Question

有两张桌子。一个表包含玩家信息，如姓名和等级。另一个表包含游戏替换信息。

输出应如下所示：

Rank| Player Name | Period 1 position | Period 2 Position| Period 3 Position |

玩家表格样本

|ID| Name | Rank    |
|1 | Fred | 10      |
|2 | Dan  | 12      | 
|3 | Mike | 15      |

职位表样本

|ID| Game_ID | Player_ID | Period | Position     | 
|1 | 12      | 1         | 1      | Striker Right| 
|2 | 12      | 1         | 2      | Mid Center   | 
|3 | 12      | 2         | 1      | Striker Left |

以下是从MySQL中获取它的代码

我尝试了各种方法来转换返回的数据集。

$players_sql = "SELECT position.id as position_id,
player.rank, 
player.first_name, 
position.period, 
position.position
FROM player
INNER JOIN position
on player.id = position.player_id
ORDER BY player.rank, position.period;";

if ($result = $mysqli->query($players_sql)) {
//    printf("Select returned %d rows.\n", mysqli_num_rows($result));
/* fetch associative array */
while ($row = $result->fetch_assoc()) {
    $subs[$row["position_id"]] = [
        "rank" => $row["rank"],
        "first_name" => $row["first_name"],
        "period" => $row["period"],
        "position" => $row["position"]
    ];
}
mysqli_free_result($result);
}

// collapse players This works
foreach ($subs as $sub) {
    $players_by_rank[$sub["rank"]] = $sub["first_name"];
}

稍后数据显示数据以用户可以提交的形式显示。下面的代码与上面的代码不匹配。下面的代码使用静态定义的数组，其中预先填充的数据不是来自数据库。它提供了一种可视化输出需求的方法。在下面的代码示例中，总共有4个时段分为4列，末尾有汇总列。必须有一种简单的方法来展平这些数据！

<?php
for ($sub = 1; $sub < $sub_info . length; $sub++) {
    ?>
    <tr>
        <td><?php echo $sub_info[sub] ?></td>
        <td><input type = "text" 
                   name = "p<?php echo $player; ?>" 
                   value = "<?php echo $names[$player - 1]; ?>"
                   id = "p<?php echo $player; ?>" ></td>
            <?php
            for ($period = 1; $period < $periods + 1; $period++) {
                ?>
            <td>
                <div class="form-group">
                    <select onchange="calTotals(<?php echo $player . ',' . $period; ?>)"
                            id="<?php echo "p{$player}p{$period}"; ?>" 
                            name="<?php echo "p{$player}p{$period}"; ?>">
                                <?php
                                $output = "";
                                for ($i = 0; $i < $players_on_field + 1; $i++) {
                                    $output = "<option";
                                    if ($i == 0) {
                                        $output .= ' selected';
                                    }
                                    $output .= ">$i</option>";
                                    echo $output;
                                }
                                ?>
                    </select>
                </div>
            </td>
        <?php } ?> <!-- Periods -->
        <td><!-- total periods in game using JavaScript -->
            <input type="text" 
                   name="<?php echo "periods{$player}"; ?>"
                   id="<?php echo "periods{$player}"; ?>">
        </td>
    </tr>
<?php } ?> <!-- players -->

Answer 1

您可以使用join连接表格，一个玩家表格实例和三个位置（每个句点一个）

SELECT pl.rank, pl.name, 
  po1.position as period_1_position, 
  po2.position as period_2_position, 
  po3.position as period_3_position from 
  player as pl 
  inner join position as po1 on pl.id = po1.player_id and po1.period=1  
  inner join position as po2 on pl.id = po2.player_id and po2.period=2  
  inner join position as po3 on pl.id = po3.player_id and po3.period=3

如何压扁MySQL结果

1 个答案: