は、私は、このソリューション
--------------------------------
と一緒に行きました
1.selectすべての分+最大LAST_UPDATE、合計（interval_avg）、合計（numb_update）とカウント（重複）foreachのホスト名
//This will get the interval_avg value
//(summarize is ok, since all except min(id) will be zero),
//give a count of how many duplicates there are per hostname,
//and will also summarize numb_updates
SELECT
MIN(id) as id,
hostname,
SUM(numb_updates) as total_numb_updates,
SUM(interval_avg) as total_interval_avg,
MAX(last_update) as last_update_max,
MIN(last_update) as last_update_min,
COUNT(*) as hostname_count
FROM
hostname_table
GROUP BY
hostname
HAVING
COUNT(*)>1
//Get all last_update from each duplicate hostname(including the original)
//Dont do this in a seperate query, you only need first+last+rowcount to figure
//out the interval average. It took me a while to realize this, so I tried many
//varieties with little success(took too long with +600k rows)
//
// --- I will include the solution I didn't go for, ---
// --- so others wont do the same mistake ---
//
// START DONT USE THIS
// 2.63sec @ 10000 rows
$sql = "SELECT
id,
".$db_table.".hostname,
last_update
FROM
".$db_table."
INNER JOIN (
SELECT
hostname,
COUNT(*)
FROM
".$db_table."
GROUP BY
hostname
HAVING
COUNT(*)>1
) as t2
ON
".$db_table.".hostname = t2.hostname";
$resource = mysql_query($sql,$con);
// END DONT USE THIS (below is a 94% improvement)
//
// START THIS IS BETTER, BUT DONT USE THIS
// 0.16 sec @ 10000 rows
//Select everything from the table
$sql = "SELECT id
FROM ".$db_table;
$resource = mysql_query($sql,$con);
$array_id_all = array();
while($assoc = mysql_fetch_assoc($resource)){
array_push($array_id_all, $assoc['id']);
}
//This will select the ID of all the hosts without duplicates
$sql = "SELECT
MIN(id) as id,
hostname
FROM
".$db_table."
GROUP BY
hostname
HAVING
COUNT(*)=1";
$resource = mysql_query($sql,$con);
$array_id_unique = array();
while($assoc = mysql_fetch_assoc($resource)){
array_push($array_id_unique, $assoc['id']);
}
$array_id_non_unique = array_diff($array_id_all, $array_id_unique);
$id_list_non_unique = implode(", ", $array_id_non_unique);
//Select everything from the table when the IDs are IN $id_list_non_unique
$sql = "SELECT *
FROM ".$db_table."
WHERE id IN (".$id_list_non_unique.")";
$resource = mysql_query($sql,$con);
$array_duplicates = array();
$i=0;
while($assoc = mysql_fetch_assoc($resource)){
$array_duplicates[$i] = array($assoc['id'], $assoc['hostname'], $assoc['interval_avg'], $assoc['last_update'], $assoc['numb_updates']);
$i++;
}
// END THIS IS BETTER, BUT DONT USE THIS
（https://stackoverflow.com/a/877051/1248273 @ニックフォーテスキューのおかげ）
2.update分（ID）でinterval_avg、分（ID）で更新numb_updates、MAX（ID）の値を持つ分（ID）で更新LAST_UPDATE
//update the interval_avg, last_update and numb_update value of the min(id)
//of each duplicate hostname.
// --- I will include the solution I didn't go for, ---
// --- so others wont do the same mistake ---
//
// START DONT USE THIS
// 167 secs @ 500k rows
UPDATE hostname_table
SET interval_avg = CASE id
WHEN 1 THEN 25
//etc
END,
last_update = CASE id
WHEN 1 THEN "2012-04-25 20:22:36"
//etc
END,
numb_update = CASE id
WHEN 1 THEN 3
//etc
END
WHERE id IN (1)
// END DONT USE THIS
//
// START USE THIS
// 5.75 secs @ 500k rows (96.6% improvement)
INSERT INTO hostname_table (id,interval_avg,last_update,numb_updates)
VALUES
('1','25','2012-04-25 20:22:36','3'),
//etc
ON DUPLICATE KEY UPDATE
interval_avg=VALUES(interval_avg),
last_update=VALUES(last_update),
numb_updates=VALUES(numb_updates)
// END USE THIS
（https://stackoverflow.com/a/3466/1248273 @ミシェル・デ・マールのおかげ）
3.delete分を除くすべての複製（ID）
//delete all duplicates except min(id)
ALTER IGNORE TABLE hostname_table ADD UNIQUE (hostname)
ALTER TABLE hostname_table DROP INDEX hostname
（私が必要な最初の情報を選択する上で右方向への良いプッシュのためGolezTrolのおかげで）
@GolezTrol： 'DELETE'文は非常にefficentないかもしれません。私は、あなたがDeleteステートメントの 'WHERE'にサブクエリを持っているとき、MySQLがうまく動作しないと思います。派生したサブクエリを結合に移動する方が効率的です。 –
@ypercubeそして、あなたはどのように削除に参加しますか？ – GolezTrol
@turbopipp interval_avgに必要なものは完全にはっきりしませんが、avarageを再計算する必要があるようですので、AVG（interval_avg）またはAVG（interval_avg * numb_updates）のどちらかが必要です。 – GolezTrol