Create link that searches in Calibre

icarine · 09-29-2014, 02:05 AM

I love how you can click on a tag in book details and search Calibre for that tag. I'm trying to create a column template that does the same thing for a custom column.

I tried a column template like:

Code:

{#genre:|<a href='search:#genre:"=|"'>}{#genre:||</a>}

But I either get this error:

Code:

calibre, version 2.2.0
ERROR: Unhandled exception: <b>TypeError</b>:Non-hexadecimal digit found

Or this error:

Code:

calibre, version 2.2.0
ERROR: Unhandled exception: <b>TypeError</b>:Odd-length string

Even if I do something simple like

Code:

<a href="search:test">test</a>

I get the same error. Is there a way to do this? What's the right URL pattern?

kovidgoyal · 09-29-2014, 02:13 AM

The part after search: has to be hex-encoded. So if you want the search to be:

search:tags:text

it becomes:

search:746167733a74657874

I dont think the template language has a utility function to hex-encode strings, you could ask chaley if he is willing to add one or you could create your own in general program mode.

icarine · 09-29-2014, 02:39 AM

Thanks for your help! I got this working by adding a custom template function:

Code:

def evaluate(self, formatter, kwargs, mi, locals, value, search_term):
	search = '%s:"=%s"' % (search_term, value.replace('"', '\\"'))
	return search.encode('hex')

09-29-2014, 02:05 AM	#1
icarine Member Posts: 13 Karma: 1034 Join Date: Mar 2010 Location: NC Device: PB 360	Create link that searches in Calibre I love how you can click on a tag in book details and search Calibre for that tag. I'm trying to create a column template that does the same thing for a custom column. I tried a column template like: Code: {#genre:\|<a href='search:#genre:"=\|"'>}{#genre:\|\|</a>} But I either get this error: Code: calibre, version 2.2.0 ERROR: Unhandled exception: <b>TypeError</b>:Non-hexadecimal digit found Or this error: Code: calibre, version 2.2.0 ERROR: Unhandled exception: <b>TypeError</b>:Odd-length string Even if I do something simple like Code: <a href="search:test">test</a> I get the same error. Is there a way to do this? What's the right URL pattern?

09-29-2014, 02:39 AM	#3
icarine Member Posts: 13 Karma: 1034 Join Date: Mar 2010 Location: NC Device: PB 360	Thanks for your help! I got this working by adding a custom template function: Code: def evaluate(self, formatter, kwargs, mi, locals, value, search_term): search = '%s:"=%s"' % (search_term, value.replace('"', '\\"')) return search.encode('hex')

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09-29-2014, 02:13 AM	#2
kovidgoyal creator of calibre Posts: 45,659 Karma: 28549046 Join Date: Oct 2006 Location: Mumbai, India Device: Various	The part after search: has to be hex-encoded. So if you want the search to be: search:tags:text it becomes: search:746167733a74657874 I dont think the template language has a utility function to hex-encode strings, you could ask chaley if he is willing to add one or you could create your own in general program mode.

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