将数据库中的值输出到html表PHP中 [英] outputting values from database into html table PHP

问题描述

我有下面的代码，该代码根据用户选择的星期从表中获取数据，只有2个星期可供选择.

I have the code below which grabs data from a table based on what week the user has chosen, there are only 2 possible weeks to choose from.

它根据if语句将配方的标题放入相关表中，但似乎产生了奇怪的输出.

It puts the title of the recipe into the relevant table based on the if statement but seems to be producing weird output.

每当我尝试查看仅填充部分数据的一周时，将一堆数据打包到错误的单元格中；看这张图: 位于第一行的沙拉"应位于星期日"

When ever i try to view a week which only has part data filled in it bunches of the data onto the wrong cells; look in this image: 'Salad' located on the top row should be in 'Sunday'

这仅在表不完整时发生.

This only happens when the table is incomplete.

 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');
                        }

推荐答案

执行if语句以查看是否存在值

Do an if statement to see if a value exists if not then

   <td>& nbsp;</td>

所以它可以正确地填满你的桌子.

so it fills your table correctly.

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