znnevergiveup
2013-08-02 09:17
浏览 191

hibernate 的问题啊

用DetachedCriteria类 组装查询条件,咋个生成这样的sql语句了捏? 如果没没条件的话还行 ,一加上where条件查询,就会出 where y1_.... 然后就报错:列名 'y1_' 无效。上一段打印出的sql

[code="java"]
select top 14 this_.id as y0_, this_.CODE as y1_, this_.PROJECT_NAME as y2_, this_.PROJECT_STATE as y3_, this_.PROJECT_CLASS as y4_, this_.CUSTOMER_ID as y5_, this_.CUSTOMER_DEPARTMENT_MAIN as y6_, this_.CUSTOMER_DEPARTMENT_HELP as y7_, this_.CUSTOMER_LINKMAN as y8_, this_.CUSTOMER_LINKMAN_PHONE as y9_, this_.CUSTOMER_CODE as y10_, this_.UNIT_PLAN as y11_, this_.UNIT_PLAN_LINKMAN as y12_, this_.UNIT_PLAN_LINKMAN_PHONE as y13_, this_.SUPERVISOR as y14_, this_.SUPERVISOR_LINKMAN as y15_, this_.SUPERVISOR_LINKMAN_PHONE as y16_, this_.PROJECT_CUSTODIAN as y17_, this_.PROJECT_DEPARTMENT as y18_, this_.PROJECT_DEPARTMENT_CUSTODIAN as y19_, this_.PROJECT_MANAGER as y20_, this_.SAFETY_OFFICER as y21_, this_.QUALITY_INSPECTOR as y22_, this_.LIBRARIAN as y23_, this_.BULIDERS as y24_, this_.MATERIALMAN as y25_, this_.TECHONLOGY as y26_, this_.FILE_TRANSFER as y27_, this_.BUILD_STATE as y28_, this_.COMPANYMANAGE_ID as y29_, this_.PROJECT_DATE as y30_, this_.USER_ID as y31_, this_.ACCESSORY_NAME as y32_ from project this_ where (y1_ like ? and y2_ like ?) order by y1_ desc
[/code]

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5条回答 默认 最新

  • zyn010101 2013-08-02 09:41
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    CODE列存在么?

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  • 卧石听泉 2013-08-02 12:30

    把SQL放在sql里运行下,看报什么错,根据信息修改下。

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  • ncist_jianeng 2013-08-02 13:54

    where语句中不能使用select中的别名呀

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  • wenjie4892543 2013-08-02 15:11

    where语句中不能使用select中的别名呀 +1

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  • ss辣椒酱子 2013-08-05 12:07

    仔细检查一下你hql 写的是否正确。转成sql 在数据库执行一下吧。
    where中确实不能使用select 中定义的别名啊。

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