Question

我试图让系统显示来自用户数据库的信息，然后它会在表格中显示，但是，我遇到的问题是页面加载，我将需要其中的三个，我认为我走了最长的路线。页面将需要显示收件人和类型，现在在我的数据库中我有5种类型的信息，它现在每个都计算它们然后显示为好坏和标记。我可以使用更少的MySQL

<?php
    $conn = new mysqli($servername, $username, $password, $dbname);
    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    } 
    $result = mysqli_query($conn,"SELECT * 
        FROM `user_flags`
        ORDER BY user_flags.timestamp DESC
        LIMIT 15");
    ?>
                                <tbody>
    <?php

    while( $row = mysqli_fetch_assoc( $result ) ) {

        $GoodRepSql = mysqli_query($conn, "SELECT COUNT(*) 
            FROM user_flags 
            WHERE user_flags.recipient = {$row['recipient']} 
            AND type = 'Good Rep' ");

        if ($GoodRepSql->num_rows > 0) {
            while($goodrow = $GoodRepSql->fetch_assoc()) {
                $GoodRep = $goodrow["COUNT(*)"] ;
            }
        } else {
            $GoodRep = $goodrow["COUNT(*)"] ;
        }

        $BadRepSql = mysqli_query($conn, "SELECT COUNT(*) 
            FROM user_flags 
            WHERE user_flags.recipient = {$row['recipient']} 
            AND type = 'Bad Rep' ");

        if ($BadRepSql->num_rows > 0) {
            while($badrow = $BadRepSql->fetch_assoc()) {
                $BadRep = $badrow["COUNT(*)"] ;
            }
        } else {
            $BadRep = $badrow["COUNT(*)"] ;
        }

        $FlagSql = mysqli_query($conn, "SELECT COUNT(*) 
            FROM user_flags 
            WHERE user_flags.recipient = {$row['recipient']} 
            AND type IN ('Smurfing', 'Griefing', 'Suspicious')");
        if ($FlagSql->num_rows > 0) {
            while($flagrow = $FlagSql->fetch_assoc()) {
                $Flags = $flagrow["COUNT(*)"] ;
            }
        } else {
            $Flags = $flagrow["COUNT(*)"] ;
        }

        echo
            "<tr onclick=\"document.location = '{$url}/{$row['recipient']}';\">
            <td>{$Recipient->players->player->personaname}</td>
            <td>{$BadRep}</td>
            <td>{$GoodRep}</td>
            <td>{$Flags}</td>
            </tr>\n";
    }

?>

Answer 1

如果您需要每种类型的计数，您可以使用：GROUP BY

SELECT COUNT(*),type FROM user_flags WHERE user_flags.recipient = {$row['recipient']} GROUP BY type

由于您希望将某些类型添加到一起，因此您需要使用php添加它们。

$RepSql = mysqli_query($conn, "SELECT COUNT(*) as repcount,type FROM user_flags WHERE user_flags.recipient = {$row['recipient']} GROUP BY type ");
    $GoodRep=$BadRep=$Flags=0;   
if ($RepSql->num_rows > 0) {
            while($rep_row = $RepSql->fetch_assoc()) {
                if( $rep_row["type"]=='Smurfing'||$rep_row["type"]=='Griefing'||$rep_row["type"]=='Suspicious'){
                    $Flags+=$rep_row["repcount"];
                }else if($rep_row["type"]=="Good Rep"){
                    $GoodRep=$rep_row["repcount"];
                }else if($rep_row["type"]=="Bad Rep"){
                    $BadRep=$rep_row["repcount"];
                }
            }
        }

使用较少的MySQLi查询

1 个答案: