Choose only required data from PHP - SQL Group by array function [duplicate] - php

Question

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variable variables bad practice to use?
(2 answers)
Closed 9 months ago.
The following SQL PHP code shows count of country from database and displays it in a table.
Here is the code
foreach($dbh->query('SELECT country,COUNT(*)
FROM data_able
GROUP BY country') as $row) { $array = array();
echo "<tr>";
echo "<td>" . $row['country'] . "</td>";
echo "<td>" . $row['COUNT(*)'] . "</td>";
echo "</tr>";
}
Here is the result
Here is the result
I would like to use the data as $country1 and $count1.
For example $country1 will output "Andorra" and $country5 will be "India".
and $count5 should give "25" as result. How can I do that ?

Answer 1

Here is an example how to declare variables dynamicly:
$array = ['1', '2', '3', '4', '5'];
foreach ($array as $i) {
${'country' . $i} = $i;
}
echo $country2;
You just need to iterate through your query results with keeping some index on variable suffix (in your case a number).
// create a dynamic variables
$result = $dbh->query('SELECT country,COUNT(*) FROM data_able GROUP BY country');
$i = 1;
foreach ($result as $row) {
${'country' . $i} = $row['country'];
${'count' . $i} = $row['COUNT(*)'];
$i++;
}
// example how to use them
$i = 1;
foreach($result as $row) {
$array = array();
echo "<tr>";
echo "<td>" . ${'country' . $i} . "</td>";
echo "<td>" .${'count' . $i} . "</td>";
echo "</tr>";
$i++;
}