Question

这是我的代码：

- (void) addOrder {

    if(addStmt == nil) {

        const char *sql = "insert into item(menuid,itemName,price,quantity,spiciness) Values(?, ?, ?, ?, ?)";

        if(sqlite3_prepare_v2(database, sql, -1, &addStmt, NULL) != SQLITE_OK)
            NSAssert1(0, @"Error while creating add statement. '%s'", sqlite3_errmsg(database));
    }

    sqlite3_bind_int(addStmt, 1, [menuID integerValue]);
    sqlite3_bind_text(addStmt, 2, [itemName UTF8String], -1, SQLITE_TRANSIENT);
    sqlite3_bind_double(addStmt, 3, price );
    sqlite3_bind_int(addStmt, 4, [quantity integerValue]);
    sqlite3_bind_text(addStmt, 5, [spiciness UTF8String],-1,SQLITE_TRANSIENT);


    if(SQLITE_DONE != sqlite3_step(addStmt))
        NSAssert1(0, @"Error while inserting data. '%s'", sqlite3_errmsg(database));
    else


        menuID = sqlite3_last_insert_rowid(database);

    //Reset the add statement.
    sqlite3_reset(addStmt);
}

我的menuID是NSdecimalNumber。在else部分

menuID = sqlite3_last_insert_rowid(database);

我收到了错误

“Implicit conversion of sqlite3_int64 to NsdecimalNumber is disallowed with ARC.

我怎么能删除这个错误？请帮助..

Answer 1

正如你所说，menuID是NSDecimalNumber而不是它应该是，

menuID = [NSDecimalNumber numberWithInt:sqlite3_last_insert_rowid(database)];

Answer 2

由于它似乎是一个64位的int，而iOS是一个32位系统，你可能想要使用：

menuID = [NSDecimalNumber numberWithLongLong:sqlite3_last_insert_rowid(database)];

iPhone：NSdecimalNumber转换中的问题

2 个答案: