保持sql分页列表的顺序 [英] Keep the order of list in sql pagination

问题描述

我有一个插入顺序的列表。我想用相同的顺序对结果进行分页。正如你现在所看到的，输出将是一个不同的顺序。

  following_companies_list_data = Company.query.filter（Company.id.in _（['2'，'24'，'1' ，'7'，'373']））。paginate 
 page per_page = 10 error_out = False我在公司：
 print i.id 
 
 7 
 24 
 373 
 2 
 1 
  

相关解决方案

/ 9475755/5341877>这个答案来自相关问题

  company_ids = ['2'，'24'， '1'，'7'，'373'] 
 order_expressions = [（Company.id == i）.desc（）for company_ids] 
 query = Company.query.filter（Company。 （* order_expressions）
 following_companies_list_data = query.paginate（page，per_page = 10，error_out = F ）
  

您也可以使用 idx 函数from intarray module

  from sqlalchemy import func 
 company_ids = ['2'，'24'，'1'，'7'，'373'] 
 query = （$ name，$ company_id）（$ company_id）（$ company_id））
 following_companies_list_data = query.paginate（page，per_page = 10，error_out = False）
  

I have a list with an order of insertion. I want to paginate the results using the same order. As you can see currently the output will be a different order.

following_companies_list_data = Company.query.filter(Company.id.in_(['2', '24', '1', '7', '373'])).paginate(
            page, per_page=10, error_out=False)

companies = following_companies_list_data.items

for i in companies:
    print i.id

7
24
373
2
1

related question

解决方案

Solution based on this answer from related question

company_ids = ['2', '24', '1', '7', '373']
order_expressions = [(Company.id==i).desc() for i in company_ids]
query = Company.query.filter(Company.id.in_(company_ids)).order_by(*order_expressions)
following_companies_list_data = query.paginate(page, per_page=10, error_out=False)

Also you can use idx function from intarray module

from sqlalchemy import func
company_ids = ['2', '24', '1', '7', '373']
query = Company.query.filter(Company.id.in_(company_ids)).order_by(func.idx(company_ids, Company.id))
following_companies_list_data = query.paginate(page, per_page=10, error_out=False)

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