PHP Mysql copy a record into a different database - php

Question

I have two separate remote databases, the table in both databases is identicle and I want to copy a record from the old database to the new using PHP
While this the best way to copy new records from one database to another gives the solution as
mysqli_query($db1,"SELECT field1,field2,field3 FROM table1");
mysqli_query($db2,"INSERT INTO table1 (field1,field2,field3)");
Because of the number of fields involved I want to try and avoid naming all the fields
I was thinking of something like this...
$m = mysqli_query($db1,"SELECT * FROM table1");
****THIS IS WHERE I'M STUCK****
HOW DO I GET TO THIS FROM THE ABOVE STATEMENT?
$values = "'".implode("','",array_values($m))."'";
$columns = implode(",",array_keys($m));
So I can do this
mysqli_query($db2,"insert into table1 ($columns) values ($values)")
I'm aware I will need to change the PRIMARY KEY id to null.

Answer 1

To fetch the actual column names you could do like this:
$sql="SELECT column_name FROM information_schema.columns WHERE table_schema = 'database_name' AND table_name = table1";
Fetch the above into an array.
$result = $mysqli->query($query);
$cols = $result->fetch_array(MYSQLI_NUM);
and create comma-separated list into $columns-variable
$columns = implode(",", $cols);
For the values, just do a regular select-statement:
$sql="SELECT * FROM table1";
$result = $mysqli->query($query);
$vals = $result->fetch_array(MYSQLI_NUM);

Answer 2

in the end it was quite simple to modify my own code and create the array using mysqli_fetch_assoc
$m = mysqli_fetch_assoc(mysqli_query($db1,"SELECT * FROM table1"));
and then to make the id null
$m["id"] = 'replacethis';
$values = str_replace("'replacethis'","null","'".implode("','",array_values($m))."'");
$columns = implode(",",array_keys($m));
then finally...
mysqli_query($db2,"insert into table1 ($columns) values ($values)");