Question

我有一个表单，我正在尝试添加一些基本的安全性和验证：

我尝试创建$error变量，以便在字段留空时显示错误消息。

目前，如果我将名称字段和单选按钮留空，我只会看到与空单选按钮有关的错误消息，而不是空名称字段。

我试图让这个表单尽可能简单，但只是想添加一些基本的安全性和验证。

有人可以帮忙吗？

她是我的HTML表单：

<?php
if(@$_REQUEST['submit'] == '1') {
    include('assets/forms/rsvp.php');
}
?>

        <form action="?" method="post">
            <?php if(@$errors) :?>
                <p class="errors"><?php echo $errors; ?></p>
            <?php endif; ?>
            <input type="hidden" name="submit" value="1" />
            <div class="form-row">
                <div class="field-l">
                    <p>Name</p>
                </div>
                <div class="field-r">
                    <p>Attending?</p>
                </div>
            </div>
            <div class="form-row guest">
                <div class="field-l">
                    <input type="text" name="name[0]" id="name" value="" tabindex="1" />
                </div>
                <div class="field-r">
                    <input type="radio" name="coming[0]" id="coming-yes" class="coming-yes" value="Yes"><label for="coming-yes">Yes</label><input type="radio" name="coming[0]" id="coming-no" class="coming-no" value="No"><label for="coming-no">No</label>
                </div>
            </div>
            <a class="addguest" href="#">Add further guest</a>
            <div class="form-row">
                <button type="submit" id="rsvp-submit" tabindex="2">Submit RSVP</button>
            </div>
        </form>

这是我的表单流程代码：

<?php

//echo "<pre>" . print_r($_POST, true) . "</pre>"; die();

if (isset($_POST['name'])) {
    $name = strip_tags(trim($_POST['name']));
}
if (isset($_POST['coming'])) {
    $coming = strip_tags(trim($_POST['coming']));
}

$errors = "";
if(!@$_POST['name'])    { $errors .= "Please enter your name.<br/>\n"; }
if(!@$_POST['coming'])  { $errors .= "Please enter yes or no for attending.<br/>\n"; }

if(@$_POST['emailaddress'] != '')   { $spam = '1'; }

if (!$errors && @$spam != '1')
    {
        $to = "example@xyz.com";
        $subject = "Wedding RSVP";
        $headers = "From: noreply@adrianandemma.com";
        $body = "The following RSVP has been sent via the website.\n\n";
        for($i=0; $i < count($_POST['name']); $i++) {
            $body .= "
            Name ".($i+1)." : " . $_POST['name'][$i] . "\n
            Coming ".($i+1)." : " . $_POST['coming'][$i] ."\n\n";
        }
        $body .= "\n\nDate Received: " . date("j F Y, g:i a") . "\n";

        mail($to,$subject,$body,$headers);
    }

?>

Answer 1

使用empty()足以检查。

if(empty($_POST['field'])){
 //Your code//
}

为了更安全，请使用strlen()

if(strlen($_POST['field']) == 0){
 //Your code//
}

合并if(strlen($field) < 0 || empty($field)){//code}

Answer 2

您的问题是您正在发送这些POST变量as an array，但将它们视为字符串。您需要单独loop through each one：

<?php
$errors = "";
foreach ($_POST["name"] as $index=>$name) {
    if (empty($name)) {
        $errors .= "Name missing from entry $index.<br/>\n";
    }
}
foreach ($_POST["coming"] as $index=>$coming) {
    if (empty($coming)) {
        $errors .= "Please enter attendance for entry $index.<br/>\n";
    }
}

if ($errors === "") {
    $to = "example@xyz.com";
    $subject = "Wedding RSVP";
    $headers = "From: noreply@adrianandemma.com";
    $body = "The following RSVP has been sent via the website.\n\n";
    foreach($_POST["name"] as $i=>$name) {
        $coming = $_POST["coming"][$i];
        $num = $i + 1;
        $body .= "Name $num : $name\nComing $num : $coming\n\n";
    }
    $body .= "\n\nDate Received: " . date("j F Y, g:i a") . "\n";

    mail($to,$subject,$body,$headers);
}

Answer 3

if（strlen（$ _ POST ['name']）== 0）{

$ errors =“请输入您的姓名”;

}
// strlen（）== 0表示名称字段为空

无法添加一些简单的PHP表单验证来检查空表单字段？

3 个答案: