查詢從SQL中刪除重復項

Question

我在桌子上叫做距離。 它有4列。 id，start_from，end_to和distance 。

我有一些重復的記錄。 在這個意義上重復記錄，

start_from   |   end_to    | distance
Chennai        Bangalore     350
Bangalore      Chennai       350
Chennai        Hyderabad     500
Hyderabad      Chennai       510

在上表中， 欽奈到班加羅爾和班加羅爾與欽奈都有相同的距離 。 所以我需要查詢在select上刪除該記錄。

我想要一個外出的東西

start_from   |   end_to    | distance
Chennai        Bangalore     350
Chennai        Hyderabad     500
Hyderabad      Chennai       510

Answer 1

您可以使用以下查詢來查找重復項：

SELECT LEAST(start_from, end_to) AS start_from, 
       GREATEST(start_from, end_to) AS end_to, 
       distance
FROM mytable 
GROUP BY LEAST(start_from, end_to), GREATEST(start_from, end_to), distance
HAVING COUNT(*) > 1

輸出：

start_from,   end_to,  distance
--------------------------------
Bangalore,    Chennai, 350

現在，您可以使用上述查詢作為派生表來過濾掉重復項：

SELECT t1.*
FROM mytable AS t1
LEFT JOIN (
    SELECT LEAST(start_from, end_to) AS start_from, 
           GREATEST(start_from, end_to) AS end_to, 
           distance
    FROM mytable 
    GROUP BY LEAST(start_from, end_to), GREATEST(start_from, end_to), distance
    HAVING COUNT(*) > 1
) AS t2 ON t1.start_from = t2.start_from AND 
           t1.end_to = t2.end_to AND 
           t1.distance = t2.distance    
WHERE t2.start_from IS NULL

WHERE子句謂詞t2.start_from IS NULL ，過濾掉重復的記錄。

輸出：

start_from  end_to     distance
--------------------------------
Chennai     Bangalore  350
Chennai     Hyderabad  500
Hyderabad   Chennai    510

Answer 2

如果Chennai to Bangalore或Bangalore to Chennai沒有區別，您可以試試這個：

select
    max(`start_from`) as `start_from`,
    min(`end_to`) as `end_to`,
    `distance`
from yourtable
group by
    case when `start_from` > `end_to` then `end_to` else `start_from` end,
    case when `start_from` > `end_to` then `start_from` else `end_to` end,
    `distance`

這是rextester中的演示 。

即使Chennai to Hyderabad也是350也有作品演示 。

如果你想讓Bangalore to Chennai留Bangalore to Chennai ，你可以改變max和min ：

select
    min(`start_from`) as `start_from`,
    max(`end_to`) as `end_to`,
    `distance`
from yourtable
group by
    case when `start_from` > `end_to` then `end_to` else `start_from` end,
    case when `start_from` > `end_to` then `start_from` else `end_to` end,
    `distance`

也是一個演示 。

並且大多數數據庫兼容的case when 。

Answer 3

在查詢中設置字段順序（使用值）有助於獲得唯一的行：

select distinct
    case when start_from  > end_to then end_to     else  start_from end as _start,
    case when start_from  > end_to then start_from else  end_to     end as _end,
    distance
from distance;

經過測試我得到：

+-----------+-----------+----------+
| _start    | _end      | distance |
+-----------+-----------+----------+
| Bangalore | Chennai   |      350 |
| Chennai   | Hyderabad |      500 |
| Chennai   | Hyderabad |      510 |
+-----------+-----------+----------+

Answer 4

假設你的表喜歡

id  start_from              end_to                  distance
0   Chennai                 Bangalore               350
1   Bangalore               Chennai                 350
2   Chennai                 Hyderabad               500
3   Hyderabad               Chennai                 510

然后您可以使用查詢與id進行比較。

Select 
    O.start_from,
    O.end_to,
    O.distance 
From 
    distance O
Left Join
    distance P
On 
    1 = 1
    and O.start_from = P.end_to 
    and O.end_to = P.start_from
Where 
    1 = 1
    and O.distance <> P.distance 
    or(O.distance = P.distance and O.id < P.id)