How to show a SQL virtual/temporary table from associative array - php

Question

I need a simple SQL query to show a virtual/temporary table without creating it in the database.
I'm using PHP to create that query string with the data.
My current PHP code is,
$array = [
['id' => 1, 'name' => 'one'],
['id' => 2, 'name' => 'two'],
['id' => 3, 'name' => 'three']
];
$subQuery = "SELECT {$array[0]['id']} AS col1, '{$array[0]['name']}' AS col2";
for ($i=1; $i < count($array); $i++) {
$subQuery .= " UNION ALL SELECT {$array[$i]['id']}, '{$array[$i]['name']}'";
}
$sql = "WITH cte AS
(
{$subQuery}
)
SELECT col1, col2 FROM cte;";
echo $sql;
Its output is,
WITH cte AS
(
SELECT 1 AS col1, 'one' AS col2 UNION ALL SELECT 2, 'two' UNION ALL SELECT 3, 'three'
)
SELECT col1, col2 FROM cte;
// Output table from the SQL
col1 col2
1 one
2 two
3 three
I got the idea for this query from here.
But the problem with this query is that,
if I have 100 data in the $array, the UNION ALL part is getting included 100 times in the SQL. I feel like it is not a better SQL because it's like UNION 100 tables at the same time.
I also can create a temporary table (CREATE TEMPORARY TABLE table_name) instead of this WITH clause, but it is not a single query because I need another query to INSERT the records to that temporary table.
Can someone please help me to simplify this query in a better way?

Answer 1

When you use MySQL 8 you can use json_table expression like:
<?php
$array = [
['id' => 1, 'name' => 'one'],
['id' => 2, 'name' => 'two'],
['id' => 3, 'name' => 'three']
];
$data = json_encode($array);
$sql = "SELECT tbl.*
FROM JSON_TABLE(
'{\"data\":$data}',
'$.data[*]' COLUMNS (
id VARCHAR(40) PATH '$.id',
name VARCHAR(100) PATH '$.name')
) tbl";
echo $sql;
PHP online editor
The result of above query in MySQL 8.0 is:
+====+=======+
| id | name |
+====+=======+
| 1 | one |
+----+-------+
| 2 | two |
+----+-------+
| 3 | three |
+----+-------+