我正在研究 sqlite 数据库,试图插入数据库,但它给出错误“插入时出错”数据库被锁定。我搜索了一些帖子,还写了 reset 和 finalize 语句,但它给出了同样的错误。
这是我的代码
if(addStmt == nil)
{
//const char *sql = "insert into Limittabel (limitAmt) Values(?)";
//const char *sql="INSERT INTO Limittabel (limitAmt, Profileid) VALUES(?, ?)";
const char *sql="insert into Limittabel (limitAmt,Profileid) Values (?,?)";
// Insert into Limittabel (limitAmt,Profileid) Values ($500,1)
if(sqlite3_prepare_v2(database, sql, -1, &addStmt, NULL) != SQLITE_OK)
NSAssert1(0, @"Error while creating add statement. '%s'", sqlite3_errmsg(database));
}
sqlite3_bind_text(addStmt, 1, [limitAmt UTF8String], -1, SQLITE_TRANSIENT);
// NSString *str_id_poi = [[NSString alloc] initWithFormat:@"%d", [Profileid integerValue]];
// sqlite3_bind_text(addStmt, 2, [str_id_poi UTF8String], -1, SQLITE_TRANSIENT);
//sqlite3_bind_text(addStmt, 2, [ UTF8String], -1, SQLITE_TRANSIENT);
NSLog(@"***Storing END on Database ***");
if(SQLITE_DONE != sqlite3_step(addStmt))
{
NSAssert1(0, @"Error while inserting data. '%s'", sqlite3_errmsg(database));
} // sqlite3_finalize(addStmt);
else{
//sqlite3_last_insert_rowid;
limitid = sqlite3_last_insert_rowid(database);
NSLog(@"YES THE DATA HAS BEEN WRITTEN SUCCESSFULLY");
}
sqlite3_finalize(addStmt);
}
sqlite3_reset(addStmt);
//sqlite3_finalize(addStmt);
}
如果我没有打开数据库,则会出现数据库锁定错误,如果打开数据库则会出现“没有此类表”错误,请帮助我。谢谢