A practical situation? Lexicographical applications and full-text
applications against text corpora require indexed substring searches,
including ends-with searches. (The FTS extension is not always a good fit.)
I am glad that only the LIKE operator has been overridden in Adobe's
version and in
Tim,
>But did I say that GLOB uses an index if it has been overloaded? No. I
>wrote that if LIKE has been overloaded, queries that contain LIKE
>won't use
>the index. Typically, GLOB won't have been overridden too just
>because LIKE
>has been overridden: the rationale for overriding the
Jean-Christophe,
But did I say that GLOB uses an index if it has been overloaded? No. I
wrote that if LIKE has been overloaded, queries that contain LIKE won't use
the index. Typically, GLOB won't have been overridden too just because LIKE
has been overridden: the rationale for overriding the
Tim,
>Queries using GLOB do use the index on the column in question (i.e.
>optimization is attempted)
>Queries using LIKE do not use that index if the LIKE operator has been
>overridden.
Sorry but GLOB doesn't use an index either if LIKE/GLOB has been
overloaded. This is consistent with the
Just to complete the thread, I decided for the following:
SELECT substr(normalized,1,1) AS letter, COUNT(*) from entry group by
letter order by letter;
Thank you ALL!
2010/4/26 Alberto Simões :
> Hello
>
> Thank you all for the answers.
>
> On Mon, Apr 26, 2010 at 12:59 PM,
Hello
Thank you all for the answers.
On Mon, Apr 26, 2010 at 12:59 PM, Black, Michael (IS)
wrote:
> When you say "running on the fly" do you mean running from an sqlite3 command
> prompt?
I mean somebody will query it and will be waiting for the answer.
> Or are you
Edit: I meant to type "Firefox" not Firebird.
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I am not quite sure what it is, and why it is, that you are doubting,
Jean-Christophe.
Queries using GLOB do use the index on the column in question (i.e.
optimization is attempted)
Queries using LIKE do not use that index if the LIKE operator has been
overridden.
You could confirm this claim
At 14:31 26/04/2010, you wrote:
>If the implementation of SQLite you are using overrides the LIKE operator
>(as more than a few do), then SQLite will not make use of an index on the
>column in question. Use the GLOB operator instead.
I doubt it. GLOB is absolutely nothing more or less than an
Yes. If the OP's [normword] column contains proper nouns, he must normalize
to lower case in order to get accurate results from GLOB.
Or, if his lexicon contains proper nouns in upper case and normal nouns in
lower case, then he could always leave the case intact and use GLOB to get a
count of
Tim Romano wrote:
> If the implementation of SQLite you are using overrides the LIKE operator
> (as more than a few do), then SQLite will not make use of an index on the
> column in question. Use the GLOB operator instead.
>
> For example, I have a lexicon containing 263,000 words:
>
> select
If the implementation of SQLite you are using overrides the LIKE operator
(as more than a few do), then SQLite will not make use of an index on the
column in question. Use the GLOB operator instead.
For example, I have a lexicon containing 263,000 words:
select count(*) from lexicon where
on Systems
From: sqlite-users-boun...@sqlite.org on behalf of Alberto Simões
Sent: Sun 4/25/2010 3:39 PM
To: General Discussion of SQLite Database
Subject: [sqlite] Searching with like for a specific start letter
Hello
I am running on the fly a query to count the number of word
On Sun, 25 Apr 2010 21:39:43 +0100, Alberto Simões
wrote:
>Hello
>
> I am running on the fly a query to count the number of
> words starting with one of the 26 letters.
>
> I am doing the usual SELECT COUNT(term) from dictionary WHERE normword
> LIKE "a%" (for the 26
On 25 Apr 2010, at 9:39pm, Alberto Simões wrote:
> One idea is to add a column named 'letter' and SELECT COUNT(letter)
> from dictionary WHERE letter = 'a'.
That will be the simplest way to make a fast lookup, though it will slow down
your INSERT function. You could speed it up a tiny bit
Alberto Simões wrote:
> I am running on the fly a query to count the number of words starting
> with one of the 26 letters.
>
> I am doing the usual SELECT COUNT(term) from dictionary WHERE normword
> LIKE "a%" (for the 26 letters)
>
> normword is the term normalized without accents and the
Hello
I am running on the fly a query to count the number of words starting
with one of the 26 letters.
I am doing the usual SELECT COUNT(term) from dictionary WHERE normword
LIKE "a%" (for the 26 letters)
normword is the term normalized without accents and the like
Is there any way to make
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