为什么这个验证码不能按预期工作？答案

【问题标题】：Why this validation code does not work as expected?为什么这个验证码不能按预期工作？
【发布时间】：2015-06-09 13:21:33
【问题描述】：

我有一个表单，for m 的操作是同一页。我正在尝试：

表单提交成功后显示感谢信息
在检测到错误的字段旁边显示错误消息
以上所有内容必须显示在同一页面中。

我的代码是：

<?php
    $errors = array();
    if (isset($_POST["Ask_the_Question"])) {
        $guest_name = $_POST["guest_name"];
        $title = $_POST["faq_title"];
        $description = $_POST["faq_desc"];
        $title = $_POST["faq_title"];

        /* validation */
        if (empty($guest_name)) {
            $errors['guest_name'] = "Please type your name!";
        }   

        if(!empty($errors)){ echo '<h1 style="color: #ff0000;">Errors!</h1><h6 style="color: #ff0000;">Please check the fields which have errors below. Error hints are in Red.</h6>';}

        if(empty($errors)){     
            echo 'Thanks, We have received your feed back';
        }
    }

    else {
?>
            <form action="index.php" method="post" class="booking_reference">
                <input type="text" name="guest_name" placeholder="Your Name" value="<?PHP if(!empty($errors)) { echo $guest_name;} ?>" />
                <?php if(isset($errors['guest_name'])) { echo '<span style="color: red">'.$errors['guest_name'].'</span>'; } ?>
                <input type="email" name="guest_email" placeholder="Your email" pattern="[a-z0-9._%+-]+@[a-z0-9.-]+\.[a-z]{2,4}$" required />
                <input type="text" name="faq_title" placeholder="FAQ Title"/>
                <input type="text" name="faq_desc" placeholder="FAQ Description"/>
                <input type="submit" name="Ask_the_Question" value="Ask the Question" />
            </form>
<?php
        }
?>

我已限制验证并仅显示此问题的第一部分。

当我提交此表单时，如果没有任何错误，我会收到消息谢谢，我们已收到您的反馈 这很好，可以按预期工作。

当存在错误/Guest name 字段为空时，我在表单提交期间收到消息错误！请检查以下有错误的字段。错误提示为红色。 也不错。

但是当我收到上述消息时，我的表单就消失了。为什么？我还想在字段旁边显示请输入您的姓名！。

【问题讨论】：

标签： php arrays forms validation

【解决方案1】：

只需删除else 条件，因为实际上如果设置了$_POST["Ask_the_Question"]，您的表单将不会显示

<?php
    $errors = array();
    if (isset($_POST["Ask_the_Question"])) {
        $guest_name = $_POST["guest_name"];
        $title = $_POST["faq_title"];
        $description = $_POST["faq_desc"];
        $title = $_POST["faq_title"];

        /* validation */
        if (empty($guest_name)) {
            $errors['guest_name'] = "Please type your name!";
        }   

        if(!empty($errors)){ echo '<h1 style="color: #ff0000;">Errors!</h1><h6 style="color: #ff0000;">Please check the fields which have errors below. Error hints are in Red.</h6>';}

        if(empty($errors)){     
            echo 'Thanks, We have received your feed back';
        }
    }
            <form action="index.php" method="post" class="booking_reference">
                <input type="text" name="guest_name" placeholder="Your Name" value="<?PHP if(!empty($errors)) { echo $guest_name;} ?>" />
                <?php if(isset($errors['guest_name'])) { echo '<span style="color: red">'.$errors['guest_name'].'</span>'; } ?>
                <input type="email" name="guest_email" placeholder="Your email" pattern="[a-z0-9._%+-]+@[a-z0-9.-]+\.[a-z]{2,4}$" required />
                <input type="text" name="faq_title" placeholder="FAQ Title"/>
                <input type="text" name="faq_desc" placeholder="FAQ Description"/>
                <input type="submit" name="Ask_the_Question" value="Ask the Question" />
            </form>

【讨论】：

【解决方案2】：

原因在这里：

<?php
if (isset($_POST["Ask_the_Question"])) {
        $guest_name = $_POST["guest_name"];
        $title = $_POST["faq_title"];
        $description = $_POST["faq_desc"];
        $title = $_POST["faq_title"];

