Say I have an enum which is just public enum Blah { A, B, C, D } and I would like to find the enum value of a string, for example "A" which would be Blah.A . How would it be possible to do this? Is the Enum.valueOf() the method I need? If so, how would I use this? Yes, Blah.valueOf("A") will give you Blah.A . Note that the name must be an exact match, including c
假设我有一个枚举 public enum Blah { A, B, C, D } 我想找到一个字符串的枚举值，例如"A" ，它将是Blah.A 怎么可能做到这一点？ Enum.valueOf()是我需要的方法吗？ 如果是这样，我将如何使用它？ 是的， Blah.valueOf("A")会给你Blah.A 请注意，名称必须完全匹配，包括case： Blah.valueOf("a")和Blah.valueOf("A ")都会抛出IllegalArgumentException 。 静态方法valu
Is there anyway to check if an enum exists by comparing it to a given string? I can't seem to find any such function. I could just try to use the valueOf method and catch an exception but I'v been taught that catching runtime exceptions is not good practice. Anybody have any ideas? I don't think there's a built-in way to do it without catching exceptions. You could instead u
无论如何，通过将枚举与给定的字符串进行比较来检查枚举是否存在？ 我似乎无法找到任何这样的功能。 我可以尝试使用valueOf方法并捕获一个异常，但我被告知捕获运行时异常并不是好的做法。 任何人有任何想法？ 我不认为有一种内置的方法可以在不捕捉异常的情况下执行此操作。 你可以改用这样的东西： public static MyEnum asMyEnum(String str) { for (MyEnum me : MyEnum.values()) { if (me.name().equalsI
I'm using Jackson JSON library to convert some JSON objects to POJO classes on an android application. The problem is, the JSON objects might change and have new fields added while the application is published, but currently it will break even when a simple String field is added, which can safely be ignored. Is there any way to tell Jackson to ignore newly added fields? (eg non-existing o
我使用Jackson JSON库将一些JSON对象转换为android应用程序上的POJO类。 问题是，JSON对象可能会更改，并且在应用程序发布时添加新的字段，但是当前，即使添加了简单的字符串字段，它也会中断，这可以安全地忽略。 有什么办法可以告诉杰克逊忽略新增的字段吗？ （例如POJO对象上不存在）？ 全球忽视会很好。 Jackson提供了一个可以在课堂上使用的注释（JsonIgnoreProperties）。 将以下内容添加到班级的顶部（而不是单
I read about sorting ArrayLists using a Comparator but in all of the examples people used compareTo which according to some research is a method for Strings. I wanted to sort an ArrayList of custom objects by one of their properties: a Date object ( getStartDay() ). Normally I compare them by item1.getStartDate().before(item2.getStartDate()) so I was wondering whether I could write something l
我阅读了关于使用比较器对ArrayLists进行排序的方法，但在所有使用compareTo的例子中，根据某些研究，这是针对Strings的一种方法。 我想通过它们的一个属性对自定义对象的ArrayList进行排序：Date对象（ getStartDay() ）。 通常我通过item1.getStartDate().before(item2.getStartDate())比较他们item1.getStartDate().before(item2.getStartDate())所以我想知道我是否可以写如下所示： public class CustomComparator {
I have a Company entity which has a legalName: String property and also an aliases: Set<String> property (so a One-to-many) Company would be the parent each alias String in the aliases collection would be the children I successfully made a query that returns all companies (parents) which match the parent legalName property with a search query string (LIKE), but I also want to return compa
我有一个Company实体，它有一个legalName: String属性，还有一个aliases: Set<String>属性（所以一对多） Company将是父对象aliases集合中的每个alias字符串都是子对象 我成功地进行了一个查询，它返回所有匹配父母legalName属性的公司（父母）和一个搜索查询字符串（LIKE），但我也想返回具有与搜索查询字符串匹配的别名（子女）的公司（父母） （所有与法定名称匹配的公司，或者至少一个别名，或两者兼有的公司） 我
