sql - how to get full outer join in left and right join order -


कृपया नीचे दी गई स्क्रिप्ट देखें 

  declare @ table1 तालिका (col1 int) घोषित करें @ टेबल 2 @ टेबल 1 मान (2) में @ टेबल 1 मान (3) डालें @ table1 मानों (5) में डालें @ टेबल 1 मूल्यों में सम्मिलित करें (7) @ टेबल 2 मानों में डालें @ टेबल 2 मूल्यों में डालें (2) @ टेबल 2 मूल्यों में डालें (2) @ टेबल 2 मानों में डालें (2) @ टेबल 2 मानों में डालें (6) @ टेबल 2 मानों में सम्मिलित करें (4) @ टेबल 2 मानों में डालें (7)   <1> 
 मामला 1:  
  चुनें @ से तालिका 1 बाएं बाहरी जुड़ें @ टेबल 2 बी पर a.col1 = b.col2 आदेश col1   
 परिणाम 1:  
  col1 col2 ----------- ----------- | 1 | 1 | | 2 | नल | | 3 | 3 | | 3 | 3 | | 5 | नल | | 7 | 7 | ---------------------------   
 केस 2:  
 <कोड > तालिका 2 से दाएं बाहरी सहभागी @ टेबल 2 बी पर कॉल करें।   
 परिणाम 2:  
  col1 col2 ----------- ----------- | 1 | 1 | | 3 | 3 | | 3 | 3 | | नल | 4 | | नल | 6 | | 7 | 7 | ---------------------------   
 वास्तविक केस:  
  तालिका 1 से एक पूर्ण बाहरी जोड़ने @ तालिका 2 बी पर a.col1 = b.col2 चुनें> वास्तविक परिणाम:  
  col1 col2 --- चुनें -------- ----------- | 1 | 1 | | 2 | नल | | 3 | 3 | | 3 | 3 | | 5 | नल | | 7 | 7 | | नल | 6 | | नल | 4 | ---------------------------   
 अपेक्षित परिणाम:  
  कॉल 1 कॉल 2 ----------- ----------- | 1 | 1 | | 2 | नल | | 3 | 3 | | 3 | 3 | | नल | 4 | | 5 | नल | | नल | 6 | | 7 | 7 | ---------------------------   
 मैंने सभी के साथ बाएं और दायां जुड़ने की क्वेरी की कोशिश की लेकिन यह परिणाम सेट में युगल। क्या ऐसा कोई तरीका है जो मुझे यह अनुमानित आउटपुट प्राप्त हो सकता है।  
 धन्यवाद, एसेन।   
 
 < P> आप उपयोग कर सकते हैं  
  SELECT * FROM @ table1 एक पूर्ण बाहरी सम्मिलित करें @ table2b पर a.col1 = b.col2 ORDER द्वारा Isnull (col1, col2)   
 अपनी वांछित ऑर्डर प्राप्त करने के लिए बिना  ORDER BY  कोई आदेश की गारंटी नहीं है।

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