Pagination keeps showing the same portion of SQL data

Question

I have a really large data set from SQL that i need to paginate.

I have an issue with my pagination code. The code does show the page number in the URL

Answer 1

You should change

$sql='SELECT * FROM ETF';
$result = mysqli_query($con, $sql);
$number_of_results = mysqli_num_rows($result);

to something like this

$count = mysqli_fetch_assoc(mysqli_query($con,"SELECT COUNT(*) AS RC FROM ETF"));
$number_of_results=$count['RC'];

because it is faster to get count from mysql (index if its a large table) instead of fetching all table data (because SELECT *...) for looping trough it just for row counting.

// retrieve selected results from database and display them on page
$sql='SELECT * FROM ETF LIMIT ' . $this_page_first_result . "," .$results_per_page;

You using good query, but using wrong function to get its results.

Query should be

$sql='SELECT * FROM ETF LIMIT ' . $results_per_page . "," .$this_page_first_result;

or:

$sql='SELECT * FROM ETF LIMIT ' . $results_per_page . " OFFSET " .$this_page_first_result;

Both of these work.

Change

while($row = mysqli_fetch_array($result)) {
  echo $row['ETF'] . ' ' . $row['ETF NAME']. '
';
}

to

while($row = mysqli_fetch_assoc($result)) {
  echo $row['ETF'] . ' ' . $row['ETF NAME']. '
';
}

because "mysqli_fetch_array" fetch array width numbered indexes (for example 0, 1, 2...) not a text ones you are trying to use, but "mysqli_fetch_assoc" does that you need.

Full code should be

';
}
// display the links to the pages
for($page=1;$page<=$number_of_pages;$page++)
{
    echo''.$page.' ';
}
mysqli_close($con);
    ?>