Question about how to create complex virtual libraries

[sbin]

Hi there!
I don't even know if this is possible, but I'd like to have a virtual library showing me the books still to read in the same series of the book I'm currently reading.

It should work like this:
1) the book I'm reading: tags:"=currently reading"
2) all the books in the same series with column #read:false

Is there a way to do it?
(The difficult part is the "books in the same series" bit)

[BetterRed]

@sbin - perhaps you can devise a solution that uses ==>> QuickView and/or Similar Books (its in the right click menu)

BR

[dgparryuk]

I keep looking at this and thinking it might be handy - i don't think it would work as a VL but you need to do something like

if book1.tag = currently reading then show book1.series = bookx.series

*thinks about wondering off an learning python*

[theducks]

Originally Posted by dgparryuk

I keep looking at this and thinking it might be handy - i don't think it would work as a VL but you need to do something like

if book1.tag = currently reading then show book1.series = bookx.series

*thinks about wondering off an learning python*

I don't think a standard search will work because the needed data {series} (the Name of) is not available until the {tags} search has been completed
(The second search can easily ignore any book with a Has_Been_Read status )

[sbin]

@BetterRed yes, right now I'm using QuickView to show me all the books in the series I'm reading, and it's certainly useful. But I'd like to distinguish which ones I've already read and which ones I haven't yet.

I read a lot of series that include novellas, short stories and online addenda (not always published in chronological order), and sometimes the situation can get a bit hairy


@dgparryuk, @theducks It doesn't necessarily have to be a virtual library, any method that could get me a detailed list (author, title, number in the series and tags) will do

[chaley]

Originally Posted by sbin

Hi there!
I don't even know if this is possible, but I'd like to have a virtual library showing me the books still to read in the same series of the book I'm currently reading.

It should work like this:
1) the book I'm reading: tags:"=currently reading"
2) all the books in the same series with column #read:false

Is there a way to do it?
(The difficult part is the "books in the same series" bit)

Multi-step but it should work.

Prerequisite: ensure the "Mark books" action is in the "context menu for the books in the library".

1) select the book being currently read.
2) right click on the book, select Mark Books / clear all marks
3) right click again, select Mark Books / books in selected series
4) search for (or have a virtual library set up for) marked:true and #read:false

There are variants that depending on your work flow might be easier. For example you could have a VL showing #read:false as long as that is true if tags contains "=currently reading".

[theducks]

I just tested my idea
Quickview honors a VL

So... simply make a: 'To be Read' VL (since I keep 'Empty books', my TBR VL does exclude those)
Then you can Search on your 'Now Reading'
In Quickview: Click the Series header to sort by those

Note Quickview only works on Authors
If your series is in a shared universe, they need to all include a common author

Larry Niven & <lots >

[chaley]

Originally Posted by theducks

Note Quickview only works on Authors
If your series is in a shared universe, they need to all include a common author

Quickview works on almost any text-like column. Try clicking in a Series cell then pressing Q.

[theducks]

Originally Posted by chaley

Quickview works on almost any text-like column. Try clicking in a Series cell then pressing Q.

Would help if I RTFM

[sbin]

Originally Posted by theducks

simply make a: 'To be Read' VL (since I keep 'Empty books', my TBR VL does exclude those)
Then you can Search on your 'Now Reading'
In Quickview: Click the Series header to sort by those

That is a very nice workaround and will do the trick, thank you!

When you already know the solution, every riddle seems simple.
And this apply to this rule for sure