CODEWITHSUNDEEP
phpimagerating
psandhu
psandhu

Reputation: 65

Rating images from a database

I am trying to create a rating system with my gallery. I can get all the images to appear on the page with a button which says "like this image". The aim is that a user can only like an image once and that once they click the "like" button, the value 1 will be added to the image in the database. I have a a database table called images with ImageID, image and rating.The initial value of rating is zero when someone uploads an image. Here is the code i used to grab the images from the database table to appear with a button:

<?php
        mysql_connect("localhost","root","");
        mysql_select_db("pardeepsandhu");


        $res= mysql_query("select * from images");
        $row=mysql_fetch_array($res);
?>
<?php

?>
<div id="w">
  <div id="content">
      <div id="images"><?php while ($row=mysql_fetch_array($res)){?>
        <div id="pic" style="display:inline-block">
        <a href="<?php echo $row["Image"]; ?>"><img src="<?php echo $row["Image"]; ?>" height="135" width="135" border="5" alt="turntable" /></a>
        <form id="form1" name="form1" method="post" action="">
        <input type="submit" name="rate" id="rate" value="Like this image"/>
        </form>
        </div>
<?php } ?>
</div>
  </div>
</div>
<script type="text/javascript">
$(function() {
    $('#images a').lightBox();
});
</script>

I know that i will need a if(isset($_POST['rate'])){} initially to get the ball rolling, however i don't know how to make it. If a user clicks a button under an image, the rate column in the database will increase by one and the user can only do this once on an image. Can someone help? Thanks!

Upvotes: 0

Views: 716

Answers (1)

Jeff
Jeff

Reputation: 799

I recommend a slightly different approach. In order to track who's rated what images, make a new DB table with columns userID and imageID. To get an image's score, you can just count the number of rows where imageID is the current image's ID, and to check if a user has already rated the image, check if a row exists such that imageID = current image and userID = current user.

Getting an image's rating

// $curID is the current image ID
$query = "SELECT * FROM `ratings` WHERE `imageID` = $curID";
// use mysqli->num_rows

Checking if a user has rated the current image

// $curUser, $curID are current user ID, current image ID
$query = "SELECT * FROM `ratings` WHERE `imageID` = $curID AND `userID` = $curUser;";
// check if mysqli->num_rows > 0

Inserting a rating

$query = "INSERT INTO `ratings` (`userID`, `imageID`) VALUES ($curID, $curUser);";

Upvotes: 3

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