2015-10-15 42 views
1

我有數據如下表:內查詢以從另一個表匹配 'LIKE' 結果

FILTER表

**id filter**  
4638 Aabe  
4639 Aaby  
4640 Aadl  
4641 Aaga  
4642 Aake 

姓氏表

**surnames**  
Aaberge  
Aabehut  
Aabyuti  
Aabytis  
Aadlit  
Aagastha  
Aaker 

我我試圖編寫一個內部查詢來檢索匹配過濾器的計數表

下面是我根據不同的SELECT語句

  1. select * from filter
  2. select count(*) from surnames where last_name like 'Aabe%' or last_name like 'Aaby%' group by last_name;

我如何寫一個內部查詢的這個條件,因爲記錄計數是大?

+0

'...上SURNAMES.surnames LIKE CONCAT(FILTER.filter, '%')' – Mihai

+0

SELECT * FROM姓氏,過濾器,其中filters.id> 4215 JOIN上SURNAMES.last_name LIKE CONCAT(濾波器JOIN .filter,'%')??糾正我,如果我錯了。非常感謝你!! –

+0

'select *,COUNT(*)從surname上加入過濾器SURNAMES.last_name LIKE CONCAT(filters.filter,'%')where filters.id> 4215 group by last_name' – Mihai

回答

0
select *, 
COUNT(*) from surnames join filters 
on SURNAMES.last_name LIKE CONCAT(filters.filter,'%') 
where filters.id > 4215 
group by last_name 
+1

非常感謝。 –

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