我在php中創建了一個網站來記錄我的學校的乒乓分數,目前贏得的玩家將記錄WinnerID, LoserID, PointsFor, PointsAgainst
。我有兩個表格,具有以下關係。需要幫助計算點得分的勝利和損失
表:用戶
- USER_ID(PK)
- 用戶名
- 的Elo
表:遊戲
- game_id(PK)
- WinnerID(FK )
- LoserID(FK)
- PointsFor
- PointsAgainst
在php文件我的INSERT語句是:
INSERT INTO games(WinnerID,LoserID,PointsFor,PointsAgainst) VALUES('$Winner_ID','$Loser_ID','$userscore','$oppscore')"
以下是我已經試過了,但它不顯示分數正確。
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
你可以通過上面的圖片看,對總計爲點和點對不上號。目前,它只顯示誰是贏家的統計數據。這意味着如果PlayerA贏得21-5,它將從select語句中顯示出來,但PlayerB不會顯示5-21的分數。任何幫助表示讚賞。
的頁面中輸入分數的PHP代碼:
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
}
}
?>
所以你真正的問題是如何讓失敗者出現在你的查詢中,目前只顯示贏家? – Sean
你能向我們展示一下你的'SELECT'現在返回的例子,以及你想要返回的結果嗎? –
@Sean正好。它只顯示WinnerID的信息,我相信,糾正我,如果我錯了。我的問題是,積分和積分不加起來。他們會是平等的。勝利和損失也是一樣。 –