[solved] Variable number of dictionnaries as argument in def()

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[solved] Variable number of dictionnaries as argument in def()

ndc85430
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Apr-20-2021, 05:28 AM (This post was last modified: Apr-20-2021, 05:29 AM by ndc85430.)

(Apr-19-2021, 05:04 PM)paul18fr Wrote: Dictionnaries names and numbers are variables

Again, avoid doing this because that code is more difficult to read and harder to maintain. If you have a load of things that you need to use, then put those things in a collection as buran shows.

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[solved] Variable number of dictionnaries as argument in def() - by paul18fr - Apr-19-2021, 02:49 PM

RE: Variable number of dictionnaries as argument in def() - by ndc85430 - Apr-19-2021, 04:52 PM

RE: Variable number of dictionnaries as argument in def() - by buran - Apr-19-2021, 05:03 PM

RE: Variable number of dictionnaries as argument in def() - by paul18fr - Apr-19-2021, 05:04 PM

RE: Variable number of dictionnaries as argument in def() - by ndc85430 - Apr-20-2021, 05:28 AM

RE: Variable number of dictionnaries as argument in def() - by paul18fr - Apr-20-2021, 07:29 AM

RE: Variable number of dictionnaries as argument in def() - by buran - Apr-19-2021, 05:18 PM

RE: [solved] Variable number of dictionnaries as argument in def() - by paul18fr - Apr-19-2021, 08:17 PM

RE: [solved] Variable number of dictionnaries as argument in def() - by buran - Apr-20-2021, 09:03 AM

RE: [solved] Variable number of dictionnaries as argument in def() - by paul18fr - Apr-20-2021, 10:07 AM

RE: [solved] Variable number of dictionnaries as argument in def() - by buran - Apr-20-2021, 10:47 AM

RE: [solved] Variable number of dictionnaries as argument in def() - by paul18fr - Apr-20-2021, 11:15 AM

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[solved] Variable number of dictionnaries as argument in def()

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