#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);
    sqlite3_exec(pDb,"INSERT INTO Test(a) VALUES(1)",NULL);

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

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

    sqlite3_exec(pDb,"DETACH Other;",&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),它可以工作.

为什么不相关数据库上的开放语句会锁定“其他”数据库？我在文档中找不到解释.