python - How to propagate changes from child widgets if any of the child have changes? -


मेरे पास एक QWidget है। और QWidget के अंदर, मेरे पास बहुत से अन्य QWidget s, जैसे: - QComboBox , QLineEdit , आदि ....

अब, मैं जानना चाहता हूं कि उपयोगकर्ता ने कोई भी परिवर्तन किया है ( उदा: - QLineEdit में कुछ लिखें, या < कोड> QComboBox , आदि .... ।)

क्या ऐसा करने का एक आसान तरीका है? या मुझे प्रत्येक विजेट के सिग्नल पर कब्जा करने की आवश्यकता है, जैसे ... 

  QObject.connect (लाइनएडिट, सिग्नल ("टेक्स्टवर्ल्ड ()"), doSomething) QObject.connect (कॉम्बोबॉक्स, सिग्नल ("वर्तमान इंडेक्स चेंगड (इंट)"), डॉट्सिंग) ... ... ...    
 
 
 हाँ, आपको मैन्युअल रूप से संकेतों को कनेक्ट करना 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