Question

我有两个不同的表，由Load DATA命令创建，我正在尝试运行以下代码：

select t.name as Name, sum(t.amount) as Total from 
(select t2.name, t1.vendorname, t1.amount  from bco_ifs_payment_data t1
join VendorName t2 on t1.vendorname like concat('%',t2.name,'%' )) as t 
group by t.name;

当我使用INSERT INTO命令将数据插入VendorName表时，代码工作并返回行，但是当我使用LOAD DATA命令在VendorName表中插入数据时，即使插入了相同的数据，也不会在JOIN中返回任何行在表格中

Vendor_Names CSV只有一列。 LOAD数据查询如下：

create Table vendorname(name nvarchar(100));
LOAD DATA LOCAL INFILE 'F:\Payments\_Global Payment\\Data\\Data\\Vendor_Names.csv' 
INTO TABLE vendorname FIELDS TERMINATED BY ',' ENCLOSED BY '"' LINES TERMINATED BY '\n'; 
update vendorname SET name = UPPER(name);

INSERT查询如下：

 INSERT INTO vendorname VALUES ('DELOITTE');
 INSERT INTO vendorname VALUES ('CBRE');

Answer 1

你可以发布两个查询，可能是csv中的值序列与插入查询的内容不同

Answer 2

知道了，必须包含以下代码：

select t.name as Name, sum(t.amount) as Total from 
(select t2.name, t1.vendorname, t1.amount  from bco_ifs_payment_data t1
join vendorname t2 on t1.vendorname like concat('%',REPLACE(t2.name,'\r','') ,'%' )) as t 
group by t.name;

MySQL中INSERT INTO和LOAD DATA之间的区别

2 个答案: