將數據庫中的值輸出到html表PHP中

Question

我下面的代碼根據用戶選擇的星期從表中獲取數據，只有2種可能的星期可供選擇。

它根據if語句將配方的標題放入相關表中，但似乎產生了奇怪的輸出。

每當我嘗試查看僅填充部分數據的一周時，將一堆數據填充到錯誤的單元格中； 看這張圖： 最上面一排的“沙拉”應位於“星期日”

這僅在表不完整時發生。

 if(!empty($_POST['selectweek'])) {
                        $selectweek = mysql_real_escape_string($_POST['selectweek']);

                        function ouptutMeal($selectweek, $mealtime, $mealname) {
                            $sqlmeasurement2 = mysql_query("SELECT title, dayid
                                                            FROM recipe
                                                            JOIN menu ON recipe.recipeid = menu.recipeid
                                                            WHERE menu.weekid = '$selectweek'
                                                            AND menu.mealtimeid = '$mealtime'
                                                            ORDER BY dayid");

                            echo "<br/>
                                <table>
                                    <td></td>
                                    <td><strong>Monday</strong></td>
                                    <td><strong>Tuesday</strong></td>
                                    <td><strong>Wednesday</strong></td>
                                    <td><strong>Thursday</strong></td>
                                    <td><strong>Friday</strong></td>
                                    <td><strong>Saturday</strong></td>
                                    <td><strong>Sunday</strong></td>
                                <tr>
                                   <td><strong>$mealname</strong></td>";
                                while($info2 = mysql_fetch_array( $sqlmeasurement2 )) {
                                    if($info2['dayid'] == '1') {
                                        echo '
                                              <td>', $info2['title'], '</td>';
                                    }

                                    elseif($info2['dayid'] == '2') {
                                        echo '
                                              <td>', $info2['title'], '</td>';
                                    }

                                     elseif($info2['dayid'] == '3') {
                                        echo '
                                              <td>', $info2['title'], '</td>';
                                    }

                                    elseif($info2['dayid'] == '4') {
                                        echo '
                                              <td>', $info2['title'], '</td>';
                                    }

                                    elseif($info2['dayid'] == '5') {
                                        echo '
                                              <td>', $info2['title'], '</td>';
                                    }

                                    elseif($info2['dayid'] == '6') {
                                        echo '
                                              <td>', $info2['title'], '</td>';
                                    }

                                    else {
                                        echo '
                                              <td>', $info2['title'], '</td>';
                                    }
                                } 
                            echo '</tr>
                                </table>';
                            }
                        ouptutMeal($selectweek, 1, 'Breakfast');
                        ouptutMeal($selectweek, 2, 'Lunch');
                        ouptutMeal($selectweek, 3, 'Evening Meal');
                        ouptutMeal($selectweek, 4, 'Pudding');
                        ouptutMeal($selectweek, 5, 'Supper & Snacks');
                        }

Answer 1

做一個if語句，看是否存在一個值，否則

   <td>& nbsp;</td>

因此它可以正確地填滿您的桌子。

Answer 2

您還可以在運行查詢之前將變量聲明為空，例如：

if(!empty($_POST['selectweek'])) {
    $info1 = '';
    $info2 = '';
    $selectweek = mysql_real_escape_string($_POST['selectweek']);

    function ouptutMeal($selectweek, $mealtime, $mealname) {
        $sqlmeasurement2 = mysql_query("SELECT title, dayid
                                        FROM recipe
                                        JOIN menu ON recipe.recipeid = menu.recipeid
                                        WHERE menu.weekid = '$selectweek'
                                        AND menu.mealtimeid = '$mealtime'
                                        ORDER BY dayid");

        echo "<br/>
            <table>
                <td></td>
                <td><strong>Monday</strong></td>
                <td><strong>Tuesday</strong></td>
                <td><strong>Wednesday</strong></td>
                <td><strong>Thursday</strong></td>
                <td><strong>Friday</strong></td>
                <td><strong>Saturday</strong></td>
                <td><strong>Sunday</strong></td>
            <tr>
               <td><strong>$mealname</strong></td>";
            while($info2 = mysql_fetch_array( $sqlmeasurement2 )) {
                if($info2['dayid'] == '1') {
                    echo '
                          <td>', $info2['title'], '</td>';
                }

                elseif($info2['dayid'] == '2') {
                    echo '
                          <td>', $info2['title'], '</td>';
                }

                 elseif($info2['dayid'] == '3') {
                    echo '
                          <td>', $info2['title'], '</td>';
                }

                elseif($info2['dayid'] == '4') {
                    echo '
                          <td>', $info2['title'], '</td>';
                }

                elseif($info2['dayid'] == '5') {
                    echo '
                          <td>', $info2['title'], '</td>';
                }

                elseif($info2['dayid'] == '6') {
                    echo '
                          <td>', $info2['title'], '</td>';
                }

                else {
                    echo '
                          <td>', $info2['title'], '</td>';
                }
            } 
        echo '</tr>
            </table>';
    }
    ouptutMeal($selectweek, 1, 'Breakfast');
    ouptutMeal($selectweek, 2, 'Lunch');
    ouptutMeal($selectweek, 3, 'Evening Meal');
    ouptutMeal($selectweek, 4, 'Pudding');
    ouptutMeal($selectweek, 5, 'Supper & Snacks');
}