Question

我正在开发一个PHP项目，我要提交以下表单：

<h2>Create Subject</h2>
   <form action="create_subject.php" method="post">
       <p>Subject name:
           <input type="text" name="menu_name" value="" />
       </p>
       <p>Position:
           <select name="position">
               <?php
               $subject_set = find_all_subjects();
               $subject_count = mysqli_num_rows($subject_set);
               for ($count=1; $count <= ($subject_count + 1); $count++) {
                   echo "<option value=\"{$count}\">{$count}</option>";
               }
               ?>
           </select>
       </p>
       <p>Visible:
           <input type="radio" name="visible" value="0" /> No
           &nbsp;
           <input type="radio" name="visible" value="1" /> Yes 
       </p>
       <input type="submit" name="submit" value="Create Subject" />
    </form>`

在create_subject.php（表单操作发生的地方）中，我有一些验证，如下所示：

if(isset($_POST['submit'])) {
    // Process the form
    $menu_name = mysql_prep($_POST["menu_name"]);
    $position = (int) $_POST["position"];
    $visible = (int) $_POST["visible"];

    //validations
    $required_fields = array("menu_name", "position", "visible");
    validate_presences($required_fields);

    $fields_with_max_lengths = array("menu_name" => 30);
    validate_max_lengths($fields_with_max_lengths);

    if(!empty($errors)) {
        $_SESSION["errors"] = $errors;
        redirect_to("new_subject.php");
    }

其中验证状态应该用作检查字段是否为空并且如下所示：

function validate_presences($required_fields) {
    global $errors;
    foreach ($required_fields as $field) {
        $value = trim($_POST[$field]);
        if (!has_presence($value)) {
            $errors[$field] = fieldname_as_text($field)." can't be blank";
        }
    }
}

但是当我提交缺少数据的表单时，我收到以下错误消息，而不是重定向回到上一页并列出会话中存储的所有错误：

注意：未定义的索引：在第10行的/Users/eak/Sites/widget_corp/public/create_subject.php中可见



注意：未定义的索引：在/Users/eak/Sites/widget_corp/includes/validation_functions.php上可见   第22行


警告：无法修改标头信息 - 已经发送的标头（输出从
开始   /Users/eak/Sites/widget_corp/public/create_subject.php:10)in
  第4行/susers/eak/Sites/widget_corp/includes/functions.php

因此输出开始于$ _POST [＆＃34;可见＆＃34;]被检测为未定义的位置。这里有什么解决方案？

Answer 1

解决方案非常简单。正如您所说$_POST["visible"]未定义，只需使用以下代码块来处理相同的内容。

if (isset($_POST["visible"])) {
    //Handle the null condition
}

编辑：根据您的评论，您有一个处理上述问题的功能，但在您使用该值后调用它。检查以下评论：

    $menu_name = mysql_prep($_POST["menu_name"]);
    $position = (int) $_POST["position"];
    $visible = (int) $_POST["visible"];  // YOU USED THIS VALUE - ERROR ALREADY OCCURRED HERE

    //validations
    $required_fields = array("menu_name", "position", "visible");  //AND THEN YOU PERFORMED VALIDATION-- CALL THIS BEFORE HAND
    validate_presences($required_fields);

Answer 2

您好，请用这些替换代码。请先进行验证

if(isset($_POST['submit'])) { 
//validations
$required_fields = array("menu_name", "position", "visible");
validate_presences($required_fields);

if(!empty($errors)) {
    $_SESSION["errors"] = $errors;
    redirect_to("new_subject.php");
}

// Process the form 
$menu_name = mysql_prep($_POST["menu_name"]); 
$position = (int)$_POST["position"]; 
$visible = (int) $_POST["visible"];



$fields_with_max_lengths = array("menu_name" => 30);
validate_max_lengths($fields_with_max_lengths);

同时更新这些函数validate_presences： -

function validate_presences($required_fields) {
    global $errors;
    foreach ($required_fields as $field) {
        if(isset($_POST[$field])){
              $value = trim($_POST[$field]);
              if (!has_presence($value)) {
                  $errors[$field] = fieldname_as_text($field)." can't be blank";
              }
        } else {
                  $errors[$field] = fieldname_as_text($field)." can't be blank";
        }
    }
}

$ _POST [＆＃34;可见＆＃34;]标记为未定义的索引

2 个答案: