HSQLDB:找不到列

dba*_*baq 3 java sql functional-testing left-join hsqldb

我正在使用HSQLDB为我的应用程序编写一些功能测试.我的应用程序生成以下SQL请求:

SELECT 
CLIENT.account_id, 
CLIENT.client_code, 
SHIPPINGADDRESS10A9.account_id, 
SHIPPINGADDRESS10A9.address_code
FROM CLIENT 
LEFT JOIN ADDRESS SHIPPINGADDRESS10A9 
    ON SHIPPINGADDRESS10A9.address_code = CLIENT.shipping_address_code AND (SHIPPINGADDRESS10A9.account_id = 4) 
WHERE CLIENT.account_id = 4
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此请求有效,可以使用我的代码(JDBC + MySQL).当我尝试使用HSQLDB执行我的功能测试时,我得到以下异常:

resultSet.findColumn("CLIENT.account_id"); //Works
resultSet.findColumn("SHIPPINGADDRESS10A9.account_id"); //Exception

Caused by: java.sql.SQLException: Column not found: SHIPPINGADDRESS10A9.account_id
    at org.hsqldb.jdbc.JDBCUtil.sqlException(Unknown Source)
    at org.hsqldb.jdbc.JDBCUtil.sqlException(Unknown Source)
    at org.hsqldb.jdbc.JDBCUtil.sqlException(Unknown Source)
    at org.hsqldb.jdbc.JDBCResultSet.findColumn(Unknown Source)
    at com.inventis.bw.data.DataBaseUtil.loadBean(DataBaseUtil.java:419)
    ... 33 more
Caused by: org.hsqldb.HsqlException: Column not found: SHIPPINGADDRESS10A9.account_id
    at org.hsqldb.error.Error.error(Unknown Source)
    at org.hsqldb.error.Error.error(Unknown Source)
    ... 36 more
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知道我在这里做错了什么吗?

编辑: 看起来问题来自表别名.

logger.debug("address account id => " + resultSet.findColumn("ADDRESS.account_id")); // Works
logger.debug("address account id => " + resultSet.findColumn("SHIPPINGADDRESS10A9.account_id")); //Exception
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使用表名访问列值可以很好地工作,但如果我使用表别名,则抛出异常.任何的想法?

谢谢你的帮助.

Gor*_*off 5

我想你的问题是两列具有相同的别名"account_id",这会引起混淆.尝试给他们不同的名字:

SELECT CLIENT.account_id as client_account_id,
       CLIENT.client_code, 
       SHIPPINGADDRESS10A9.account_id as shipping_account_id
       SHIPPINGADDRESS10A9.address_code
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只是好奇,你能写这样的查询吗?

SELECT c.account_id as client_account_id,
       c.client_code, 
       a.account_id as shipping_account_id
       a.address_code
FROM CLIENT c LEFT JOIN
     ADDRESS a 
     ON a.address_code = c.shipping_address_code AND 
        a.account_id = c.account_id
WHERE c.account_id = 4;
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这应该是等效的,并且清楚地表明您不需要返回account_id两次select.