Question

自从我来到这里已经有一段时间了。

我很难理解如何查询1个表（位置）和加入（左，右或内）不确定。

表：位置字段：locationID，locationParentID，locationName

locationID | locationParentID | LOCATIONNAME

1 | 0 |北爱尔兰 2 | 0 |英国 3 | 0 |苏格兰 4 | 1 |唐郡 5 | 1 |郡阿马 6 | 2 |伦敦 7 | 2 |利兹 8 | 3 |格拉斯哥 9 | 3 |爱丁堡 10 | 4 |贝尔法斯特 11 | 4 |纽敦纳兹

任何人都可以指出我正确的方向，以便我可以在1张桌子上为这个3级位置结构的某种连接执行1次查询吗？

非常感谢任何帮助。

SELECT * FROM locations（JOIN EACH SUB LOCATION AS ['sub']匹配LOCATION_PARENT_ID）ORDER BY locationParentID，locationName ASC。

SELECT * FROM _locations 
JOIN _locations as parent_location 
ON _locations.locationParentID = parent_location.locationID
JOIN _locations AS super_parent_location 
ON parent_location.locationParentID = super_parent_location.locationID


Array
(
[0] => 117
[locationID] => 1
[1] => 19
[locationParentID] => 0
[2] => Hull
[locationName] => England
[3] => hull
[locationUri] => 
[4] => 2
[locationLevel] => 1
[5] => 19
[6] => 1
[7] => East Yorkshire
[8] => 
[9] => 2
[10] => 1
[11] => 0
[12] => England
[13] => 
[14] => 1
)

我想试试

Array
(
[locationID] => 1
[locationParentID] => 0
[locationName] => England
[locationUri] => 
[locationLevel] => 1
[sub'] => Array (
                [locationID] => 55
                [locationParentID] => 1
                [locationName] => Leeds
                [locationUri] => 
                [locationLevel] => 2
                [sub'] => Array (
                                [locationID] => 144
                                [locationParentID] => 55
                                [locationName] => Chorley
                                [locationUri] => 
                                [locationLevel] => 3))
)

现在真的很想让这个工作，如果有人能提供任何进一步的消息，我将非常感激。

我当前的查询是

SELECT * FROM _locations 
JOIN _locations as parent_location 
ON _locations.locationParentID = parent_location.locationID
JOIN _locations AS super_parent_location 
ON parent_location.locationParentID = super_parent_location.locationID

输出所有字段，但我想使用mysqli_fetch_array（$ q，MYSQL_ASSOC）返回3级深度数组;

再次感谢

Answer 1

尝试这样的事情

SELECT * FROM location 
JOIN location as parent_location 
  ON location.locationParentID = parent_location.locationID
JOIN location AS super_parent_location 
  ON parent_location.locationParentID = super_parent_location.locationID

基本上你只是一遍又一遍地加入同一个表并使用别名。

Answer 2

SELECT * FROM _locations 
JOIN _locations as parent_location 
ON _locations.locationParentID = parent_location.locationID
JOIN _locations AS super_parent_location 
ON parent_location.locationParentID = super_parent_location.locationID


Array
(
[0] => 117
[locationID] => 1
[1] => 19
[locationParentID] => 0
[2] => Hull
[locationName] => England
[3] => hull
[locationUri] => 
[4] => 2
[locationLevel] => 1
[5] => 19
[6] => 1
[7] => East Yorkshire
[8] => 
[9] => 2
[10] => 1
[11] => 0
[12] => England
[13] => 
[14] => 1
)

我想试试

Array
(
[locationID] => 1
[locationParentID] => 0
[locationName] => England
[locationUri] => 
[locationLevel] => 1
[sub'] => Array (
                [locationID] => 55
                [locationParentID] => 1
                [locationName] => Leeds
                [locationUri] => 
                [locationLevel] => 2
                [sub'] => Array (
                                [locationID] => 144
                                [locationParentID] => 55
                                [locationName] => Chorley
                                [locationUri] => 
                                [locationLevel] => 3))
)

MySQLi查询使用1表连接

2 个答案: