Question

我已经创建了一个预订表格。用户在从下拉菜单中选择住宿选项时可以选择不选择住宿选项，因此值为“选择一个...”。如果他们没有选择住宅区，一旦他们提交了表格，就会将他们带到另一个页面，根据他们的班级和偏好给他们提供他们为住宿区域选择的下拉列表。表格的结果打印在第三页上，其名称，性别，居住区域等等。问题是我使用$ _POST [“宿舍”]检索居住区域。这在用户确实选择了居住区域时有效，但如果没有，则$ _POST似乎不会更新并显示为“选择一个......”。

以下是我为该用户提供居住区选项的页面代码：

<html>
<?php
require 'sql_helper.php';
echo " <form action = \"results2.php\" method = \"post\">\n<br> ";

//Read the information entered on the index page
$specialNeeds = isset($_POST["specialNeeds"]);
$laundry = isset($_POST["laundry"]);
$fullyEquippedKitchen = isset($_POST["fullyEquippedKitchen"]);
$class = $_POST["class"];
$dorm = $_POST["dorm"];

//If user did not choose a dorm
if ($dorm == "Select One...") { 
    $sql = "SELECT name, roomsAvailable FROM $dormTable WHERE class = $class || class = 'Select One...'"; 
        if($result = mysqli_query($conn, $sql)) {
        $numRows = mysqli_num_rows($result);
        echo " <strong>Residence Areas</strong>\n ";
        echo " <select name = dorm> \n";
        for ($i = 0; $i < $numRows; $i++){
            $aDorm = mysqli_fetch_assoc($result);
            $dormName = $aDorm['name'];
            $dormAvailable = $aDorm['roomsAvailable'];
        if($dormName != 'Select One...'){
            echo "<option value = \"$dormName\" > $dormName ($dormAvailable)</option>\n";
        }
        elseif ($dormAvailable == 0 && $dormName != 'Select One...'){
            echo "<option value = \"$dormName\" disabled=\"disabled\"> $dormName </option> \n";
        }

        elseif ($dormName == 'Select One...'){
            echo "<option value = \"$dormName\" disabled=\"disabled\" selected = \"true\"> $dormName </option> \n";
        }
            }
            echo "</select>\n<br><br>";
    } 
   foreach ($_POST as $k => $v){
       echo"<input type = hidden name = $k value = \"$v\" ><?php echo print_r($_POST) ?>";
   }

   echo " <input type = \"submit\" value = \"Submit\"> \n <br> ";
} 
//If user did choose a dorm
else { 
    $sql = "SELECT * FROM $dormTable WHERE name = '$dorm'";
    if ($result = mysqli_query($conn, $sql)) {
        $dormRecord = mysqli_fetch_assoc($result); 
        if ($class == $dormRecord['class']) {
            unset($errorString);

            // check preferences
            if ($fullyEquippedKitchen) {
                if (!$dormRecord['fullyEquippedKitchen']) {
                    $errorString = "$dormRecord[name] does not have a kitchen.\n";
                     echo $errorString;
                     echo 'Go back to the previous page to re-enter your preferences.<br><br>';
                     echo '<a href = "index2.php">Go back</a><br><br>';
                }
                else{
                    foreach ($_POST as $k => $v){
                        echo"<input type = hidden name = $k value = \"$v\"> <?php echo print_r($_POST) ?>";
                     }
                    echo "Your preferences are a match!<br>";
                    echo " <input type = \"submit\" value = \"Submit\"> \n <br> ";
                }
            }
            else{
                foreach ($_POST as $k => $v){
                    echo"<input type = hidden name = $k value = \"$v\"> <?php echo print_r($_POST) ?>";
                }
                echo "Your preferences are a match!<br>";
                echo " <input type = \"submit\" value = \"Submit\"> \n <br> ";
            }
        } 
        else {
            $errorString = "You cannot live in $dormRecord[name] because of your class status.\n";
            echo $errorString;
            echo 'Go back to the previous page to re-enter your preferences.<br><br>';
            echo '<a href = "index2.php">Go back</a><br><br>';
        }

    } 
    else {
        echo "something is wrong: ".mysqli_error($conn);
        die;
    }
    ?>
</form>
<?php
}
?>

帖子的价值不会更新

0 个答案: