Turning Data into URLs?

Posted on 16th Feb 2014 07:03 pm by admin

This may seem very basic to you, but I'm having a hard time figuring out how to do this since I don't even know what the proper term for it is:

Background info:
Animals Table - animalid, name, photo, achievementname, price, resale, levelid, description
Achievement Table - achievementid, achievementname, photo, xprequired, rewardname, description
Levels Table - levelsid, name, xprequired, description
Rewards Table - rewardid, rewardname, desciption

What I want to do is have my animals.php page pull up the MySQL Animals Table (already done) and have all the animal's achievementsid values link to a new page detailing that specific achievement (achievement.php). In turn, that achievement page would show the rewardid value that would link to a page detailing that specific reward (reward.php)

Example:
Animals.php shows a table with:
1-Horse--horse.png-Yee Haw!-1,000$-250$-1-A horse is a horse of course!

I want Yee Haw! to display as a link that would open up a php page that would show information relating to that specific achievement, ignoring all other achievements.

This would work in the same vein as this website: http://farmville.aweblog.net/almanac/animals/all-animals

I assume in being correct that I would have to call the result of that value into a variable that would be passed on as a cookie to another php file but I have not idea where or how to start on that.

My loop looks like this (shortened due to repetition):
while($row = mysql_fetch_array($result)){
echo "<tr>";
echo "<td>" .$row['idAnimal']. "</td>";
echo "<td><img src='".$url.$row['photoAnimal']."' height='50px' width='50px'></td>";
echo "<td>" .$row['nameAnimal']. "</td>";
echo "<td>" .$row['nameCategory']. "</td>";
echo "</tr>";
}


Assuming I wanted each animal name to link to a detailed animal page, what would I do? Can anyone give me some pointers or point me to a good resource for this kind of information? Googling is pointless as I don't know what terms I should be using...
Thanks

Your Answer:

Login to answer

Other forums

Why is this query failing?
Why is this not working?
$query = "SELECT * FROM `users` WHERE `userid` = " . $USERID

PHP Thumbnail Creation
Ok so i use this function to create thumbnails:

Code: function createthumb($name,$filename,$n

Images in email problem
I found this script for sending emails, it works but if the email has <img src="image.jp

bit of help needed
Im about to sort out my registration page for my website by customising a "registration" s

Giving the below syntax error
Dear Experts,

I have used 2 buttons in gui status as display and cjhange.When the user wi

Export hangs
Hi all, please help

I have a Oracle 10.2.0.3.0 database. When I want to query the dba_segmen

Help with explandable category tree
I have the below query:

SELECT l1.id as lev1_id,l2.id as lev2_id,l3.id as lev3_id,l1.categ

Date Question
I'm using the TIMESTAMP() function within mysql to set the date/time for certain events. However, ho

Is it possible to put an entire 500-page book in a database with PHP?
I am working on an intranet and I was wondering if its possible to code php with mysql to enter a fu

cURL and Sessions
Ohai.

So, I'm trying to cURL a bunch of things off of a page, and put them into a session. I'