【发布时间】：2019-06-05 04:35:02
【问题描述】：
到目前为止，我有以下 MySQL 查询：
SELECT CONCAT('program:', program_pk) AS global_id,
program_name AS name,
NULL AS parent_global_id
FROM program
UNION ALL
SELECT CONCAT('theme:', theme_pk) AS global_id,
theme_name AS name,
CONCAT('program:', program_pk) AS parent_global_id
FROM theme CROSS JOIN program
UNION ALL
SELECT DISTINCT
CONCAT('theme:', theme_fk, ',strand:', strand_name) AS global_id,
strand_name AS name,
CONCAT('theme:', theme_fk) AS parent_global_id
FROM strand
UNION ALL
SELECT CONCAT('strand_year:', strand_pk) AS global_id,
strand.year AS name,
CONCAT('theme:', theme_fk, ',strand:', strand_name) AS parent_global_id
FROM strand
UNION ALL
SELECT CONCAT('strand_year:', strand_pk, ',unit:', unit_pk) AS global_id,
CONCAT(unit.unit_code, ' ', unit.unit_name) AS name,
CONCAT('strand_year:', strand_pk) AS parent_global_id
FROM strand LEFT JOIN unit ON strand.year = unit.year
表程序
+------------+--------------+
| program_pk | program_name |
+------------+--------------+
表格主题
+----------+------------+
| theme_pk | theme_name |
+----------+------------+
表链
+-----------+-------------+----------+------+
| strand_pk | strand_name | theme_fk | year |
+-----------+-------------+----------+------+
表格单元
+---------+-----------+-----------+--------+------+----------+
| unit_pk | unit_code | unit_name | points | year | theme_fk |
+---------+-----------+-----------+--------+------+----------+
关系是：
程序 -> 主题 -> 链 -> 年份 -> 单元
我现在需要将表learning_event添加到查询中
表学习事件
+-------------------+---------------------+---------+-----------+----------------+
| learning_event_pk | learning_event_name | unit_fk | strand_fk | core_condition |
+-------------------+---------------------+---------+-----------+----------------+
从家长 unit 那里分支学习活动给：
程序 -> 主题 -> 链 -> 年份 -> 单元 -> 学习活动
请注意，对于给定的链和单元，仅应显示与链相关的学习事件。
我玩过这个，但真的不确定如何让它与单元和链相关的学习事件一起工作。
更新
在 JSON 格式中，我现有的查询是这样的：
{
"name": "MD",
"children": [{
"name": "Professional",
"children": [{
"name": "Professional Behavours",
"children": [{
"name": "Year 1",
"children": [{
"name": "IMED4443 Integrated Medical Sciences 1"
}, {
"name": "IMED4444 Integrated Medical Sciences 2"
}]
}
我正在寻找的新输出是这样的：
"name": "MD",
"children": [{
"name": "Professional",
"children": [{
"name": "Professional Behavours",
"children": [{
"name": "Year 1",
"children": [{
"name": "IMED4443 Integrated Medical Sciences 1"
}, {
"name": "IMED4444 Integrated Medical Sciences 2",
"children": [{
"name": "Lecture - CVS"
}, {
"name": "Lecture - Type 1 Diabetes"
}...
并且学习活动应该只显示这与单元和链的关系。
仅供参考，关系是通过以下方式处理的：
$result = $connection->query($query);
$data = array();
while ($row = $result->fetch_object()) {
$data[$row->global_id] = $row;
}
$roots = array();
foreach ($data as $row) {
if ($row->parent_global_id === null) {
$roots[]= $row;
} else {
$data[$row->parent_global_id]->children[] = $row;
}
unset($row->parent_global_id);
unset($row->global_id);
}
$json = json_encode($roots);
新更新
这个查询由 Jonathan Willcock 提供并由我修改，很接近，它显示了所有主题和链的年份，以及第一个主题“专业”的单元和学习事件，但单元没有显示对于任何其他主题。
SELECT CONCAT('program:', program_pk) AS global_id,
program_name AS name,
NULL AS parent_global_id
FROM program
UNION ALL
SELECT CONCAT('theme:', theme_pk) AS global_id,
theme_name AS name,
CONCAT('program:', program_fk) AS parent_global_id
FROM theme
UNION ALL
SELECT
CONCAT('theme:', theme_fk, ',strand:', strand_name) AS global_id,
strand_name AS name,
CONCAT('theme:', theme_fk) AS parent_global_id
FROM strand
UNION ALL
SELECT
CONCAT('theme:', theme_fk, ',strand:', strand_name, ',strandyear:', strandyear_name) AS global_id,
strandyear_name AS name,
CONCAT('theme:', theme_fk, ',strand:', strand_name) AS parent_global_id
FROM strandyear sy
INNER JOIN strand s ON s.strand_pk = sy.strand_fk
UNION ALL
SELECT
CONCAT('theme:', theme_fk, ',strand:', strand_name, ',strandyear:', strandyear_name, ',unit:', unit_name) AS global_id,
unit_name AS name,
CONCAT('theme:', theme_fk, ',strand:', strand_name, ',strandyear:', strandyear_name) AS parent_global_id
FROM unit u
INNER JOIN strandyear sy ON u.strandyear_fk = sy.strandyear_pk
INNER JOIN strand s ON s.strand_pk = sy.strand_fk
UNION ALL
SELECT
CONCAT('theme:', theme_fk, ',strand:', strand_name, ',strandyear:', strandyear_name, ',unit:', unit_name, ',learning_event:', learning_event_name) AS global_id,
learning_event_name AS name,
CONCAT('theme:', theme_fk, ',strand:', strand_name, ',strandyear:', strandyear_name, ',unit:', unit_name) AS parent_global_id
FROM learning_event le
INNER JOIN unit u ON u.unit_pk = le.unit_fk
INNER JOIN strandyear sy ON u.strandyear_fk = sy.strandyear_pk
INNER JOIN strand s ON s.strand_pk = sy.strand_fk
注意，parent_global_id 需要和前面的 global_id 相同。
最后更新
上面的查询工作正常！问题是单位表。更新db-fiddle
【问题讨论】：
-
您确定 UNION ALL 是建立您所描述的关系的正确方法吗？很难说出你想要做什么，因为没有关于你想要的结果的信息。在您的情况下，似乎加入会更合适。如果您可以共享预期最终输出的样本，那么提供一些帮助会更容易，因为 UNION 将您的所有层次结构都放在同一级别，如果不应用一些，您将无法确定之后是什么围绕您的全局父 ID 的某种逻辑。
-
查看更新的 OP...
-
为什么单元有一个theme_fk？
-
很高兴你捡到它...它是作为测试放入的，它没有被使用。
-
嗯？没有名为
children的列。向我们展示SELECT的输出。