to get the thing to show so people can access the data. Thanks

Thank you very much, works like a charm now.

The only things I changed was the host, user, password database and the tablename in the select statement and I am getting the following error Warning: mysql_query(): 7 is not a valid MySQL-Link resource in /home/garrett/public_html/links.php on line 155 line 155 is $query = mysql_query("SELECT * from links", $con) or die(mysql_error());

The part about fetching and echoing it is what I don't have. I have connected to the db. And once I select * from links WHERE approved = Approved I am stuck.

Okay, so I'm more of a newbie today than most days, just when you think you are figuring it out. I am trying to make a links directory, I got my form working to insert the data into the database, now I am trying to get the results to layout in a table. My only fields are Name, Description, and URL and I can't seem to get it to layout. Can someone please help me a little bit? Thanks

Perfect solution, that worked great, thank you very much. Thanks, Donnie

Yes, but with the other fields showing below each image.

Considering I am very new to PHP and just learning, this is what I have now. <?php $con = mysql_connect("localhost","username","password"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("my_db", $con); $result = mysql_query("SELECT * FROM puppies WHERE approved='yes'"); echo "<table border='1'>"; while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td width=\"150\" align=\"center\"><img src=\"http://www.americanmi-kiregistryassociation.com/images/puppy_pics/" . $row['photo'] . "\"><br />"; echo "Date of Birth: " . $row['dob'] . "<br />"; echo "Date Available: " . $row['doa'] . "<br />"; echo "Puppy Number: " . $row['nbr'] . "</td>"; echo "</tr>"; } echo "</table>"; mysql_close($con); ?> The table fields had to change some to fit my needs. I need to make this read 3 wide rather than 1. Thanks, Donnie

I have a mysql table written for images. The fields are id, image, title, comment. I am needing to write a table that will lay the images in a page 3 wide and I am having trouble getting them to layout. I can get the data to lay out one image wide but not 3. I understand I do this with an "i" variable but have no idea how to lay this out and can't find a tutorial for this anywhere on the web. Can someone please help me with this? Thanks, Donnie