If the value in key is a blob, then like matches it and = does not.
Because like has to do a string conversion on key, it also doesn't use
the index.
Try:
SELECT * FROM recordings WHERE "key" =
cast('4df0247ce1a97685a782d2cb051b48ed952e666c' as blob);
Or try inserting the key as text in the first place.
Or use a binary blob instead of a hex dump of it, and use (note the
"x"):
SELECT * FROM recordings WHERE "key" =
x'4df0247ce1a97685a782d2cb051b48ed952e666c';
--David Garfield
Eric Anderson writes:
> The below statement returns records:
>
> SELECT * FROM recordings WHERE "key" LIKE
> '4df0247ce1a97685a782d2cb051b48ed952e666c';
>
> But this one does not:
>
> SELECT * FROM recordings WHERE "key" =
> '4df0247ce1a97685a782d2cb051b48ed952e666c';
>
> The only difference is that = and LIKE have been swapped.
>
> I realize "key" is a keyword but it is quoted properly from what I can
> tell. I'm on sqlite version 3.7.4 (standard one with Mint 11 Linux).
> Any ideas? The only thing I can think of is some sort of encoding
> issue that the LIKE operator is getting around somehow. Or perhaps the
> fact that it is a keyword?
>
> Any ideas on how to debug?
>
> --
> http://saveyourcall.com - Easily record phone calls from any phone
> http://pixelwareinc.com - Atlanta-based web development and design
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