Remote User's hostname + App Engine Python -


इसी प्रकार:

रिमोट-एडीडी - रिमोट यूजर के आईपी एड्रेस। < / Blockquote> 

  self.request.remote_addr   
 इसके बजाय, क्या दूरस्थ उपयोगकर्ता के होस्टनाम को प्राप्त करने का कोई तरीका है? आधिकारिक दस्तावेज के अनुसार, रिमोट प्राप्त करने का विकल्प नहीं है, ऐसा लगता है कि ऐसा कुछ ऐसा ही हो सकता है,  
  self.request.hostname   
 उपयोगकर्ता का होस्टनाम  
 मैं मेजबाननाम + आईपी के संयोजन के साथ एक मशीन की विशिष्ट रूप से पहचान करने की कोशिश कर रहा हूं, क्योंकि कुछ उपयोगकर्ता इंटरनेट गेटवे साझा कर सकते हैं, इस प्रकार ऐसा एक ही आईपी पता संभव है और इससे आईपी पता नहीं होगा पर्याप्त अद्वितीय।  
 इस पर कोई सलाह? या कम से कम, क्या किसी अज्ञात पहचान वाली मशीन की विशिष्ट रूप से पहचान करने के लिए वैकल्पिक तरीका सुझा सकता है?   
 
 
 टी यह HTTP पर संभव है लगता है आम तौर पर, इस प्रयोजन के लिए कुकीज का उपयोग किया जाता 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