Display mysql in separate tables - php

Question

Hi I am just a beginner with phpmysql and trying to display result in separate table by a row.
I have this query:
$albania = mysqli_query($con,"SELECT * FROM `data` ORDER BY citid ASC");
while($row = mysqli_fetch_array($albania))
I know that this will output all results onto 1 table. I want to put results onto separate tables dependent on the $row['citizenship']
I have about 40 different results in that row, so I want to separate it by that row and output separately.

Answer 1

If you order by the field you want to split on (citizenship) and keep track of the current value (done below with $currentCitizenship), then you should be able to just look for changes in that value and start a new table when the next section starts.
$albania = mysqli_query($con,"SELECT * FROM `data` ORDER BY citizenship ASC, citid ASC");
$currentCitizenship = ""
while($row = mysqli_fetch_array($albania)){
//handle the first time through the loop and start the first table
if ($currentCitizenship = ""){
//Start the first table
echo "<table>";
$currentCitizenship = $row['citizenship'];
}elseif ($currentCitizenship != $row['citizenship']){
//change to next citizenship table
echo "</table><table>";
$currentCitizenship = $row['citizenship'];
}
/*
Your current row printing code
*/
}
//close off the last table
echo "</table>"

Answer 2

What you wrote will not output anything yet. It will just fetch all rows from the database. You need to add code in PHP to diplay it in the screen.
Something like this. Need to put the correct php syntax:
$albania = mysqli_query($con,"SELECT * FROM `data` ORDER BY citid ASC");
while($row = mysqli_fetch_array($albania))
{
if($row['citizenship'] equals 'valuea')
print in one table
else
print in another table
}