Why I cannot create standalone object of HttpURL in pydantic?

Question

from pydantic import BaseModel, Field, HttpUrl
from typing import Optional

class TestClass(BaseModel):
    url:Optional[HttpUrl] = None

Creating object TestClass with url="https://www.test.com" works.

Here the imported HttpUrl or BaseModel are class. When I try to create httpurl object standalone it gives typeerror e.g. below.

from pydantic import HttpUrl

myurl = HttpUrl("https://www.test.com")

Why it cannot be used to convert string to http object like above. It results errors like: need keyword-only args if that is provided then 2 positional provided required 3

Thanks ewong for looking in this thread traceback was quite common like typing error: keyword argument required or less number of arguments. — avs_2823, Jun 12 '22 at 22:33

score 10 · Accepted Answer · answered Aug 14 '22 at 23:22

10

You can use parse_obj_as method. This way there will be fewer dependencies:

from pydantic import parse_obj_as, HttpUrl

url=parse_obj_as(HttpUrl, "https://www.test.com")

answered Aug 14 '22 at 23:22

alexandrmoroz

337
4
12

score 1 · Answer 2 · answered Jun 11 '22 at 20:05

1

pydantic types are valid only when used as a class variables inside BaseModel derivatives.

Under the hood, pydantic fires validate method of AnyUrl class which is inconvenient for external usage.

Maybe urllib library will come handy for you?

from urllib.parse import urlparse
myurl = urlparse("https://www.test.com")

answered Jun 11 '22 at 20:05

bc291

1,071
11
23

Thanks, I observed that. However we still create stand alone object with HttpUrl with mandatory parameters like. HttpUrl(url=url,schema='',host='') – avs_2823 Jun 12 '22 at 22:25

Why I cannot create standalone object of HttpURL in pydantic?

2 Answers2