我有以下数据结构和数据:
CREATE TABLE `parent` (
`id` int(11) NOT NULL auto_increment,
`name` varchar(10) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1;
INSERT INTO `parent` VALUES(1, 'parent 1');
INSERT INTO `parent` VALUES(2, 'parent 2');
CREATE TABLE `other` (
`id` int(11) NOT NULL auto_increment,
`name` varchar(10) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1;
INSERT INTO `other` VALUES(1, 'other 1');
INSERT INTO `other` VALUES(2, 'other 2');
CREATE TABLE `relationship` (
`id` int(11) NOT NULL auto_increment,
`parent_id` int(11) NOT NULL,
`other_id` int(11) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1;
INSERT INTO `relationship` VALUES(1, 1, 1);
INSERT INTO `relationship` VALUES(2, 1, 2);
INSERT INTO `relationship` VALUES(3, 2, 1);
我想找到父母双方的记录1和2。
我就是这么想的,但我想知道是否有更好的方法:
SELECT p.id, p.name
FROM parent AS p
LEFT JOIN relationship AS r1 ON (r1.parent_id = p.id)
LEFT JOIN relationship AS r2 ON (r2.parent_id = p.id)
WHERE r1.other_id = 1 AND r2.other_id = 2;
结果是1,“父级1”,这是正确的。问题是,一旦得到一个5+连接的列表,它就会变得混乱,并且随着关系表的增长,它会变得缓慢。
有更好的办法吗?
我正在使用MySQL和PHP,但这可能是非常通用的。
最佳答案
好吧,我测试过了。从最好到最坏的问题是:
查询1:连接(0.016s;基本上是即时的)
SELECT p.id, name
FROM parent p
JOIN relationship r1 ON p.id = r1.parent_id AND r1.other_id = 100
JOIN relationship r2 ON p.id = r2.parent_id AND r2.other_id = 101
JOIN relationship r3 ON p.id = r3.parent_id AND r3.other_id = 102
JOIN relationship r4 ON p.id = r4.parent_id AND r4.other_id = 103
查询2:存在(0.625s)
SELECT id, name
FROM parent p
WHERE EXISTS (SELECT 1 FROM relationship WHERE parent_id = p.id AND other_id = 100)
AND EXISTS (SELECT 1 FROM relationship WHERE parent_id = p.id AND other_id = 101)
AND EXISTS (SELECT 1 FROM relationship WHERE parent_id = p.id AND other_id = 102)
AND EXISTS (SELECT 1 FROM relationship WHERE parent_id = p.id AND oth
查询3:聚合(1.016s)
选择p.id,p.name
来自父p
式中(从关系中选择COUNT(*),其中parent_id=p.id和other_id IN(100101102103))
查询4:联合聚合(2.39s)
SELECT id, name FROM (
SELECT p1.id, p1.name
FROM parent AS p1 LEFT JOIN relationship as r1 ON(r1.parent_id=p1.id)
WHERE r1.other_id = 100
UNION ALL
SELECT p2.id, p2.name
FROM parent AS p2 LEFT JOIN relationship as r2 ON(r2.parent_id=p2.id)
WHERE r2.other_id = 101
UNION ALL
SELECT p3.id, p3.name
FROM parent AS p3 LEFT JOIN relationship as r3 ON(r3.parent_id=p3.id)
WHERE r3.other_id = 102
UNION ALL
SELECT p4.id, p4.name
FROM parent AS p4 LEFT JOIN relationship as r4 ON(r4.parent_id=p4.id)
WHERE r4.other_id = 103
) a
GROUP BY id, name
HAVING count(*) = 4
实际上上面的数据是错误的,所以要么是错误的,要么是我做错了。不管是什么情况,以上只是一个坏主意。
如果不是很快,那么您需要查看查询的解释计划。你可能只是缺少合适的指标。试试看:
CREATE INDEX ON relationship (parent_id, other_id)
在沿着聚集路线前进(选择COUNT(*)FROM…)之前,您应该阅读SQL Statement - “Join” Vs “Group By and Having”。
注:上述计时基于:
CREATE TABLE parent (
id INT PRIMARY KEY,
name VARCHAR(50)
);
CREATE TABLE other (
id INT PRIMARY KEY,
name VARCHAR(50)
);
CREATE TABLE relationship (
id INT PRIMARY KEY,
parent_id INT,
other_id INT
);
CREATE INDEX idx1 ON relationship (parent_id, other_id);
CREATE INDEX idx2 ON relationship (other_id, parent_id);
以及近80万条记录的创建:
<?php
ini_set('max_execution_time', 600);
$start = microtime(true);
echo "<pre>\n";
mysql_connect('localhost', 'scratch', 'scratch');
if (mysql_error()) {
echo "Connect error: " . mysql_error() . "\n";
}
mysql_select_db('scratch');
if (mysql_error()) {
echo "Selct DB error: " . mysql_error() . "\n";
}
define('PARENTS', 100000);
define('CHILDREN', 100000);
define('MAX_CHILDREN', 10);
define('SCATTER', 10);
$rel = 0;
for ($i=1; $i<=PARENTS; $i++) {
query("INSERT INTO parent VALUES ($i, 'Parent $i')");
$potential = range(max(1, $i - SCATTER), min(CHILDREN, $i + SCATTER));
$elements = sizeof($potential);
$other = rand(1, min(MAX_CHILDREN, $elements - 4));
$j = 0;
while ($j < $other) {
$index = rand(0, $elements - 1);
if (isset($potential[$index])) {
$c = $potential[$index];
$rel++;
query("INSERT INTO relationship VALUES ($rel, $i, $c)");
unset($potential[$index]);
$j++;
}
}
}
for ($i=1; $i<=CHILDREN; $i++) {
query("INSERT INTO other VALUES ($i, 'Other $i')");
}
$count = PARENTS + CHILDREN + $rel;
$stop = microtime(true);
$duration = $stop - $start;
$insert = $duration / $count;
echo "$count records added.\n";
echo "Program ran for $duration seconds.\n";
echo "Insert time $insert seconds.\n";
echo "</pre>\n";
function query($str) {
mysql_query($str);
if (mysql_error()) {
echo "$str: " . mysql_error() . "\n";
}
}
?>
所以再次加入扛天。
关于sql - 您如何执行与联接的AND?,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/599461/