如何从关联数组中显示 SQL 虚拟/临时表

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 |
+----+-------+