criteria

在系统中创建一个张记录数据字典类型： 数据字典类型表 创建一张表记录数据字典明细： 数据字典明细表 将上边变化灵活的配置项叫做：“普通配置项” 将上边固定的配置项（每个配置顶都有一个代码）：“业务代码” 普通配置项存储： 普通配置项名称存储在DICTINFO表中info字段 普通配置项对应的类型id存储在DICTINFO表中TYPECODE 业务代码存储： 业务代码对应的名称存储在DICTINFO表中info字段 业务代码对应的类型id存储在DICTINFO表中TYPECODE 业务代码存储在DICTINFO表中DICTCODE(是和普通配置顶的区别) 2、数据字典的使用 3、 小结 普通配置项：对业务数据进行简单的归类，这些归类受用户要求变化较灵活，将这些配置项作普通配置项配置数据字典表。 业务代码：系统运行所必须的，在系统设计时定义的固定代码，这些代码可能需要在程序代码进行硬编码。 4、系统中的应用 public List findDictinfoByType (String typecode) throws Exception { DictinfoExample dictinfoExample = new DictinfoExample(); DictinfoExample.Criteria criteria = dictinfoExample.createCriteria(

I am basically trying to create this query with NHibernate ICriteria interface: SomeTable 1:n AnotherTable SomeTable has columns: PrimaryKey, NonAggregateColumn AnotherTable has columns: PrimaryKey, ForeignKey, AnotherNonAggregate, YetAnotherNonAggregate SELECT table1.NonAggregateColumn, subquery.SubQueryAggregate1, subquery.SubQueryAggregate2 FROM SomeTable AS table1 LEFT JOIN ( SELECT table2.ForeignKey, COUNT(table2.AnotherNonAggregate) AS SubQueryAggregate1, AVG(table2.YetAnotherNonAggregate) AS SubQueryAggregate2 FROM AnotherTable AS table2 GROUP BY (table2.ForeignKey) ) AS subquery ON

Normally when ordering ascending by a field, you get the NULL values first, and then the more interesting values. Often, you want NULL values last. In MySQL, you can do this with: SELECT * FROM people ORDER BY ISNULL(name), name; However, I'm using Grails with Hibernate criteria, and I have absolutely no idea how to do this there. Is this even supported in any way? Is there some way to order by a custom SQL expression? I'd hate to rewrite all my criteria to plain SQL just to get it to sort correctly. In hibernate you can try with this below code : Criteria c = ...; c.addOrder(Order.asc("name")

I have some problems with projection in Grails. Could you please help me review them and suggest solutions for me? I want to query data on many tables which has many-to-one relationship and projection on some properties on both of them. For example: class Person { int id String name String address static hasMany = [cars : Car] } class Car { int id String brand long price Person owner static belongsTo = [owner : Person] } So, how can I use just one query withCriteria (apply projection) to get information of a specified car (include brand, price, and the owner name)? Is it possible to use: Car

I am trying to set up a Criteria according to the Doctrine Docs . Unfortunately they don't tell you how to access attributes of an related Object. Let me give you an example. I have an ArrayCollection of Products. Every Product has a Category. I want to filter the ArrayCollection for a Category Name. Now I am trying to set up a Criteria as follows: $criteria = Criteria::create() ->where(Criteria::expr()->eq("category.name", "SomeCategoryName")); Now I get the following Exception: An exception has been thrown during the rendering of a template ("Unrecognized field: category.name") How can I

I have my entity class available via a method. I'm trying to figure out, how via the JPA JPQL or Criteria API's I could issue a truncate or delete from. I think that the criteria api is more natural for working with classes, and truncate is a faster operation so these are prefered. This is what I put together so far, but not sure what to add/change about it. CriteriaBuilder cb = this._em().getCriteriaBuilder(); cb.createQuery( _entityClass() ).from( _entityClass() ); note: _entityClass returns MyEntity.class , I have no other references to MyEntity this is a more generalized implementation.