outputting values from database into html table PHP

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伪装坚强ぢ
伪装坚强ぢ 2020-12-22 01:06

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.

It puts the title of the

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  •  太阳男子
    2020-12-22 02:06

    You could also declare the variables as empty before the query is run, example:

    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 "
    "; while($info2 = mysql_fetch_array( $sqlmeasurement2 )) { if($info2['dayid'] == '1') { echo ' '; } elseif($info2['dayid'] == '2') { echo ' '; } elseif($info2['dayid'] == '3') { echo ' '; } elseif($info2['dayid'] == '4') { echo ' '; } elseif($info2['dayid'] == '5') { echo ' '; } elseif($info2['dayid'] == '6') { echo ' '; } else { echo ' '; } } echo '
    Monday Tuesday Wednesday Thursday Friday Saturday Sunday
    $mealname', $info2['title'], '', $info2['title'], '', $info2['title'], '', $info2['title'], '', $info2['title'], '', $info2['title'], '', $info2['title'], '
    '; } ouptutMeal($selectweek, 1, 'Breakfast'); ouptutMeal($selectweek, 2, 'Lunch'); ouptutMeal($selectweek, 3, 'Evening Meal'); ouptutMeal($selectweek, 4, 'Pudding'); ouptutMeal($selectweek, 5, 'Supper & Snacks'); }

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