persistence

Using Persistent Store in BlackBerry

由 烈酒焚心 提交于 2019-12-04 07:08:20
问题 I am developing a BlackBerry application. I want to store the details of multiple users in my mobile. I have to store data like username, first name, last name ,email id ,phone number for each user. Can any one please provide me a sample code for persistent store using which I can store all this data in a vector and retrieve later. 回答1: This link should answer most of what you need to know - http://www.miamicoder.com/post/2010/04/13/How-to-Save-BlackBerry-Application-Settings-in-the

Entity framework 4 many-to-many insertion?

由 ♀尐吖头ヾ 提交于 2019-12-04 06:24:28
问题 I'm not very familiar with the many-to-many insertion process using Entity Framework 4, POCO. I have a blog with 3 tables: Post, Comment, and Tag. A Post can have many Tags and a Tag can be in many Posts . Here are the Post and Tag models: public class Tag { public int Id { get; set; } [Required] [StringLength(25, ErrorMessage = "Tag name can't exceed 25 characters.")] public string Name { get; set; } public virtual ICollection<Post> Posts { get; set; } } public class Post { public int Id {

one to one mapping using primary key join column

由 僤鯓⒐⒋嵵緔 提交于 2019-12-04 06:19:47
问题 I have employee and employee detail classes mapped (bi directional) using primary key join column (employee_id) @Entity @Table(name="employee") public class Employee { @Id @GeneratedValue @Column(name="employee_id") private Long employeeId; @Column(name="firstname") private String firstname; @Column(name="lastname") private String lastname; @Column(name="birth_date") @Temporal(value = TemporalType.DATE ) private Date birthDate; @Column(name="cell_phone") private String cellphone; @OneToOne

Serialization vs Embedded Database for simple standalone application

由 帅比萌擦擦* 提交于 2019-12-04 06:11:49
问题 Say, I have a pretty simple Java application that needs the way to store some user settings. XML is not a really good solution, since I want to store them in binary form. So, what would be the best solution in this case, embedded database (such as Apache Derby) or just plain old serialization? I know that these are two completely different things, but both allow to persist some application state. So what would you chose, and why? Edit As far as storing simple user preferences go, .properties

Persisting objects in SugarORM

由 有些话、适合烂在心里 提交于 2019-12-04 05:54:52
问题 I have a Book class: public class Book extends SugarRecord { private String mBookName; private String mAuthorName; private List<Page> mPageList; public Book() { } public Book(String bookname, String authorName) { mBookName = bookname; mAuthorName = authorName; mPageList = new ArrayList<>(); } public String getAuthorName() { return mAuthorName; } public void setAuthorName(String authorName) { mAuthorName = authorName; } public String getBookName() { return mBookName; } public void setBookName

Firebase 3.0 session persistance

由 試著忘記壹切 提交于 2019-12-04 05:50:22
It seems impossible to use session persistence in firebase 3.0. This was possible in the previous version: https://www.firebase.com/docs/web/guide/login/password.html authWithPassword() takes an optional third parameter which is an object containing any of the following settings: remember - String If not specified - or set to default - sessions are persisted for as long as you have configured in the Login & Auth tab of your App Dashboard. To limit persistence to the lifetime of the current window, set this to sessionOnly. A value of none will not persist authentication data at all and will end

what are some common problems with Hibernate/persistence?

由 心已入冬 提交于 2019-12-04 05:20:36
I have an application I'd like to test-proof against possible problems related to Hibernate and/or persistence. What other problems? How do I reproduce them (literally)? And how do you recover from them? To make it clear: I'm talking about multi-threaded cluster environment (the most complex one). My one: org.hibernate.StaleObjectStateException: Row was updated or deleted by another transaction (or unsaved-value mapping was incorrect) Reproduce: Load the object. UPDATE using HQL. Try to update (save) the loaded object. Handle: Not sure... Lazy loading is one the big issues you'll encounter,

Getting all mapped Entities from EnitityManager

由 不羁岁月 提交于 2019-12-04 03:31:15
问题 I have a piece of maintenance code that should grant select privileges to a certain user at certain points of time: grant select on A_DB.A_TABLE to READ_ONLY_USER; I want to do this for all tables. I could use select * from tab in Oracle or show tables in MySQL to get the complete list and then move on like that. But since I already have the javax.persistence.EntityManager Object at hand, I wondered if there is another way to get at all the mapped Entities, the Manager knows about (I am using

org.hibernate.HibernateException: save is not valid without active transaction

由 六月ゝ 毕业季﹏ 提交于 2019-12-04 03:21:28
I am creating JSF application and using some hibernate stuff in it. All I want to do is to save the entity into the database but I keep getting this exception: org.hibernate.HibernateException: save is not valid without active transaction At first I was getting this exception: org.hibernate.HibernateException: No CurrentSessionContext configured! Then I found that I need to add this into my hibernate configuration: <property name="hibernate.current_session_context_class">thread</property> This solved this issue but now the above one appears. I am saving entity into database like this: public

Unable to Import persistence.xml within applicationContext.xml file

由 一个人想着一个人 提交于 2019-12-04 02:21:48
问题 I'm using eclipse juno IDE I have Java application which have src folder. within the folder I have: 1) applicationContext.xml 2) persistence.xml I also have DBInterface and i implemented it with JPA. Now in the applicationContext.xml file I have a bean for the JPA implemention. When I tried to inject the bean i got an excpetion that said something like "No Persistence Provider was found". So I tried to import the persistence file within the applicationContext file but i'm getting another
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