ejb 3.0 - How do i do a like statement in ejb-ql? -


संभव डुप्लिकेट:

मुझे एक उदाहरण से यह कोड मिला है, मैं ईबीजे क्यूएल के साथ एक खोज कैसे करूं? 

  SELECT OBJECT (पी) व्यक्ति पी WHERE से (पी.नाम की तरह? 1)   
 मैं यहाँ से कैसे जारी रखूं? मैं परिणाम वापस करना चाहूंगा जहां p.name में स्ट्रिंग "टेस्ट" होता है   
 
 
 मुझे समाधान मिला यहाँ एक उदाहरण है।  
  क्वेरी क्वेरी = em.createQuery ("SELECT c से TestEntity c WHERE c.titleLIKE: searchString"); Query.setParameter ("खोज-स्ट्रिंग", "%" + "परीक्षण" + "%");

Comments

Post a Comment

Popular posts from this blog

Python SQLAlchemy：AttributeError: Neither 'Column' object nor 'Comparator' object has an attribute 'schema' -

I tried to create a new database in my project, but when I run the script, I get this error, I have another project, this type of definition worked before, but now it gets a single error. I am using Python 2.7.8 and the version of SQLAlchemy module is 0.9.8. By the way, using a project flask- SQLAlchemy, it works well, I'm confused. Traceback information is as follows: Traceback (most recent call final): File "D: / Projects / IOM / db_create.py", line 4, & lt; Module & gt; From the models the base file "D: \ Projects \ OO-IM \ models.py", in line 15, in the & lt; Module & gt; Column ( 'followed_id', int (), ForeignKey ( 'user.id')) file "C: \ Python27 \ lib \ site-packages \ SQLAlchemy \ SQL \ schema.py", line 369, __new__ schema = metadata. Schema file "C: \ Python27 \ lib \ site-packages \ SQLAlchemy \ SQL \ elements.py", line 662, AttributeError in __getattr__ key): Neither 'column' object nor ...
Read more

java - How not to audit a join table and related entities using Hibernate Envers? -

I use Eners to be hibernate to audit my institutions. I have an audit entity, Foo , which includes a list & lt; Bars & gt; However, I do not want to audit the bar institutions, in the form of properties, I have written that: @inti @edicated public class Foo {@JoinTable (name = "T_FOO_BAR", joinColumns = @JoinColumn (name = "FOO_ID"), inverseJoinColumns = @JoinColumn (Name = "BAR_ID")) @ManyToMany (Cascade = Persist) @ Indited (targetAuditMode = RelationTargetAuditMode.NOT_AUDITED ) Public listing & lt; Bars & gt; GetBars () {Return bars; }} Now, I want to retrieve an amendment of foo : AuditReader reader = AuditReaderFactory.get (GetEntityManager ()); FU modification = (FU) reader.CreativeTeA () .entensEtrevision (Foo Class, 42) .getSingleResult (); Unfortunately, when I want to retrieve all the data (i.e. when it loads lazy times ), error me ORA-00942: table or view is not present , because Tried to query this: Sel...
Read more

mongodb - CakePHP paginator ignoring order, but only for certain values -

While pagging my user model in kppp, I can sort with some values, but not others. For example, I can order results from created or email , for example, but username or reputation seems to come back order by arbitrary command for example with a list of users: $ this-> paginate = array ('condition' => array ( 'User.is_active' = & gt; true), 'border' => 24, 'order' => array ('User.created' => DESC)); Works as expected, but $ this - & gt; Paged = array ('conditions' = & gt; array (' user.isIactive '= & gt; true),' border '= & gt; 24,' command '= & gt; array (' user repeats' = ' & Gt; DESC ')); No, no. I first thought that this could be a database issue, but when I search the database directly, sorted as expected. Note: I am using Mangodebi with the Mongo DB plugin of Ichikawa for the KPPHP; Pagnetizing users are used to work properl...
Read more