PHP Loop 跳过 if 语句

Question

我通过个人项目学习 PHP。我花了最后 2 个小时试图弄清楚为什么我的 IF 语句在我的循环中被跳过，我是一个更有经验的编码员可以在这里为我提供一些帮助，因为我目前处于僵局。

我想做什么

用户在体育赛事中进行一轮选择，轮次结束后根据正确选择的数量计算用户名次。

除了我(极其)难以计算正确选择的数量外，一切都按预期进行。

IMAGE1:代码应包含以下内容

IMAGE2:代码当前显示以下内容(错误)

我的代码

$sql ="SELECT picks.*, results.*, users.*
        FROM picks
        inner join results on picks.gameID = results.gameID
        inner join users on picks.userID = users.userID 
        WHERE picks.tournament = 'SixNations' AND picks.weekNum = '3'
        order by users.lastname, users.firstname";
        $stmnt = $db->prepare($sql);

        //JUST DISABLING FOR DEBUGING PURPOSES
        /*
        $stmnt->bindValue(':tournament', $tournament);
        $stmnt->bindValue(':weekNum', $round);*/
        $stmnt->execute()
        $picks = $stmnt->fetchAll();
        $totalCorrectPicks = 0;
        echo'<table class="table table-responsive table-striped">';
        echo'<thead>';
        echo'
        <tr>
        <th>FirstName</th>
        <th>Picked Team</th>
        <th>Winning Team</th>
        <th>Margin</th>
        <th>TOTAL CORRECT PICKS</th>
    </tr>
    </thead>
    <tbody>';
   $i = 1; //USED TO COUNT NR GAMES
        foreach($picks as $pick){
            $user = $pick['firstname'];
            $picked = $pick['selectedTeam'];
            $result = $pick['won'];
            $nrGames = $i;

            echo '<tr>';
            echo'<td>';
            echo $user;
            echo'</td>';
            echo'<td>';
            echo $pick['selectedTeam'];
            echo'</td>';
            echo'<td>';
            echo $pick['won'];
            echo'</td>';

        if($picked == $result){
            //USER PICKED CORRECT TEAM
            $totalCorrectPicks = $totalCorrectPicks +1;
                echo'<td>';
                    $margin = abs($pick['pts'] - $pick['points']);
                    echo $margin;
                echo'</td>';
        }//if
        else if($picked != $result){
            //user PICKED INCORRECT TEAM
            echo'<td>';
            $totalCorrectPicks += 0;
            $margin = '<span style="color:red">WRONG PICK</span>';
            echo $margin;
            echo'</td>';    
        }//else if
         ##THIS IF STATMENT IS GETTING IGNORED##
         if($user != $pick['firstname']){
             echo'<td>';
        echo $output = 'Total Correct Picks'. $totalCorrectPicks.' / '.$i; 
            echo'</td>';
            echo'</tr>';
         }//if   
          $i++; //keep track of number games  
      }//foreach
      echo'</tbody>';
      echo'</table>';

如图 2 所示，上面的代码 应该打印每个用户选择的正确游戏总数的 IF 语句被跳过/忽略。 关于原因和/的任何建议/帮助非常欢迎我如何解决这个问题。

Answer 1

希望这可能有所帮助:

代码 1

// Assign Variable
$user = $pick['firstname'];
$nrGame = 1;

// Check user are the same from previous row or not
if($tempUser <> $user){
    $points = 0;
    $nrGame = 1;
}

// Pick correct team
if ($picked == $result){
    $points += 1;
}

// Last Column
echo "<td>".$points." / " .$nrGame."</td>";

// Assign temporary user
$tempUser = $user;

$nrGame++;

CODE 2(仅在最后一行显示点数)

// Assign variable
$maxGame = 3;

// Last Column
if ($nrGame == $maxGame)
    echo "<td>".$points." / " .$nrGame."</td>";
else
    echo "<td></td>";