Reorder / reset auto increment primary key

Question

I have a MySQL table with an auto increment primary key. I deleted some rows in the middle of the table. Now I have, for example, something like this in the ID column: 12, 1

Answer 1

You may simply use this query

alter table abc auto_increment = 1;

Answer 2

You could drop the primary key column and re-create it. All the ids should then be reassigned in order.

However this is probably a bad idea in most situations. If you have other tables that have foreign keys to this table then it will definitely not work.

Answer 3

Even though this question seems to be quite old, will post an answer for someone who reaches in here searching.

SET @count = 0;
UPDATE `users` SET `users`.`id` = @count:= @count + 1;

If the column is used as a foreign key in other tables, make sure you use ON UPDATE CASCADE instead of the default ON UPDATE NO ACTION for the foreign key relationship in those tables.

Further, in order to reset the AUTO_INCREMENT count, you can immediately issue the following statement.

ALTER TABLE `users` AUTO_INCREMENT = 1;

For MySQLs it will reset the value to MAX(id) + 1.

Answer 4

SELECT * from `user` ORDER BY `user_id`; 

SET @count = 0;

UPDATE `user`  SET `user_id` = @count:= @count + 1;

ALTER TABLE `user_id` AUTO_INCREMENT = 1;

if you want to order by

Answer 5

To reset the IDs of my User table, I use the following SQL query. It's been said above that this will ruin any relationships you may have with any other tables.

ALTER TABLE `users` DROP `id`;
ALTER TABLE `users` AUTO_INCREMENT = 1;
ALTER TABLE `users` ADD `id` int UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY FIRST;

Answer 6

You can remove the primary key auto increment functionality of that column, then every time you update that column run a query before hand that will count all the rows in the table, then run a loop that iterates through that row count inserting each value into the respective row, and finally run a query inserting a new row with the value of that column being the total row count plus one. This will work flawlessly and is the most absolute solution to someone trying to accomplish what you are. Here is an example of code you may use for the function:

$table_row_count = mysql_result(mysql_query("SELECT COUNT(`field_1`) FROM `table`"), 0);
$viewsrowsdata = mysql_query("
    SELECT `rank`, `field1`, `field2`, `field3`, `field4`
        FROM (SELECT (@rank:=@rank+1) as `rank`, `field1`, `field2`, `field3`, `field4`
            FROM (SELECT * FROM `views`) a
            CROSS JOIN (SELECT @rank:=0) b
            ORDER BY rank ASC) c
");
while ($row = mysql_fetch_assoc($viewsrowsdata)) {
    $data[] = $row;
}
foreach ($data as $row) {
    $new_field_1 = (int)$row['rank'];
    $old_field_1 = (int)$row['field1'];
    mysql_query("UPDATE `table` SET `field_1` = $new_field_1 WHERE `field_1` = $old_field_1");
}
mysql_query("INSERT INTO `table` (`field1`, `field2`, `field3`, `field4`) VALUES ('$table_row_count' + 1, '$field_2_value', 'field_3_value', 'field_4_value')");

Here I created an associative array which I had appended on a rank column with the query within a select query, which gave each row a rank value starting with 1. I then iterated through the associative array.

Another option would have been to get the row count, run a basic select query, get the associative array and iterate it through the same way but with an added variable that updates through each iteration. This is less flexible but will accomplish the same thing.

$table_row_count = mysql_result(mysql_query("SELECT COUNT(`field_1`) FROM `table`"), 0);
$viewsrowsdata = mysql_query("SELECT * FROM `table`");
$updated_key = 0;
while ($row = mysql_fetch_assoc($viewsrowsdata)) {
    $data[] = $row;
}
foreach ($data as $row) {
    $updated_key = $updated_key + 1;
    mysql_query("UPDATE `table` SET `field_1` = '$updated_key' WHERE `field_1` = '$row['field_1']'");
}
mysql_query("INSERT INTO `table` (`field1`, `field2`, `field3`, `field4`) VALUES ('$table_row_count' + 1, '$field_2_value', 'field_3_value', 'field_4_value')");