sqlite> create table bird (id integer primary key autoincrement not null,swing int);//创建一个新表，id是整数，自增和主键。 sqlite> .schema bird//查看表的架构 CREATE TABLE bird (id integer primary key autoincrement not null,swing int); sqlite> insert into bird(swing) values(abs(random())%100);//插入一条新记录，因为id是自增所以只赋值swing字段即可，使用每次一个100以内的随机数（绝对值） sqlite> insert...
所以出于某种原因,我在select语句中使用占位符时遇到错误.def get_id(table_name, id_name):db = sqlite3.connect('test_db')max_id = db.execute('''SELECT max(?) FROM ?''', (id_name, table_name)).fetchone()[0]if not max_id:primary_id = 0else:primary_id = max_id + 1此函数返回此错误：File "test.py", line 77, in get_id max_id = db.execute('''SELECT max(?) FROM ?''', (id_name, table_name)).fetchone()[0] sqlit...
我正在尝试如何最好地在PHP中准备我的SQLite SQL字符串. SQLite3类附带了一个escapeString()函数,但这是我的问题： 尝试1)$sql = "INSERT INTO items ('id','content','title','created') VALUES ('4e7ce7c18aac8', 'Does this work', NULL, '2011-09-23T16:10:41-04:00');"; $sql = SQLite3::escapeString( $sql ); echo ($sql);这导致一个字符串全部被抬起：INSERT INTO items (”id”,”content”,”title”,”created”) VALUE...