使用SQL * Loader更新表中的列? [英] Update a column in table using SQL*Loader?

问题描述

我写了一个具有以下查询的SQL脚本. 查询工作正常.

I have written a SQL Script having below query. Query works fine.

update partner set is_seller_buyer=1 where id in (select id from partner 
where names in 
(
'A','B','C','D','E',... // around 100 names.
));

但是现在，我不想在查询本身中编写大约100个名称，而是要从CSV文件中获取所有名称. 我在互联网上读到有关SQL * Loader的信息，但在更新查询上却得不到很多. 我的csv文件仅包含名称.

But now instead of writing around 100 names in a query itself , I want to fetch all the names from the CSV file. I read about SQL*Loader on internet but i did not get much on update query. My csv file contain only names.

我尝试过

  load data
  infile 'c:\data\mydata.csv'
  into table partner set is_wholesaler_reseller=1
  where id in (select id from partner 
  where names in 
  ( 
  'A','B','C','D','E',... // around 100 names.
  ));
  fields terminated by "," optionally enclosed by '"'         
  ( names, sal, deptno )

我该如何做到这一点? 预先感谢.

How i can achieve this? Thanks in advance.

推荐答案

SQL * Loader不执行更新，仅执行插入.因此，您应该将名称插入一个单独的表中，例如names，然后从该表运行更新:

SQL*Loader does not perform updates, only inserts. So, you should insert your names into a separate table, say names, and run your update from that:

update partner set is_seller_buyer=1 where id in (select id from partner 
where names in 
(
select names from names
));

您的加载程序脚本可以更改为:

Your loader script can be changed to:

load data
  infile 'c:\data\mydata.csv'
  into table names
  fields terminated by "," optionally enclosed by '"'         
  ( names, sal, deptno )

对此的一种替代方法是使用外部表，该表允许Oracle将平面文件当作表一样对待.可以在此处找到示例.

An alternate to this is to use External Tables which allows Oracle to treat a flat file like it is a table. An example to get you started can be found here.

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