在事务中为两个表插入相同的键（1. primary 2. foreign）

I have in this code 2 queries. (in my real code I have 6 queries and I need transaction).

I don't know how to get variable $category_id cause that category isn't putted yet in database (it should be inserted in same time - all or nothing)

code:

try {
    $this->mysqli->begin_transaction();

    $this->mysqli->query("INSERT INTO `category` (`name`) VALUES ('$category')");
    $this->mysqli->query("INSERT INTO `subcategory` (`name`,`category_id` ) VALUES ('$subcategory','$category_id')");

    $this->mysqli->commit();

}
catch (Exception $e) {
    echo $e;
    $this->mysqli->rollBack();
}

mysql tables:

    category:
    ---------
    |id|name|

    subcategory:
    |id|name|category_id|

So I need some solution how to know before query what is the value of $category_id, or how to modify query so category_id in database is filed.

LAST_INSERT_ID() is what you want here.

try {
    $this->mysqli->begin_transaction();

    $this->mysqli->query("INSERT INTO `category` (`name`) VALUES ('$category')");
    $this->mysqli->query("INSERT INTO `subcategory` (`name`,`category_id` ) VALUES ('$subcategory', LAST_INSERT_ID())");

    $this->mysqli->commit();

}
catch (Exception $e) {
    echo $e;
    $this->mysqli->rollBack();
}

P.S. Look into prepared statements, instead of concatenating variables into your query.