Database not taking value

[FruitFaster]

The value I want storing is not being stored ($final) in the database which is staying at 0 or being updated as 0. Please can someone hint me in the right direction.

<?php

$con2=mysqli_connect("host.com","uname","pwd","dbase");

  // Check connection
if (mysqli_connect_errno($con2))
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  
  
  }
  $value=$_POST['rating'];
  $id=$_POST['id'];
  
  
   mysqli_query  ($con2, "UPDATE table SET totalraters = totalraters + 1 WHERE id='$id' ");
   mysqli_query  ($con2, "UPDATE table SET ratingsum = ratingsum + '$value' WHERE id='$id' ");
   
  
  
  
  $query = "SELECT ratingsum, totalraters FROM table WHERE id='$id' ";
  $result2=mysqli_query($con2, $query);
  while ($row2= mysqli_fetch_array($result2))
  {
  
  
  

  $ratingsum=$row2['ratingsum']; 
  $totalraters=$row2['totalraters'];
  
   
   IF ($totalraters != 0)  {
  $rating = ($ratingsum/$totalraters);
  $final = (int)round($rating);
  }
  mysqli_query ($con2, "INSERT INTO table rating VALUES '$final' WHERE id='$id' ");
  
  
  
  include ('index.html');


}

?>  

The table has a field called 'rating' which is set to int (6) NOT NULL.

[jcbones]

Being that there is obviously more than one column in the `table rating` table, then you would need to specify the column that you want the value stored in. 

 

You should always add de-bugging into your scripts, it will help you with these time consuming errors:

change

mysqli_query ($con2, "INSERT INTO table rating VALUES '$final' WHERE id='$id' ");

 

to

mysqli_query ($con2, "INSERT INTO table rating VALUES '$final' WHERE id='$id' ") or trigger_error(mysqli_error($con2));

 

 

[FruitFaster]

Thankyou!

I changed that and realised what was wrong with that line (there's no WHERE for INSERT INTO) so I changed it. I checked my database but the number in the rating field is wrong. It updated it to 1 and it should be 4.

 

The line I changed is now:

 mysqli_query ($con2, "UPDATE table SET rating='$final' AND id='$id'  ") or trigger_error(mysqli_error($con2));

Edited May 5, 2013 by FruitFaster

[FruitFaster]

Please can someone give me a hint as why my database is being updated wrongly as explained ^^^^

[jcbones]

You will need to detail the expected values, AND your table structures.  

 

these two

 

mysqli_query ($con2, "UPDATE table SET totalraters = totalraters + 1 WHERE id='$id' ");
mysqli_query ($con2, "UPDATE table SET ratingsum = ratingsum + '$value' WHERE id='$id' ");

 

Can be combined

mysqli_query($con2,"UPDATE table SET totalraters = totalraters + 1, ratingsum = ratingsum + $value WHERE id = $id");

 

I'm still not sure what `table rating` looks like, nor how it ties into the other tables, so I won't put a query here for it as it may lead down the wrong rabbit hole, until structure and usage is addressed.