Question

Can I use a CASE statement within substr and filter?

Here is my requirement:

>> value = '00021050430' #here value is a database column
>> query.filter((func.substr(value,case([(func.length(value) > 7,func.length(varying_value)-7+1))],else_=1),7)=='1050430')

Output I am expecting is:

>> query.filter(func.substr(value,6,7))

The above throws an error.

Answer 1

You do not need to use a CASE() statement here. Just use Python:

query.filter(func.substr(value, 1 if len(value) > 7 else 1, 7) == '1050430')

This is called a conditional expression; it’s constructed as true_expr if test_expr else false_expr, where true_expr is used if the test_expr evaluates to True, otherwise the false_expr part is used.

If value is not a constant but a column reference, then you do need to use a CASE() statement. Use sqlalchemy.sql.expression.case() for that:

from sqlalchemy.sql.expression import case

query.filter(func.substr(value, case([(func.length(column) > 7, func.length(column) - 7 + 1)], else_=1), 7) == '1050430')

but you need to make sure your ( and ) parenthesis are balanced (you had one too many).