[Flask] Problem with sqlite3 LIKE clause

[Tom]

Thanks Anthony! A good 'Doh' moment there. Spent so much time trying to 
get the parameter substitution bent to my will it looks like I was even 
trying to throw 'C' at it in the end!
Tom
On 10/01/17 16:30, Anthony Ford wrote:
> I think the issue is it's not seeing the % as the wildcard. If you add 
> the wildcard chars to the search string before you supply it to 
> execute, it works.
>
> There's a few Stack Overflow questions on this same topic: 
> http://stackoverflow.com/questions/3105249/python-sqlite-parameter-substitution-with-wildcards-in-like
>
> Try:
>
>     searchstring = '%'+request.form['searchstr']+'%'
>
>     cur = g.db.execute('select * from entries where description like
>     (?) order by id desc',(searchstring,))
>
> The only change is adding the '%' surrounding your search string.
>
>
> Anthony Ford,
> KF5IBN,
> ford.anthonyj at gmail.com <mailto:ford.anthonyj at gmail.com>
>
> On Tue, Jan 10, 2017 at 5:37 AM, Tom via Flask <flask at python.org 
> <mailto:flask at python.org>> wrote:
>
>     I'm trying to modify the flaskr tutorial to make a full blown wiki
>     using ckeditor and its mostly working ok - bar the search.
>     There is a field 'description' which is meant to contain words and
>     phrases for searching.
>     I wish to search a text field in the database but cannot get
>     anything other than the following to parse:
>
>         searchstring=request.form['searchstr'];
>         cur = g.db.execute('select * from entries where description
>     like (?) order by id desc',(searchstring,))
>
>     any attempts to add %'s etc seem lead to failure. I can build the
>     sql string myself but would like to make it safe.
>     Any ideas how to do this?
>     Tom
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