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SQLite: how to use PRAGMA application_id?

While

PRAGMA application_id = <integer>;

doesn't fail it seems like it doesn't do something like

PRAGMA application_id;

always returns an empty set.

I tried using python (sqlite3 and APSW) and sqlite3 (1) command line

note: I am running debian 3.7.13 sqlite version

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3 answers

Application_id was introduced in SQLite 3.7.17 .

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PRAGMA application_id; PRAGMA application_id = integer; Application_id PRAGMA 32- " ", 68 . , SQLite , ID , , (1), , " SQLite3". , magic.txt SQLite.

http://www.sqlite.org/pragma.html#pragma_application_id

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Note that the integer must be within the range of the signed 32-bit value (i.e., in the range -2 32 to +2 32 -1, which is -2147483648 - +2147483647). If the provided integer value is out of range, then the operator will be accepted without errors, however, the application identifier will be set to zero.

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