[英]Chain of SQL subqueries within large query of JOINS
我正在尝试使用复杂的嵌套选择操作来构造SQL查询!
我原来的SQL查询成功加入了约30个表,可以按需检索数据! 但是,这30个表中的每个获取的记录在另一个表中(注释)都有很多记录! 我想做的是按ID将(Comments表)中的每条记录分配给其他30表中的记录,并在一个查询中将它们全部检索到。 但这不是唯一的挑战,除了(Comments表)中的记录外,这30个表中的一些还有另一个称为(extras)的表中的更多记录,因此我正在LEFT JOIN内的主子查询中寻找其他子查询外部主要查询。
使想法更清晰; 没有子查询的脚本将如下所示:
$query = $mysqli->query("
SELECT
parent1.parent1_id,
parent1.child1_id,
parent1.child2_id,
parent1.child3_id,
parent2.parent2_id,
parent2.child1_id,
parent2.child2_id,
parent2.child3_id,
child1.child1_id,
child1.child1_content,
child2.child2_id,
child2.child2_content,
child3.child3_id,
child3.child3_content
FROM
parent1
LEFT JOIN child1
ON child1.child1_id = parent1.child1_id
LEFT JOIN child2
ON child2.child2_id = parent1.child2_id
LEFT JOIN child3
ON child3.child3_id = parent1.child3_id
LEFT JOIN followers
ON parent1.user_id = followers.followed_id
AND parent1.parent1_timestamp > followers.followed_timestamp
AND parent1.parent1_id NOT IN (SELECT removed.isub_rmv FROM removed)
AND parent1.parent1_hide = false
WHERE
followers.follower_id = {$_SESSION['info']}
{$portname_clause}
ORDER BY
parent1.parent1_timestamp DESC
LIMIT
{$postnumbers}
OFFSET
{$offset}
")
// Now fetching and looping through the retrieved data
while($row = $query->fetch_assoc()){
echo $row['child1_content'];
$subquery1 = $mysqli->query("SELECT extras.child1_id,
extras.extrasContent FROM extras WHERE extras.child1_id =
{$row['child1_id']}");
while($row1 = $subquery1->fetch_assoc()){
echo $row1['extrasContent'];
}
echo $row['child2_content'];
$subquery2 = $mysqli->query("SELECT extras.child2_id,
extras.extrasContent FROM extras WHERE extras.child2_id =
{$row['child2_id']}");
while($row2 = $subquery2->fetch_assoc()){
echo $row2['extrasContent'];
}
echo $row['child3_content'];
$subquery3 = $mysqli->query("SELECT extras.child3_id,
extras.extrasContent FROM extras WHERE extras.child3_id =
{$row['child3_id']}");
while($row3 = $subquery3->fetch_assoc()){
echo $row3['extrasContent'];
// Here i need to run additional query inside the subquery 3 to retrieve the (Comments table) data beside (extras table)
$subquery4 = $mysqli->query("SELECT comments.comment_id, comments.comment FROM comments WHERE comments.child3_id = {$row['child3_id']} OR comments.child3_id = {$row3['child3_id']}");
while($row4 = $subquery4->fetch_assoc()){
echo $row4['comment'];
}
}
} // No sane person would make such code
因为上面的代码完全是多余的,所以我寻找了一种更好的方法来执行它,这就是我遇到子查询概念的地方,但是我对子查询一无所知,并且在研究了它之后不久,我想到了这个混乱的地方代码,请在下面检查!
我不在这里发布origianl代码,因为它太长了,我在其中包含了要对其应用查询以演示过程的表的虚拟示例。
SELECT
parent1.parent1_id,
parent1.child1_id,
parent1.child2_id,
parent1.child3_id,
parent2.parent2_id,
parent2.child1_id,
parent2.child2_id,
parent2.child3_id
FROM
parent1
LEFT JOIN
( SELECT
child1.child1_id,
child1.child1_content
FROM
child1
WHERE
child1.child1_id = parent1.child1_id ) child1
( SELECT extras.extrasID, extras.extrasContent
FROM
extras
WHERE
extras.child1_id = child1.child1_id )
ON parent1.child1_id = child1.child1_id
LEFT JOIN child2
( SELECT
child2.child2_id,
child2.child2_content
FROM
child2
WHERE
child2.child2_id = parent1.child2_id )
( SELECT
extras.extrasID,
extras.extrasContent
FROM
extras
WHERE
extras.child2_id = child2.child2_id )
ON parent1.child2_id = child2.child2_id
LEFT JOIN child3
( SELECT
child3.child3_id,
child3.child3_content
FROM
child3
WHERE
child3.child3_id = parent1.child3_id )
( SELECT
extras.extrasID,
extras.extrasContent
FROM
( SELECT
comments.comment_id,
comments.comment
FROM
comments
WHERE
comments.child3_id = extras.child3_id ) extras
JOIN child3
ON extras.child3_id = child3.child3_id )
ON parent1.child3_id = child3.child3_id
LEFT JOIN followers
ON parent1.user_id = followers.followed_id
AND parent1.parent1_timestamp > followers.follower_timestamp
AND parent1.parent1_id NOT IN (SELECT removed.isub_rmv FROM removed)
AND parent1.parent1_hide = false
WHERE
followers.follower_id = {$_SESSION['info']}
{$portname_clause}
ORDER BY
parent1.parent1_timestamp DESC
LIMIT
{$postnumbers}
OFFSET
{$offset} // <-- Sorry for the bad code formatting!
我正在使用MySql 5.6.37
我还没有掌握子查询的概念,坦率地说,我在学习它时迷失了自己,感到困惑,并且由于下面的注释中也提到了另一个原因。
注意:我事先表示歉意,因为我住的地方没有电,ADSL或电话,而且我的USB调制解调器几乎没有信号,所以我可能不会立即得到答复,我每天只有两个小时的平均电耗是三个小时Desil发生器。 我给笔记本电脑充电并检查互联网,剩下的一两个小时用于其他生活。 我知道我在开玩笑,因为我正在开发一个没有电力或永久性互联网的Web项目。 但是生活并不能提供一切! 大声笑。
这就是我解决问题的方法!
SELECT
parent1.parent1_id,
parent1.child1_id,
child1.child1_id,
child1.child1_content,
comments.comment_id,
comments.comment,
comments.child1_id
FROM parent1 LEFT JOIN
(
SELECT comments.comment_id, comments.comment, comments.child1_id
FROM
(
SELECT comments.comment_id,comments. comment, comments.child1_id
FROM comments
) comments JOIN child1
ON comments.child1_id = child1.child1_id
) comments
ON child1.child1_id = comments.
它需要一些别名,然后就可以了。
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