Hi guys! Great to see Apache 2 license and 4 spaces indentation!:)

Hi guys! Great to see Apache 2 license and 4 spaces indentation!:)

Thanks! see more to come;)

Как мир тесен )))

Как мир тесен )))

That is not a coincidence:)

whut?

i don't speak russian, sorry :(

Как мир тесен )))

Translation: What a small world

well yeah, but i don't get the joke this time

Translation: What a small world

Thanks bro??

It seems that several members of this chat know each other for a long time

well yeah, but i don't get the joke this time

A couple of Russian guys who know each other by professional interests on the same professional chat

That is what you typically expect to happen, imo

It seems that several members of this chat know each other for a long time

Yeah! I remember you pitching me PonyORM concept on the job interview:)

Yeah :)

oh

awww

sweet

Yeah! I remember you pitching me PonyORM concept on the job interview:)

How was it?)

It was in Netrika

Yeah, I figured

How was it?)

Interview was great. Though I cannot say the same about the concept

?

Interview was great. Though I cannot say the same about the concept

What can you say about the concept then?)

I felt like I don't swing that way

I felt like I don't swing that way

Didn't get what you mean

Didn't get what you mean

I mean it looked like that is not the way I do things

I mean it looked like that is not the way I do things

Ah, I see

There were the second and the third layers of the joke to be honest...:)

You know me. I'm almost complete jerk in such a jokes

hehe)

Hello Telegram Group :D

Hello Telegram Group :D

Hey Luckydonald, welcome :)

Can I bother you guys with helping me write queries?

Sure, go ahead

https://editor.ponyorm.com/user/luckydonald/RssFeedStuff

I am very new to pony and aren't so good with it. I want to retrieve all Messages of a FeedEntry with status "UNREAD"

If I modelled correctly, the feed entry can have multible links, each have a own message. And the feed entry has his own message too

So the message is either Feed or Link->Feed

feed_entry = FeedEntry[1] # instance of Feed Entry select(m for m in Message if m.feed_entry == feed_entry and m.feed_entry.status == 'UNREAD')

If I modelled correctly, the feed entry can have multible links, each have a own message. And the feed entry has his own message too

That's correct

Oh, we need to check the other relationship in Message too, hold on

I assume FeedEntry[1] is the FeedEntry with ID 1?

yes

I am very new to pony and aren't so good with it. I want to retrieve all Messages of a FeedEntry with status "UNREAD"

feed_entry = FeedEntry[1] select(m for m in Message if (m.feed_entry == feed_entry and m.feed_entry.status == 'UNREAD') or (m.link.feed_entry == feed_entry and m.link.feed_entry.status == 'UNREAD'))

here we check both relationships of the Message entity

And together with doing that for each feed_entry it would look like this? for m in select( m for m in AMessage if ( m.feed_entry == feed_entry or m.link.feed_entry == feed_entry ) and feed_entry.status == 'UNREAD' for feed_entry in FeedEntry ): do_post(m)

if you want to use two 'for' this is going to be this way

select(m for m in Message for feed_entry in FeedEntry if (m.feed_entry == feed_entry or m.link.feed_entry == feed_entry) and feed_entry.status == 'UNREAD')

in the first case: select(m for m in Message if (m.feed_entry == feed_entry and m.feed_entry.status == 'UNREAD') or (m.link.feed_entry == feed_entry and m.link.feed_entry.status == 'UNREAD')) you iterate over the Message entity and check both relationships

actually, I think you should use the first case scenario

in the first case: select(m for m in Message if (m.feed_entry == feed_entry and m.feed_entry.status == 'UNREAD') or (m.link.feed_entry == feed_entry and m.link.feed_entry.status == 'UNREAD')) you iterate over the Message entity and check both relationships

But with this one I am only addressing a single FeedEntry (feed_entry = FeedEntry[1])?

correct if you want to retriewe messages for all of them, you need to remove this condition

Right, I want to get all of them.

select(m for m in Message if m.feed_entry.status == 'UNREAD' or m.link.feed_entry.status == 'UNREAD')

Oh

you check Message.feed_entry and Message.link.feed_entry relationships

This is so short. I guess I was thinking way to complex

each instance of Message can have one of those two relationships, right?

Yep

so you need to select the messages where one of the related FeedEntry instances has the status equal to 'UNREAD'

And it won't get None errors on m.feed_entry.*?

Nope

it will select only those messages, which have this relationship

Or right it is translating it first. So not real python properties. Always forget that.

Cool

Many thanks!

Or right it is translating it first. So not real python properties. Always forget that.

correct)

Many thanks!

you are welcome)

I assume FeedEntry[1] is the FeedEntry with ID 1?

AFAIK it could be any primary key

AFAIK it could be any primary key

Correct, this was an example of extracting an instance of FeedEntry. You could retirew it in any other way, for example using FeedEntry.get() https://docs.ponyorm.com/api_reference.html#Entity.get

Guys it looks like questions like this could be collected on a special web page with the answers and examples

This would be very handy

Guys it looks like questions like this could be collected on a special web page with the answers and examples

There is such a place https://docs.ponyorm.com/working_with_entity_instances.html

Hi is talking about community provided faq I guess

Hi is talking about community provided faq I guess

something like that. this page looks a bit tutorialish

may be a wiki or something like that

or... this is great and hacky:) you may post questions on StackOverflow and answer them, even if they wasn't originally asked there

this will allow to google an answer and spread PonyORM on SO at the same time:)

this will allow to google an answer and spread PonyORM on SO at the same time:)

Good catch)

@lig11 You just want to create stackoverflow

oh, I miss “you may post questions on StackOverflow and answer them”

Use the github wiki?

Welcome!

Hi guys!

To guys

Really happy to join you all

Welcome

Thanks buddy

Hey guys I have a problem with the display of the ORM online editor. I found no good place to file an issue, so I went for the doc repo: https://github.com/ponyorm/pony-doc/issues/1