这段代码运行良好:
def rps_tournament_winner(tournament)
if tournament[0][0].kind_of?(String)
puts game_winner tournament
else
for i in 0..tournament.length-1
rps_tournament_winner tournament[i]
end
end
end
然而,此代码给出以下错误语法错误,意外的tIDENTIFIER,预期的kDO或'{‘或'(’
def rp
我已经使用INNER JOIN连接了多个表,并得到了类似如下的输出:
SPORT COUNTRY LEAGUE MATCH
SPORT COUNTRY LEAGUE MATCH
SPORT COUNTRY LEAGUE MATCH
我希望SPORT只在第一次在查询中找到时才显示。国家和联赛也是一样,所以他们是比赛的头球。如下所示:
SPORT
COUNTRY
LEAGUE1
MATCH
MATCH
MATCH
LEAGUE2
MATCH
MATCH
MATCH
代码:
if ($db_found) {
$SQL ="
select sportname,
tournament_t
DAL.TournamentsDataContext tdc = new SchoolSports.DAL.TournamentsDataContext();
var tournamentTable = tdc.GetTable<DAL.Tournament>();
var tournamentRecord = (from rec in tournamentTable
where rec.TournamentId == TournamentId
我正在使用Java Spring和JDBC开发CRUD应用程序。我看过教程,但我坚持使用"U“letter of the crUd :)。我做了一些事情,但如果你能帮我,我将不胜感激。
以下是我做过的事情:
TournamentDAO:
public void edit(Tournament tournament){
String sql = "update TOURNAMENT.TOURNAMENT set name = ? , location = ? , date = ? , where id = ?";
m_jdbcTemplate.updat
我在解决“石头,纸,剪刀”的游戏。输入数组并递归地输出优胜者。这是我的代码:
class RockPaperScissors
# Exceptions this class can raise:
class NoSuchStrategyError < StandardError; end
def self.winner(player1, player2)
strategy = player1[1]+player2[1]
raise NoSuchStrategyError.new("Strategy must be one of R,P,S&
SELECT player.name,player.handicap,
SUM(tournament_player.points) as total_points,
COUNT(tournament_player.player_id) as attendances
FROM player
INNER JOIN tournament_player ON player.id=tournament_player.player_id
GROUP BY player.id
ORDER BY SUM(tournament_player.points) DESC
上面的工作很完美,但它也选择没有积分的球
我正在试验SQLAlchemy中的relationship功能,但是我还没能破解它。下面是一个简单的MRE: from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy import Column, ForeignKey, Integer, create_engine
from sqlalchemy.orm import relationship, sessionmaker
Base = declarative_base()
class Tournament(Base):
__tablenam
我希望我的玩家对象继承锦标赛中的筹码总和。我得到(试图除以0)。我认为这是因为它没有从父锦标赛对象继承筹码。为什么这不起作用?
public partial class Form1 : Form
{
public Tournament Tournament { get; set; }
public Form1()
{
InitializeComponent();
Tournament = new Tournament();
Tournament.PlayerList.Add(new Player("Hero
我有一个具有以下结构的csv文件:
team,tournament,player
Team 1,spring tournament,Rebbecca Cardone
Team 1,spring tournament,Salina Youngblood
Team 1,spring tournament,Catarina Corbell
Team 1,summer tournament,Cara Mejias
Team 1,summer tournament,Catarina Corbell
...
Team 10, spring tournament,Jessi Ravelo
我想要创建一个嵌套
我有组件,当在我的应用程序中看到垂直定位,我希望这是水平的。我应该将css属性应用到哪个组件,它将是什么?
谢谢
主成分
import React from 'react'
import Tournament from './Card.jsx'
import Tournaments from './ListTournaments.jsx'
import Carousel from './Carousel.jsx'
import '../resources/styles/grid.css'
import H
我试着理解SwiftUI,并有一个简单的应用程序来帮助我学习。我的应用程序有一个模型和两个视图,如下所示。
第一个问题是具体的:我如何使“更新比赛”工作?我不知道我应该绑定到哪个变量,如果有的话,或者这是否是正确的方法。
struct Tournament { // eventually will have more properties, such as Bool, Date, arrays, etc.
var name: String
var location: String = "Franchises"
#if DEBUG
var tourn
我想用额外的数据填充两个额外的临时属性,然后发送回响应
'use strict';
var mongoose = require('mongoose');
var express = require('express');
var app = express();
var TournamentSchema = new mongoose.Schema({
createdAt: { type: Date, default: Date.now },
deadlineAt: { type: Date }
});
var Tourna
我有两个对象团队和锦标赛,
class Tournament(Base):
name = models.CharField(db_index=True, max_length=255)
tournament = Tournament.objects.get(id=kwargs.get('tournament_id'))
class Team(Base):
name = models.CharField(db_index=True, max_length=255)
tournaments = models.ManyToManyField(Tourna
您好,我开发了一些函数来创建一个锦标赛类型的地图,但是我不知道如何创建这个地图;我对我们在map.c模块中定义的结构map_t和我想要在tournment.c模块中创建的地图感到困惑
struct node
{
MapDataElement data;
MapKeyElement key;
struct node* next;
};
struct Map_t
{
Node Head; //first node of the linked list that point of the next
copyMapDataElements copyDa
这些查询单独工作。如何编写求和所有结果的UNION?
SELECT SUM(players) FROM `tournament_players` FROM (
SELECT COUNT(*) as `players` FROM `tournament_players` WHERE `foursome_1_p1_name` IS NOT NULL AND `tournament` BETWEEN 13 AND 17 AND flight = '8 AM'
UNION
SELECT COUNT(*) as `players` FROM `tournament_players` WH
这是我的tournaments_controller文件:
class TournamentsController < ApplicationController
before_action :authenticate_user!, only: [:new, :create, :destroy]
def index
end
def show
end
def new
render action: 'new'
end
def create
self.tournament = Tournament.new(tournament_params)
i
我必须连接三张桌子,但我不太擅长这个。我有三张桌子
tournaments
id | name |
teams
id | name
teams_to_tournament
id | tournament_id | team_id
现在我想将这个连接到一个查询中,这样我就可以显示该锦标赛的所有锦标赛和以下球队
Tournament name
tournament team1,tournament team 3,...
Tournament name 2
tournament team1,tournament team 3,...
你能帮我做这个吗?
我有问题,我的“岗位创造”的行动。测试成功通过,属性有效时,但当属性无效时,播放机也被保存。这很奇怪,因为只有:invalid_player可以用无效的属性保存。例如,如果我将wins改为-1或"string",则保存属性为:invalid_player的播放机。但是,如果我为:player更改属性,比如wins = -1,验证器将阻止player被保存。
控制台输出错误消息:
Failures:
1) PlayersController user is signed in POST create with invalid attributes does not save