问题描述
我有以下设置:
@Entity
public class Function {
private String name;
@OnetoMany(mappedBy = "function",cascade = CascadeType.ALL,orphanRemoval = true)
@Where(clause = "type = 'In'") // <=== seems to cause problems for CriteriaBuilder::size
private Set<Parameter> inParameters = new HashSet<>();
@OnetoMany(mappedBy = "function",orphanRemoval = true)
@Where(clause = "type = 'Out'") // <=== seems to cause problems for CriteriaBuilder::size
private Set<Parameter> outParameters = new HashSet<>();
}
@Entity
public class Parameter {
private String name;
@Enumerated(EnumType.STRING)
private ParameterType type;
@ManyToOne(fetch = FetchType.LAZY,optional = false)
@JoinColumn(name = "function_id")
private Function function;
}
我要解决的总体问题是找到具有outParameter
且具有确切动态名称集的所有函数。查找名称正确为outParameter
的{{1}} s的所有函数
这在sql中似乎是一个“ exact relational division”问题,因此可能会有更好的解决方案,但是我这样做的方式是这样的:
('outParam1','outParam2')
我遇到的问题是List<String> paramNames = ...
Root<Function> func = criteria.from(Function.class);
Path outParams = func.get("outParameters");
Path paramName = func.join("outParameters").get("name");
...
// CriteriaBuilder Code
builder.and(
builder.or(paramNames.stream().map(name -> builder.like(builder.lower(paramName),builder.literal(name))).toArray(Predicate[]::new)),builder.equal(builder.size(outParams),paramNames.size()));
似乎没有考虑到builder.size()
注释。因为“ CriteriaBuilder代码”嵌套在适用于任何类型的实体的通用@Where
中,所以我不能简单地添加Specification
子句。
该代码在一个函数具有0个输入参数时起作用,但在有多个输入参数时不起作用。我查看了生成的sql,发现它丢失了:
query.where()
任何帮助表示赞赏(或者通过其他方法解决问题,或者通过变通方法来解决SELECT disTINCT
function0_.id AS id1_37_,function0_.name AS name4_37_,FROM
functions function0_
LEFT OUTER JOIN parameters outparamet2_ ON function0_.id = outparamet2_.function_id
AND (outparamet2_.type = 'Out') -- <== where clause added here
WHERE (lower(outparamet2_.name)
LIKE lower(?)
OR lower(outparamet2_.name)
LIKE lower(?))
AND (
SELECT
count(outparamet4_.function_id)
FROM
parameters outparamet4_
WHERE
function0_.id = outparamet4_.function_id) = 2 -- <== where clause NOT added here
无法正常工作。
解决方法
where注释在函数实体中,在子查询中您尚未使用该实体,因此操作是正确的,请尝试将函数实体用作子查询的根,或手动实现where。
对于下一个,建议您在答案中包含完整的Criteria API代码。