Projectify plugin: request for hints about managing hierarchies of Todos

#1

I am trying to learn the basics of Projectify plugin

Consider the following scenario:

* Project-1
  * Subproject-1.1
    * Subproject-1.1.1
      * Task-1.1.1.1
      * Task-1.1.1.2

* Project-2
  * Subproject-2.1
    * Subproject-2.1.1
      * Task-2.1.1.1

I would like a way to see: a list of all tasks (as in, “leaves” from all project trees), sorted by project.

By creating a couple of subproject/tasks, I’ve noticed that a “leaf”:

  1. has a todo tag
  2. does not have a SubProject tag

This means that I can solve half of my problem like this:

<<list-links "[tag[todo]!tag[SubProject]]">>

This will produce something like:

* Task-1.1.1.1
* Task-1.1.1.2
* Task-2.1.1.1

Unfortunately, I’m stuck here, because I don’t know how to include the project name. My goal is to have links (still pointing to the “leaf” task tiddlers, but with names like this:

* [Project-1] Task-1.1.1.1
* [Project-1] Task-1.1.1.2
* [Project-2] Task-2.1.1.1

Is it possible to be done in a trivial way, not involving complex programming with tree walking algorithms, which is scary for an newbie TiddlyWiki end user?

#2

Since there could be an arbitrary depth of subprojects, you will need to either write a recursive filter to find the project name, or use a custom plugin that does it for you, for example:

With the taggingtree[] operator:

<$list filter="[tag[Project]]">

!! <$link/>

{{{ 
   [{!!title}taggingtree[]tag[todo]!tag[SubProject]]
   || $:/plugins/nico/projectify/ui/todo/TodoItem
}}}

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This way you display the projects, then each tasks for each projects, even those in subprojects.

You can download taggingtree.js here: Recursive filter operators to show all tiddlers beneath a tag and all tags above a tiddler - #52 by Yaisog

I think you could do the same with the kin operator, I’m not sure which one would be the best for this use case tbh.

#3

according to https://yaisog.tiddlyhost.com taggingtree is deprecated, but thanks for pointing me in the right direction. I just had to use the descendants filter instead in your example

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