Add custom label for percolate query subscriptions

[shanbady]

What are the relevant tickets?

Closes https://github.com/mitodl/hq/issues/5598

Description (What does it do?)

This PR adds a "display label" field to percolate queries which overrides any other label for display on the frontend

Screenshots (if appropriate):

How can this be tested?

1. Checkout this branch and run migrations.
2. Subscribe/follow to a few things
3. Go to django admin and edit a percolate query you are subscribed to. Put something custom in the "Display label" field
4. go to your settings page and view your subscriptions list. note that the subscription displays your custom label

[mbertrand]

Works great! Made a suggestion for improving the new test. I thought of a few minor nitpicks that might be nice-to-haves but aren't critical and aren't mentioned in the issue:

• Prepopulate display_label for existing channel-based percolate queries (to channel.title)?
• When new percolate queries for channels are created, maybe assign the channel title as the initial display_label value
  nvm about the above two, channel title might change
• Might be nice to add display_label to PercolateQueryAdmin so they can be seen/searched in django admin:

Also left some other inline code comments, but those are probably out of the scope of this PR. Just the test change should be enough for approval.

[mbertrand]

source_description doesn't seem to be getting tested here, suggest doing something like this:

@pytest.mark.django_db
@pytest.mark.parametrize("test_label", ["new courses about cats", ""])
@pytest.mark.parametrize("is_channel_query", [True, False])
def test_percolate_query_display_labels(
    mocker, mocked_es, test_label, is_channel_query
):
    """
    Test that makes sure we display the display label for a percolate query if it is defined
    """
    mocker.patch(
        "learning_resources_search.indexing_api.index_percolators", autospec=True
    )
    mocker.patch(
        "learning_resources_search.indexing_api._update_document_by_id", autospec=True
    )
    channel = ChannelFactory.create(
        search_filter="department=physics", channel_type="department"
    )

    original_query = {"department": ["physics"]} if is_channel_query else {
        "q": "testing search filter",
        "certification": None,
        "yearly_decay_percent": None,
    }
    query = PercolateQueryFactory.create(
        original_query=original_query,
        query=original_query,
        display_label=test_label,
    )
    assert query.source_description() == (
        test_label
        if test_label
        else channel.title
        if is_channel_query
        else query.original_url_params()
    )

[shanbady]

fixed test

[mbertrand]

Delete this line, it is overriding the parameterized value:

test_label = "new courses about cats"

Also:

    original_query = {"department": ["physics"]} if is_channel_query else {
        "q": "testing search filter",
        "certification": None,
        "yearly_decay_percent": None,
    }

[mbertrand]

Q: Should something else be displayed that's more user friendly than this when display_label and channel are both None?

[mbertrand]

One thing I found a bit counterintuitive but is outside the scope of this PR: the source_description is used as the title in this list while source_label is shown underneath that. I would expect the opposite.

[shanbady]

working on another pr for sending the emails I can see about changing this there.

[mbertrand]

Outside of this scope's PR changes, but should this function also take the source_type into consideration?

    def source_channel(self):
        original_query_params = self.original_url_params()
        channels_filtered = Channel.objects.filter(search_filter=original_query_params)
        if channels_filtered.exists() and self.source_type ==  CHANNEL_SUBSCRIPTION_TYPE:
            return channels_filtered.first()
        return None

[shanbady]

potentially. - im currently working on another PR that involves sending actual emails for percolated searches so this might change there

[mbertrand]

👍