foreach循环插入所有的复选框（选中和未选中）

Question

我已被示出在下面

<p><form name="university" action="/university_handler" method="post"> 
    <fieldset> 
    <table class="table"> 
     <thead> 
     <tr> 
      <th><span class="help-block">University Department</span></th>        
     </tr> 
     </thead> 
     <tbody> 
     <tr> 
      <td><?php 
      $query = mysqli_query($db, "SELECT university_department FROM university WHERE university_id = '$university_id'") 
       or die ("Could not search!"); 
      while($row = mysqli_fetch_array($query)){ 
       $university_department = $row['university_department']; 
       $_SESSION['university_department'] = $university_department; 
       $universityDepartment = $_SESSION['university_department']; 
       echo "<label><input type='checkbox' name='university_department[]' value='{$universityDepartment}'>$universityDepartment</label><br><input type='text' value='' name='professor_name[{$universityDepartment}]' placeholder='Professor-Name'><input type='text' value='' name='class_name[{$universityDepartment}]' placeholder='Class-Name'>";} 
      ?></td> 
     </tr> 
     </tbody> 
    </table> 
    <button type="submit" name="Submit"class="btn btn-info">Submit</button> 
    </fieldset> 
</form></p> 

现在，当使用我university_handler将这些值插入到数据库中的所有的生成使用PHP HTML复选框形式复选框正在插入，而不仅仅是已被选中的复选框。我一直在尝试一系列的事情，似乎没有任何工作。这是处理程序。

<?php 
session_start(); 
include("connect.php"); 

$university_id = $_SESSION['university_id']; 

// check if share_form is submitted and not empty 
$error_message = ""; 
if(is_array($_POST['university_department']) && !empty($_POST['university_department'])){ 
    $error = array(); 
    $universityDepartment = $_POST['university_department']; 
    if (count($universityDepartment)>0){ 
    foreach (str_replace('#', '', $_POST['class_name']) as $departmentName => $stripid){ 
     $class_name_backslash = $stripid . '/'; 
     $class_name = mysqli_real_escape_string($db, $stripid); 
     print_r($class_name); 
    } 
    $query_uni = ("INSERT INTO temp_list(departmentName, class_name, professor_name) VALUE ('$departmentName','$class_name', '$professor_name')"); 
    $q_u = mysqli_query($db, $query_uni) or die ('Error posting data'); 
} 
}?> 

Answer 1

我喜欢@ Martin的回答。我想改变的唯一的事情是$_REQUEST

当使用$_REQUEST你说嘿看看帖子或变量并使用任一值。那么如果你同时使用$_POST和$_GET变量，会发生什么？一个会被使用，而另一个则不会。

因此，让我们以不同的方式使用相同的代码。

$checkBoxName = (isset($_POST['checkBoxName']) ? $_POST['checkBoxName'] : (isset($_GET['checkBoxName']) ? $_GET['checkBoxName'] : '')); 
if ($checkBoxName != '') { 
    //do stuff here 
} 

这种方式，因为你看不到，而无需使用像谷歌铬合金的开发工具的信息后，最好做这个顺序，这样你检查什么什么，你可以看到之前没有见过。

希望这有助于=）

编辑：

基于你给我的信息，我能想出一个办法只插入你检查过的类。随意做改变这一点，但这个应该工作

<?php 
$departList = ($_POST['university_department'] ? $_POST['university_department'] : ($_GET['university_department'] ? $_GET['university_department'] : array())); 
$classList = ($_POST['class_name'] ? $_POST['class_name'] : ($_GET['class_name'] ? $_GET['class_name'] : array())); 
$profList = ($_POST['professor_name'] ? $_POST['professor_name'] : ($_GET['professor_name'] ? $_GET['professor_name'] : array())); 
if (count($departList) > 0) { 
    foreach ($departList as $key => $val) { 
     $class = $classList[$val]; 
     $professor = $profList[$val]; 
     $query_uni = ("INSERT INTO temp_list(departmentName, class_name, professor_name) VALUE ('$val','$class', '$professor')"); 
     $q_u = mysqli_query($db, $query_uni) or die ('Error posting data'); 
    } 
} 

?> 

好运=）

Answer 2

您可以使用isset函数来检查复选框是否被选中。

if (isset($_REQUEST['checkBoxName'])); 
    { 

    $variable = $_REQUEST['checkBoxName']; 
    } 

使用此方法，您只会获得具有值的复选框。

希望这会有帮助