我正在用PHP创建一个网站,为我的学校记录乒乓球分数,目前获胜的玩家将记录WinnerID, LoserID, PointsFor, PointsAgainst
。我有两个表有以下关系。
表:用户
表:游戏
我在php文件中的insert语句是:
INSERT INTO games(WinnerID,LoserID,PointsFor,PointsAgainst) VALUES('$Winner_ID','$Loser_ID','$userscore','$oppscore')"
这是我尝试过的,但它没有正确显示分数。
SELECT min(u.username) 'Username', COUNT(g.WinnerID) 'Wins', sum(g.PointsFor) 'Points For', sum(g.PointsAgainst) 'Points Against', u.Elo 'Ranking'
from games g
LEFT JOIN users u
on g.WinnerID = u.user_id
Group by g.WinnerID
正如您在上图中所看到的,总计的点数和点数不会相加。目前,它仅显示获胜者的统计数据。意味着如果PlayerA赢得21-5,它将从select语句中显示,但是PlayerB将不会显示5-21的分数。任何帮助表示赞赏。
用于输入分数的页面的PHP代码:
if(isset($_POST['btn-post']))
{
$opponent = $_POST["opponent"];
//$opponent = array_key_exists('opponent', $_POST) ? $_POST['opponent'] : false;
$userscore = mysql_real_escape_string($_POST['userscore']);
$oppscore = mysql_real_escape_string($_POST['oppscore']);
if($userscore != $oppscore)
{
if($userscore > $oppscore)
{
$Winner_ID = $_SESSION['user'];
$query = mysql_query("SELECT user_id FROM users WHERE username = '".$opponent."'");
$result = mysql_fetch_array($query) or die(mysql_error());
$Loser_ID = $result['user_id'];
$query1 = mysql_query("SELECT Elo FROM users WHERE user_id=".$_SESSION['user']);
$result1 = mysql_fetch_array($query1) or die(mysql_error());
$winnerRating = $result1['Elo'];
$query2 = mysql_query("SELECT Elo FROM users WHERE user_id=".$Loser_ID);
$result2 = mysql_fetch_array($query2) or die(mysql_error());
$loserRating = $result1['Elo'];
$rating = new Rating($winnerRating, $loserRating, 1, 0);
$results = $rating->getNewRatings();
if(mysql_query("UPDATE users SET Elo = " . $results['a'] . " WHERE user_id=".$_SESSION['user']))
{
}
else
{
?>
<script>alert('There was an error while entering winners(user) ranking...');</script>
<?php
}
if(mysql_query("UPDATE users SET Elo = " . $results['b'] . " WHERE user_id=".$Loser_ID))
{
}
else
{
?>
<script>alert('There was an error while entering losers(opp) ranking..');</script>
<?php
}
}
elseif($oppscore > $userscore)
{
$Loser_ID = $_SESSION['user'];
$query = mysql_query("SELECT user_id FROM users WHERE username = '".$opponent."'");
$result = mysql_fetch_array($query) or die(mysql_error());
$Winner_ID = $result['user_id'];
//get rating from user table in database
$query1 = mysql_query("SELECT Elo FROM users WHERE user_id=".$_SESSION['user']);
$result1 = mysql_fetch_array($query1) or die(mysql_error());
$loserRating = $result1['Elo'];
$query2 = mysql_query("SELECT Elo FROM users WHERE user_id=".$Loser_ID);
$result2 = mysql_fetch_array($query2) or die(mysql_error());
$winnerRating = $result1['Elo'];
$rating = new Rating($winnerRating, $loserRating, 1, 0);
$results = $rating->getNewRatings();
$results = $rating->getNewRatings();
if(mysql_query("UPDATE users SET Elo = " . $results['b'] . " WHERE user_id=".$_SESSION['user']))
{
}
else
{
?>
<script>alert('There was an error while entering losers(user) ranking...');</script>
<?php
}
if(mysql_query("UPDATE users SET Elo = " . $results['a'] . " WHERE user_id=".$Winner_ID))
{
}
else
{
?>
<script>alert('There was an error while entering winners(opp) ranking...');</script>
<?php
}
}
if(mysql_query("INSERT INTO games(WinnerID,LoserID,PointsFor,PointsAgainst) VALUES('$Winner_ID','$Loser_ID','$userscore','$oppscore')"))
{
?>
<script>alert('Your scores were successfully entered');</script>
<?php
}
else
{
?>
<script>alert('There was an error while entering your score...');</script>
<?php
}
}
else
{
?>
<script>alert('There cannot be a tie in ping pong, please re-enter your scores...');</script>
<?php
}
}
?>
答案 0 :(得分:1)
尝试将其作为内部联接,并删除MIN()
列上的username
:
SELECT u.username, COUNT(g.WinnerID),
SUM(g.PointsFor), SUM(g.PointsAgainst), u.Elo
FROM users u, games g
WHERE u.user_id = g.WinnerID
GROUP BY u.username, u.Elo;
此外,在其他人开始执行任务之前,您应该使用mysqli
代替mysql
(或者更好,PDO
)并且您应该使用预备语句而不是动态语句SQL。
答案 1 :(得分:1)
您的查询失败,因为它没有考虑玩家输掉的行。您可以使用unions来解决此问题。以下查询应该执行您想要的操作:
SELECT username AS "Username",
SUM(wins) AS "Wins",
SUM(PF) AS "Points For",
SUM(PA) AS "Points Against",
elo AS "Ranking"
FROM (
(SELECT users.user_ID,
users.username AS username,
COUNT(games.WinnerID) AS wins,
SUM(games.PointsFor) AS PF,
SUM(games.PointsAgainst) AS PA,
users.Elo AS elo
FROM users, games
WHERE games.WinnerID = users.user_ID
GROUP BY users.user_ID)
UNION ALL
(SELECT users.user_ID,
users.username AS username,
0 AS wins,
SUM(games.PointsAgainst) AS PF,
SUM(games.PointsFor) AS PA,
users.Elo AS elo
FROM users, games
WHERE games.LoserID = users.user_ID
GROUP BY users.user_ID)
) AS t
GROUP BY username
ORDER BY user_ID;
请注意,在&#34;丢失查询&#34;字段PointsAgainst
应该作为播放器&#34;指向&#34;反之亦然。
答案 2 :(得分:0)
您
INSERT INTO games(WinnerID,LoserID,PointsFor,PointsAgainst) VALUES('$Winner_ID','$Loser_ID','$userscore','$oppscore')
查询包含可能错误顺序的$userscore
和$oppscore
值。 {/ 1}}和$Winner_ID
可能会在您的条件处理中发生变化,但$Loser_ID
和$userscore
在此时并未同样翻转。
此外,您的条件结构通常也是不必要的冗长。看起来你应该能够先确定赢家和输家的ID和得分,然后再进行一次所有的处理,而不是诉诸于容易出错的复制粘贴重复的几乎相同的代码。