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How to loop through all records? #294

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Tylersuard opened this issue Jul 11, 2024 · 3 comments
Open

How to loop through all records? #294

Tylersuard opened this issue Jul 11, 2024 · 3 comments

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@Tylersuard
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Hello, I would like to loop through every record in my database, modify it, and then update it. How can I do that? Thank you for your help.

@hemidactylus
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hemidactylus commented Jul 22, 2024

Hello, Tyler!
I think you have two options here:

  • collection method update_many (docs) will match an arbitrary number of documents and apply a certain update to all of them, such as {"$inc": {"field": 1}} to increment the value of field by 1 on each matched document. Note that for a large number of matches, this may take long (the client will proceed through pages of results from the API). Also if other processes are altering the documents while this one is running you may encounter a kind of "race condition" so to speak.
  • In case the update requires more logic than fits the update parameter of the above, you will probably have to loop through documents with a find method, compute the changes (with whatever external logic you will need to apply) and call as many update_one methods (or similar) as needed.

If the amount of documents is really large, keep in mind that you may prefer to work in (client-managed) batches and/or keep track of whether the update has been made, with the help of auxiliary metadata in the documents (e.g. {"last_update_round": <integer>} or similar. This would improve restarts if something times out and also reduce the strain associated to very-long-lasting cursors on the API side.

Let me know if this helps and if you think an explicit example will help you further!

@Aakasha063
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Assuming you are working with Django.

You can Query All Records:
records = MyModel.objects.all() retrieves all instances of the MyModel class from the database.
Then, Loop Through Each Record:
The for record in records: loop iterates through each record retrieved.
Modify and Save the Record:
You modify the fields of the record as needed, and then call record.save() to persist the changes to the database.

from myapp.models import MyModel  # Replace with your actual model
from cassandra.cqlengine.management import sync_table

sync_table(MyModel)

records = MyModel.objects.all()

for record in records:
    record.some_field = 'new_value'  # Replace with your actual modifications
      record.save()

@hemidactylus
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@Tylersuard I suspect the reply above (related to Django and based on Cassandra as opposed as the Data API of Astra) is not very relevant.
May I ask you if my earlier reply has helped you solve the issue?
Cheers

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