        /* validation */
        if (empty($guest_name)) {
            $errors['guest_name'] = "Please type your name!";
        }   

        if(!empty($errors)){ echo '<h1 style="color: #ff0000;">Errors!</h1><h6 style="color: #ff0000;">Please check the fields which have errors below. Error hints are in Red.</h6>';}

        if(empty($errors)){     
            echo 'Thanks, We have received your feed back';
        }
    } else {
      // your form code will never be called if $_POST['Ask_the_Question'] is set

要做你想做的事，你可能想做这样的事情：

<?php
    $errors = array();
    if (isset($_POST["Ask_the_Question"])) {
        $guest_name = $_POST["guest_name"];
        $title = $_POST["faq_title"];
        $description = $_POST["faq_desc"];
        $title = $_POST["faq_title"];

        /* validation */
        if (empty($guest_name)) {
            $errors['guest_name'] = "Please type your name!";
        }   

        if(!empty($errors)){ echo '<h1 style="color: #ff0000;">Errors!</h1><h6 style="color: #ff0000;">Please check the fields which have errors below. Error hints are in Red.</h6>';}
     }

     if(empty($errors)){     
          echo 'Thanks, We have received your feed back';
      } else { ?>
            <form action="index.php" method="post" class="booking_reference">
                <input type="text" name="guest_name" placeholder="Your Name" value="<?PHP if(!empty($errors)) { echo $guest_name;} ?>" />
                <?php if(isset($errors['guest_name'])) { echo '<span style="color: red">'.$errors['guest_name'].'</span>'; } ?>
                <input type="email" name="guest_email" placeholder="Your email" pattern="[a-z0-9._%+-]+@[a-z0-9.-]+\.[a-z]{2,4}$" required />
                <input type="text" name="faq_title" placeholder="FAQ Title"/>
                <input type="text" name="faq_desc" placeholder="FAQ Description"/>
                <input type="submit" name="Ask_the_Question" value="Ask the Question" />
            </form>
           <?php

        }
    }

?>

【讨论】：

【解决方案3】：

试试下面的代码。我已经删除了 else 部分并使用 true/false 设置标志来检查是否有效。

<?php
    $errors = array();
    if (isset($_POST["Ask_the_Question"])) {
        $guest_name = $_POST["guest_name"];
        $title = $_POST["faq_title"];
        $description = $_POST["faq_desc"];
        $title = $_POST["faq_title"];

        /* validation */
        $chkValidate = "true";
        if (empty($guest_name)) {
            $errors['guest_name'] = "Please type your name!";
            $chkValidate = "false";
        }   

        if(!empty($errors)){ echo '<h1 style="color: #ff0000;">Errors!</h1><h6 style="color: #ff0000;">Please check the fields which have errors below. Error hints are in Red.</h6>';
            $chkValidate = "false";
        }

        if($chkValidate == "true"){     
            echo 'Thanks, We have received your feed back';
        }
    }
    ?>
            <form action="index.php" method="post" class="booking_reference">
                <input type="text" name="guest_name" placeholder="Your Name" value="<?php if(!empty($errors) && $chkValidate != "false") { echo $guest_name;} ?>" />
                <?php if(isset($errors['guest_name'])) { echo '<span style="color: red">'.$errors['guest_name'].'</span>'; } ?>
                <input type="email" name="guest_email" placeholder="Your email" pattern="[a-z0-9._%+-]+@[a-z0-9.-]+\.[a-z]{2,4}$" required  />
                <input type="text" name="faq_title" placeholder="FAQ Title"/>
                <input type="text" name="faq_desc" placeholder="FAQ Description"/>
                <input type="submit" name="Ask_the_Question" value="Ask the Question" />
            </form>
<?php

?>

【讨论】：

【解决方案4】：

其他答案很好，但只是为了澄清会发生什么。

但是当我收到上述消息时，我的表单就消失了。为什么？

您的表单消失了，因为如果您通过第一个 if，您将无法访问您的 else。

if (isset($_POST["Ask_the_Question"])) {
    ...
} else {
    xxx;
}

这意味着如果你想看到你的表单，你必须把它放在某个地方，它可以显示为elseif（有更多限制），或者是内部或外部。

if (isset($_POST["Ask_the_Question"]) && empty($errors)) {
    ...
} elseif (isset($_POST["Ask_the_Question"]) && !empty($errors)) {
    ...
} else {
    ...
}

我还想显示请输入您的姓名！在田野旁边。

要显示您可以使用的所有错误，例如。 foreach 任何你想展示的地方。

foreach ($errors as &$error) {
    echo "Error: $error<br />\n";
}

顺便提一下empty(); function。

【讨论】：