我有三张 table 。
基本收入表:
id base_income
------------------------
1 Grants from government
2 other grants
3 Taxes
子收入表:
id base_id sub_income
---------------------------------------
1 1 Special Grants
2 1 General Grants
3 2 Local authorities
4 2 private
5 3 Professional tax
6 3 garbage tax
7 3 light tax
内部子收入表:
id sub_id inner_subincome
-------------------------------------
1 1 matching grant
2 1 XIIIth finance commission
3 1 GIA members salary Grant
4 2 Grants
5 2 Grants in Lieu
我想从三个表中获取与每个表相关的 id 的值。
我尝试了以下查询:
select base_income, sub_income,inner_subincome from base_income_tble left join
sub_income_tble ON base_income_tble.id = sub_income_tble.base_id left join
inner_subincome_tble
on sub_income_tble.id = inner_subincome_tble.sub_id OR sub_income_tble.id =
inner_subincome_tble.sub_id;
并获取值:
我想在html中显示如下格式的数据:
Grants from Government
special Grants
Matching grant
XIIIth finance commission
GIA members salary Grant
General Grants
Grants
Grants in Lieu
other grants
Local authorities
private
可以使用嵌套的foreach来渲染如上格式的数据，如果可以的话条件是什么。
最佳答案
我想到的第一个方法 -
1) 获取数据库中所有数据的列表，包括id数字
select b.id as base_id, s.id as sub_id, i.id as inner_id, b.base_income, s.sub_income, i.inner_subincome from base_income_tble b left join sub_income_tble s on b.id = s.base_id left join inner_subincome_tble i on s.id = i.sub_id
+---------+--------+----------+------------------------+-------------------+--------------------------+
| base_id | sub_id | inner_id | base_income | sub_income | inner_subincome |
+---------+--------+----------+------------------------+-------------------+--------------------------+
| 1 | 1 | 1 | Grants from government | Special Grants | Matching grant |
| 1 | 1 | 2 | Grants from government | Special Grants | XIIIth finance comission |
| 1 | 1 | 3 | Grants from government | Special Grants | GIA members |
| 1 | 2 | 4 | Grants from government | General Grants | Grants |
| 1 | 2 | 5 | Grants from government | General Grants | Grants in Lieu |
| 2 | 3 | NULL | Other grants | Local authorities | NULL |
| 2 | 4 | NULL | Other grants | Private | NULL |
| 3 | 5 | NULL | Taxes | Professional tax | NULL |
| 3 | 6 | NULL | Taxes | Garbage tax | NULL |
| 3 | 7 | NULL | Taxes | Light tax | NULL |
+---------+--------+----------+------------------------+-------------------+--------------------------+
2)将其构建为递归数组
$data = array();
while($r = mysqli_fetch_assoc($rows)){
$bId = 'id' . $r['base_id'];
$sId = 'id' . $r['sub_id'];
$iId = 'id' . $r['inner_id'];
if( !array_key_exists($bId, $data) )
$data[$bId] = array('title' => $r['base_income'], 'values' => array());
if( $r['sub_id'] && !array_key_exists($sId, $data[$bId]['values']) )
$data[$bId]['values'][$sId] = array('title' => $r['sub_income'], 'values' => array());
if( $r['sub_id'] && $r['inner_id'] )
$data[$bId]['values'][$sId]['values'][$iId] = array('title' => $r['inner_subincome']);
}
此后 $data 数组的内容 -
Array
(
[id1] => Array
(
[title] => Grants from government
[values] => Array
(
[id1] => Array
(
[title] => Special Grants
[values] => Array
(
[id1] => Array
(
[title] => Matching grant
)
[id2] => Array
(
[title] => XIIIth finance comission
)
[id3] => Array
(
[title] => GIA members
)
)
)
[id2] => Array
(
[title] => General Grants
[values] => Array
(
[id4] => Array
(
[title] => Grants
)
[id5] => Array
(
[title] => Grants in Lieu
)
)
)
)
)
[id2] => Array
(
[title] => Other grants
[values] => Array
(
[id3] => Array
(
[title] => Local authorities
[values] => Array
(
)
)
[id4] => Array
(
[title] => Private
[values] => Array
(
)
)
)
)
[id3] => Array
(
[title] => Taxes
[values] => Array
(
[id5] => Array
(
[title] => Professional tax
[values] => Array
(
)
)
[id6] => Array
(
[title] => Garbage tax
[values] => Array
(
)
)
[id7] => Array
(
[title] => Light tax
[values] => Array
(
)
)
)
)
)
title 元素存储收入的名称，values 存储应显示在其下方的条目。现在您只需要一个递归函数，它接受一个数组，输出标题并循环遍历值，为每个值调用自身。 (您也可以使用嵌套的 foreach 来做到这一点)
另一种方法是将所有数据存储在一个表中。只需有 id、parent_id 和 venue 字段，其中 parent_id 字段为 NULL对于顶级条目，相关的 id 应该属于其他所有内容。不过，有效显示此内容的代码与上面的代码有很大不同。
关于php - 使用mysql从三个不同的表中呈现php中的数据，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/47647950/