如何使用in（）函數使用從mysql存儲函數返回的值？

Question

我有一個名爲'dealBusinessLocations'的字段（在一個表'dp_deals'中），它包含以逗號分隔的格式的另一個表（dp_business_locations）的一些ID。

dealBusinessLocations 
---------------------- 
0,20,21,22,23,24,25,26 

我需要在查詢的in（）函數內使用這個值。

像

select * from dp_deals as d left join dp_business_locations as b on(b.businessLocID IN (d.dealBusinessLocations) ; 

正弦MySQL不支持任何字符串爆炸功能，我創建了一個存儲功能

delimiter // 
DROP FUNCTION IF EXISTS BusinessList; 
create function BusinessList(BusinessIds text) returns text deterministic 
BEGIN 
    declare i int default 0; 
    declare TmpBid text; 
    declare result text default ''; 
    set TmpBid = BusinessIds; 
    WHILE LENGTH(TmpBid) > 0 DO 
      SET i = LOCATE(',', TmpBid); 
      IF (i = 0) 
        THEN SET i = LENGTH(TmpBid) + 1; 
      END IF; 
      set result = CONCAT(result,CONCAT('\'',SUBSTRING(TmpBid, 1, i - 1),'\'\,')); 
      SET TmpBid = SUBSTRING(TmpBid, i + 1, LENGTH(TmpBid)); 
    END WHILE; 
    IF(LENGTH(result) > 0) 
     THEN SET result = SUBSTRING(result,1,LENGTH(result)-1); 
    END IF; 
    return result; 
END// 
delimiter ; 

該功能可以正常使用。

mysql> BusinessList('21,22') 
BusinessList('21,22') 
----------------------- 
'21','22' 

但是使用該函數的查詢也不起作用。這裏是查詢。

select * from dp_deals as d left join dp_business_locations as b on(b.businessLocID IN (BusinessList(d.dealBusinessLocations))); 

我已經使用功能argumet靜態值也嘗試過，但沒有用

select * from dp_deals as d left join dp_business_locations as b on(b.businessLocID IN (BusinessList('21,22'))); 

似乎有一些問題與使用從函數返回值。

Answer 1

首先，請閱讀本：

Is storing a comma separated list in a database column really that bad?
Yes, it is

然後，去和規範你的表。

---

現在，如果你真的不能做，否則，或直到你歸一化，使用FIND_IN_SET()功能：

select * 
from dp_deals as d 
    left join dp_business_locations as b 
    on FIND_IN_SET(b.businessLocID, d.dealBusinessLocations) 

然後，再次讀到了那篇文章。如果查詢速度慢，或者如果您有其他問題與此表，那麼你就知道爲什麼：

Is storing a comma separated list in a database column really that bad?
Yes, it is

Answer 2

我認爲你的問題是IN()並不期望得到一個字符串中有很多字段，但很多字段。

隨着你的函數你在發送這樣的：

WHERE something IN ('\'21\',\'22\''); /* i.e. a single text containing "'21','22'" */ 

，而不是預期的

WHERE something IN ('21','22'); 

Answer 3

簡單，使用find_in_set()代替。

SELECT * 
FROM dp_deals as d 
LEFT JOIN dp_business_locations as b 
     ON (FIND_IN_SET(b.businessLocID,d.dealBusinessLocations) > 0); 

參見：http://dev.mysql.com/doc/refman/5.0/en/string-functions.html#function_find-in-set

請注意，如果你刪除CSV，走出過地獄，你可以使用一個簡單的連接，如：

SELECT d.*, GROUP_CONCAT(b.dealBusinessLocation) as locations 
FROM dp_deals as d 
LEFT JOIN dp_business_location as b 
     ON (d.dealBusinessLocation = b.businessLocID); 

，這將是非常非常快，並歸爲一獎金。