我目前有以下查询(以及一些周围的vB代码)。我希望我可以将其简化为单个SELECT语句,而不是在每个页面上再运行10个以获取个人名称。
$results = $db->query_read_slave("
SELECT user.id AS steam, user.skills AS level
FROM wcsp.wcgousers AS user
WHERE race = '_wcs_'
ORDER BY ABS(user.skills) DESC LIMIT 10
");
$rank = 1;
while ($user = $db->fetch_array($results)) {
$result = $db->query_first("
SELECT old.name AS name
FROM wcsp.warn_oldnames AS old
WHERE id = '$user[steam]'
ORDER BY lasttime
DESC LIMIT 1
");
$listing[] = array("id" => $user['steam'], "level" => $user['level'], "name" => $result['name'], "rank" => $rank);
$rank += 1;
}
我尝试过LEFT JOIN,但我遇到的问题是我需要LEFT JOIN中的子查询类似于:
SELECT user.id AS steam, user.skills AS level, names.name AS name
FROM wcsp.wcgousers AS users
LEFT JOIN
(
SELECT names.name
FROM wcsp.warn_oldnames AS names
WHERE id = wcsp.wcgousers.id
ORDER BY lasttime DESC LIMIT 1
) AS names
ON names.id = users.id
WHERE users.race = '_wcs_'
由于子查询中的数据库检查不同,这将无效。
答案 0 :(得分:1)
如果我理解正确,您希望为每个用户获取最新的Name
。
SELECT a.id AS steam,
a.skills AS level,
b.name
FROM wcgousers a
INNER JOIN warn_oldnames b
ON a.ID = b.ID
INNER JOIN
(
SELECT ID, MAX(lasttime) max_date
FROM warn_oldnames
GROUP BY ID
) c ON b.ID = c.ID AND
b.lastTime = c.max_date
WHERE a.race = '_wcs_'
-- ORDER BY ABS(a.skills) DESC
-- LIMIT 10