如果同一表的其他版本中不存在ID，则将行添加到表中

Question

I searched the solution in google but I can't find (or I don't understand) the right solution. I have two tables, they are the same, but in the newest version, some rows have been loose. So I would like to be able to compare this two tables and inject the missed rows by comparing the ID. For example: If in the table v1 we have the row with the ID 12 and not in the table of the v2 then we add this row in the v2 table.

I wrote an php script but I can't go further, I don't get it how I can compare this values at the same time, maybe I'm using a bad approach, here is my php code:

<?php
    function_db_connect();
    $req_src = "SELECT * FROM `table_v1` ";

    $req_ex_src=mysql_query($req_src) or die("Erreur: ".mysql_error());

    $i =0;

    while($row = mysql_fetch_array($req_ex_src)) {
        $i = $i +1;
        $rowID[$i] = $row['ID'];
    }

    mysqli_close();

    $n = count($rowID);

    for ($j=0;$j<=$n;$j++){
       ??????????????????????
    }

I get all the ID in an array and my idea was to compare this values with the v2 table in the for loop.

That's all I have done, I'm a little confused... Do you think there's another method better than this?

Answer 1

You can use a simple mysql query to do this no need to use php unless there are other factors needing to be taken into account

To do this use insert ignore:

INSERT IGNORE INTO table_v2 (SELECT * FROM table_v1)

this will select all the rows in table_v1, and try to insert them into table_v2, if a row's keys already exist that row is ignored and the query continues on.