04-28-2024, 02:29 PM | #1 |
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date_arithmetic using a field instead of a set number
In Calibre, I have a custom column that predicts how many days it will take to finish a book based on my wpm and the word count of the book. I would love to have a secondary column that predicts the finish date of the book. The date_arithmetic function is close to what I need, I think, and would look something like this (perhaps?):
program: date_arithmetic(today(),field('#completiondate')d) where field('#completiondate') is my column that predicts how many days it would take to finish the novel. However, I don't think date_arithmetic works with a variable field instead of a set number increase. (or if it does, I am unsure where to put the single quotes) Does anyone know a way to accomplish this? |
04-28-2024, 03:13 PM | #2 |
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I tried it with this (substituting one of my own datetime columns):
Code:
program: days = days_between(today(),$$#kobodate); date_arithmetic(today(),days) |
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04-28-2024, 03:49 PM | #3 |
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I should have been more specific:
field('#completiondate') is an integer (calculated from word count/average words read per day), not a date (poor naming on my part) date_arithmetic asks for an integer followed by a specification, like '1d' for add one day. I'm looking for a way to make that integer be pulled from a separate column, which is populated with an integer, and add that to today's date. For example, the field('#completiondate') predicts that I could finish the novel in 5 days. This column would ideally add five days to today's date, and display it as the predicted finish date. This is more something I've been tinkering around with just to see if it could be done than something I need for any reason, but I'm invested in seeing if there is a solution now that I can use as a springboard for some other ideas for columns I have. |
04-28-2024, 04:08 PM | #4 |
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Does this do what you want?
Code:
program: mynum = strcat($$#completiondate, 'd'); date_arithmetic(today(),mynum) Code:
program: if !$$#completiondate then '' else mynum = strcat($$#completiondate, 'd'); date_arithmetic(today(),mynum) fi Last edited by ownedbycats; 04-28-2024 at 04:13 PM. |
04-28-2024, 06:30 PM | #5 |
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That did it! I had to round the completion date number, but that was simple, so thank you!!!
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