Question

我不知道为什么只有名为“ Member”的实体被编号。

其他实体没有。

附加的是QEntity源。

所以我的问题是

为什么编号为“成员”实体
我不想编号。我该怎么办才能解决这个问题？

import static com.querydsl.core.types.PathMetadataFactory.*;

import com.querydsl.core.types.dsl.*;

import com.querydsl.core.types.PathMetadata;
import javax.annotation.Generated;
import com.querydsl.core.types.Path;


/**
 * QMember is a Querydsl query type for Member
 */
@Generated("com.querydsl.codegen.EntitySerializer")
public class QMember extends EntityPathBase<Member> {

    private static final long serialVersionUID = -1377787676L;

    public static final QMember member = new QMember("member1"); //why numbering...

    public final NumberPath<Integer> memberIdx = createNumber("memberIdx", Integer.class);

    public QMember(String variable) {
        super(Member.class, forVariable(variable));
    }

    public QMember(Path<? extends Member> path) {
        super(path.getType(), path.getMetadata());
    }

    public QMember(PathMetadata metadata) {
        super(Member.class, metadata);
    }

}

Answer 1

因为“成员”是JPQL关键字。

https://docs.oracle.com/html/E13946_04/ejb3_langref.html#ejb3_langref_collection_member

QueryDsl包含这些列表，并通过添加数字后缀对它们进行转义：

http://www.querydsl.com/static/querydsl/4.1.4/apidocs/index.html?com/querydsl/codegen/Keywords.html

public static final Collection<String> JPA = ImmutableList.of(
        "ABS","ALL","AND","ANY","AS","ASC","AVG","BETWEEN",
        "BIT_LENGTH[51]","BOTH","BY","CASE","CHAR_LENGTH",
        "CHARACTER_LENGTH","CLASS",
        "COALESCE","CONCAT","COUNT","CURRENT_DATE","CURRENT_TIME",
        "CURRENT_TIMESTAMP",
        "DELETE","DESC","DISTINCT","ELSE","EMPTY","END","ENTRY",
        "ESCAPE","EXISTS","FALSE","FETCH",
        "FROM","GROUP","HAVING","IN","INDEX","INNER","IS","JOIN",
        "KEY","LEADING","LEFT","LENGTH","LIKE",
        "LOCATE","LOWER","MAX","MEMBER","MIN","MOD","NEW","NOT",
        "NULL","NULLIF","OBJECT","OF","OR",
        "ORDER","OUTER","POSITION","SELECT","SET","SIZE","SOME",
        "SQRT","SUBSTRING","SUM","THEN",
        "TRAILING","TRIM","TRUE","TYPE","UNKNOWN","UPDATE","UPPER",
        "VALUE","WHEN","WHERE");

鉴于您无需在代码中引用它，为什么会出现问题？

生成JPA QEntity时，将使用特定单词生成它

1 个答案: