Hi, I have the following data I want to parse through, to simply ext

Help with indentation in templates about ttp HOT 4 CLOSED

dmulyalin commented on August 16, 2024

Help with indentation in templates

from ttp.

Comments (4)

dmulyalin commented on August 16, 2024 1

Have a look at tests on how to define templates within functions using triple quotes:

ttp/test/pytest/test_anonymous_group.py

Line 28 in 8e67084

def test_anonymous_group_with_vars():

from ttp.

dmulyalin commented on August 16, 2024

Probably you now invoking ttp parser object properly, this works for me using Python 3.7.7:

import pprint
from ttp import ttp

data = """
interface Loopback0
 description Fabric Node Router ID
 ip address 192.2.101.70 255.255.255.255
 ip pim sparse-mode
 ip router isis 
 clns mtu 1400
end
interface Loopback0
 description Fabric Node Router ID
 ip address 192.2.101.71 255.255.255.255
 ip pim sparse-mode
 ip router isis 
 clns mtu 1400
end
    """
template = """{{ ignore("\s+") }}ip address {{ ip_address }} 255.255.255.255"""
parser = ttp(data, template)
parser.parse()
res = parser.result()
pprint.pprint(res)

# prints:
# [[[{'ip_address': '192.2.101.70'}, {'ip_address': '192.2.101.71'}]]]

from ttp.

aninchat commented on August 16, 2024

I had tried the above method earlier and it works. However, lets say I create the template inside a python function, where it will automatically be indented, like so:

def parse(data):
    template = """
    {{ ignore("\s+") }}ip address {{ ip_address }} 255.255.255.255 <== indented whitespaces
    """
    parser = ttp(data=data, template=template)
    parser.parse()
    print(parser.result())

This is adding extra whitespaces due to the indentation and it breaks the template parsing. For one line templates, I can declare them inline and escape this problem. But for bigger templates, it is challenges and makes readability of the template difficult.

from ttp.

aninchat commented on August 16, 2024

Thank you, this works perfectly.

from ttp.

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