I have to write a criteria query with where clause to match both first & last names in the child collection. Both names are in different rows Tried this, but does not return anything even when the matching data is present, probably because it's trying to match both restrictions on same row. Criteria criteria = getCurrentSession().createCriteria(Form.class); criteria.createAlias("respo
我必须使用where子句写一个条件查询来匹配子集合中的第一个和最后一个名字。 这两个名字都在不同的行中 尝试过，但即使匹配数据存在也不会返回任何内容，可能是因为它试图在同一行上匹配两个限制。 Criteria criteria = getCurrentSession().createCriteria(Form.class); criteria.createAlias("responses", "r"); criteria.add(Restrictions .conjunction() .add(Restrictions.eq("r.id", "
I would like to query a Parent object by a list of child objects with Hibernate Criteria Query. I know how to do it with IDs of the childs but not using entities directly. This works: List<Long> listOfChildLongChildIds = new ArrayList<Long>(); listOfChildLongChildIds.add(new Long(1)); listOfChildLongChildIds.add(new Long(2)); Criteria criteria = getSession().createCriteria(Paren
我想通过Hibernate Criteria Query的子对象列表来查询Parent对象。 我知道如何用孩子的ID做，但不直接使用实体。 这工作： List<Long> listOfChildLongChildIds = new ArrayList<Long>(); listOfChildLongChildIds.add(new Long(1)); listOfChildLongChildIds.add(new Long(2)); Criteria criteria = getSession().createCriteria(Parent.class); Criteria criteriaChilds = criteria.createCriteria("childs")
DetachedCriteria criteria = DetachedCriteria.forClass(Parent.class,"parent"); criteria.createAlias("parent.child","thechild"); criteria.add(Restrictions.eq("thechild.property", "somevalue"); I read many sample code like above, here the criteria join two tables, parent and child, but I just don't know how/where to specify a On condition on the criteria. like the following select * from pare
DetachedCriteria criteria = DetachedCriteria.forClass(Parent.class,"parent"); criteria.createAlias("parent.child","thechild"); criteria.add(Restrictions.eq("thechild.property", "somevalue"); 我阅读了上面的许多示例代码，这里的标准连接两个表，父母和孩子，但我不知道如何/在哪里指定条件的On条件。 如下所示 select * from parent inner join child on parent.childid = child.id 但在上述标准，我只是找不
I have an association mapped by the following: @Entity public class Parent { ... @Id @Column(name = "parent_id") private Long id; @OneToMany(mappedBy = "parent") @OrderBy("id") private List<Child> children; ... } @Entity public class Child { ... @Id @Column(name = "child_id") private Long id; @ManyToOne @NotFound(action = NotFoundAction.IGNORE
我有一个关联映射如下： @Entity public class Parent { ... @Id @Column(name = "parent_id") private Long id; @OneToMany(mappedBy = "parent") @OrderBy("id") private List<Child> children; ... } @Entity public class Child { ... @Id @Column(name = "child_id") private Long id; @ManyToOne @NotFound(action = NotFoundAction.IGNORE) @JoinColumn(name =
I've been working with JPA (implementation Hibernate) for some time now and each time I need to create entities I find myself struggling with issues as AccessType, immutable properties, equals/hashCode, ... . So I decided to try and find out the general best practice for each issue and write this down for personal use. I would not mind however for anyone to comment on it or to tell me whe
我一直在使用JPA（实现Hibernate）一段时间，每次我需要创建实体时，我发现自己正在与AccessType，不可变属性，equals / hashCode ......等问题挣扎。 所以我决定尝试找出每个问题的一般最佳做法，并写下来供个人使用。 但我不介意任何人对此发表评论或告诉我我错在哪里。 实体类 实现Serializable 原因：规范说你必须，但是一些JPA提供者不强制执行此操作。 作为JPA提供程序的Hibernate不会强制执行此操作，但如果Se