c - SQLite:当主数据库有一个打开的准备语句时，为什么我不能分离数据库？

Question

为什么以下代码不起作用:

#include <stdio.h>
#include <sqlite3.h>

int main(void)
{
    sqlite3 *pDb;
    sqlite3_stmt *stmt;
    char *errmsg;

    sqlite3_open(":memory:", &pDb);

    sqlite3_exec(pDb, "CREATE TABLE Test(a INTEGER)", NULL, NULL, NULL);
    sqlite3_exec(pDb, "INSERT INTO Test(a) VALUES(1)", NULL, NULL, NULL);

    sqlite3_prepare_v2(pDb, "SELECT * FROM Test", -1, &stmt, NULL);
    sqlite3_step(stmt);

    sqlite3_exec(pDb, "ATTACH 'Test.db' as Other;", NULL, NULL, NULL);

    sqlite3_exec(pDb, "DETACH Other;", NULL, NULL, &errmsg);
    printf("error: %s\n", errmsg);
    return 0;
}

输出:

error: database Other is locked

如果我在 sqlite3_step(stmt) 之后执行 sqlite3_reset(stmt)，它会起作用。

为什么不相关数据库上的 open 语句会锁定“其他”数据库？我在文档中找不到解释。

Answer 1

transaction documentation说:

An implicit transaction (a transaction that is started automatically, not a transaction started by BEGIN) is committed automatically when the last active statement finishes. A statement finishes when its prepared statement is reset or finalized.

交易总是affect all attached databases ，因此打开的事务将使所有数据库保持锁定状态。