Taking a value from a mysql table and adding 1 - php

Question

This is my first time posting in stackoverflow so please tell me if I need to add more information!
I am attempting to take a value from the mysql database and add 1 to it when the page loads. The user clicks on the link on the first page and when the linked page is loading and the php code below runs it adds 1 to the value of the column Clickcount in the entry with the Name value TheButton in the table "lots".
The following code worked when I ran it through Dreamweaver with WAMP:
<?php
$con=mysqli_connect("localhost","root","qwerty71","Lots");
mysqli_query($con,"UPDATE Lots SET Clickcount=Clickcount+1
WHERE Name='TheButton'");
$result=mysqli_query($con,"SELECT * FROM Lots where Name='TheButton'");
while($row= mysqli_fetch_array($result))
{
echo $row['Clickcount'];
}
mysqli_close($con);
?>
But when I put the same code into my website (only changing the values after the mysqli_connect to connect to the database) it doesn't work. The echo at the bottom is to show the value after it is changed, I have the same echo on the first page to show the value before it is edited by the second page. I set the value to zero in the database before I loaded the first page or clicked the link. The first page shows the value at zero, and after clicking the link the second page changes the value and displays a 1. When I try to click back to the first page it shows a 1 instead of a zero, but then both pages continue to display 1s instead of adding 1 every time the link is clicked.
I have tried multiple ways of changing the code after the SET Clickcount=
Including attempting to add a variable before query that I set equal to the Clickcount value and trying to add 1 to that.
Is there an error in my code that would prevent it from working properly, or is there an alternate way to set this up to get the desired result?
Re-explanation for clarity:
My table is as follows:
Name Clickcount
TheButton 0
I am trying to increase the value under "clickcount" by 1 every time a page is loaded. My code (posted above) functioned properly when run on a local server, I changed the parameters to match my server online.
The homepage has a portion of the code above to show the value of the Clickcount column and a link to the second page, where the code above is located.
The code functioned once, changing the value from zero to one, but then ceased to work on consecutive attempts.
Is there an error in my code? Or is there a better way to accomplish the task of increasing the value?
Further update:
The code seems to be working once every 5-10 minutes. I have left it as it is above, and it was not doing this before. I do not know what changed.
In addition, every few times that it does work, I get the result of receiving two different values when calling the following portion of the code on two separate pages:
$result=mysqli_query($con,"SELECT * FROM Lots where Name='TheButton'");
while($row= mysqli_fetch_array($result))
{
echo $row['Clickcount'];
}
mysqli_close($con);
This code is on the homepage, where it sometimes results in a number one less than what the value is recorded to be in the mysql table. Where on the page with the complete code listed above, it always shows the same value as in the table.
In my continuing effort to figure this out, I have removed all of the files off of the server where my site is hosted, and started again from scratch. After re-writing all of the code for the two pages, creating a new table in a new database, and implementing the code above on the second page, the error remains.
I have also just now signed up for a new hosting service to see if it was the server that was causing the problem. Sadly, that was not it. Does anyone have a suggestion for a solution?

Answer 1

Look to this points:
Copy your query and put in the database. It works?
In the end of your update line, put or die(mysqli_error($con)). If show errors, look to error;
There value 'TheButton' in the column Name in your remote table?
Try change in the clause where in update query, Name='TheButton' to Name LIKE 'TheButton'
Always throw messages errors in your codes. Is the best way to find mistakes. ;)

Answer 2

The page was loading a cached version instead of loading from the site itself. I added some coding to the header to remove the caching, this fixed the problem.