Question

我正在构建一个php联系表单，用于检查是否填写了某些字段（没有ajax）。表格不适用于100％。我会尽力解释。当我提交表单时，用户导航到http://mydomain.com/send.php 收到我的错误消息（如果有的话）。当我刷新send.php页面时，我收到以下错误：注意：未定义的索引：/Library/WebServer/Documents/~aledvertising/send.php中的名称

这是我的代码

html表单

<form method="post" action="send.php">
    <div id="form-top">

    </div>
    <div id="form-left">  
<label>Naam:<span class="star">*</span></label>
    <input name="name" placeholder="Uw naam">


    <label>Email:<span class="star">*</span></label>
    <input name="email" type="email" placeholder="Uw email">

    <label>Hoeveel is 2+2? (Anti-spam)<span class="star">*</span></label>
<input name="human" placeholder="Uw antwoord">

    </div>
    <div id="form-right">
    <label>Uw bericht:<span class="star">*</span></label>
    <textarea name="message" placeholder="Uw bericht"></textarea>
    </div>
        <input id="submit" name="submit" type="submit" value="Verzenden">
</form>

SEND.PHP

<?php
    $name = $_POST['name'];
    $email = $_POST['email'];
    $message = $_POST['message'];
    $from = 'From: ' . $_POST['email']; 
    $to  = 'myemail@mail.com' . ', '; 
    $to .= $_POST['email'];
    $subject = 'Uw vraag op www.aledvertising.be';
    $human = $_POST['human'];

    $body = "From: $name\n E-Mail: $email\nMessage:\n $message";
 ?>
 <?php
if ($_POST['submit']) {
    if ($name != '' && $email != '' && $message != '') {
        if ($human == '4') {                 
            if (mail ($to, $subject, $body, $from)) { 
            echo '<p class="green">Uw bericht is succesvol verzonden.</p>';
        } else { 
            echo '<p>Er iets misgelopen, probeer opnieuw aub.</p>'; 
        } 
    } else if ($_POST['submit'] && $human != '4') {
        echo '<p>2+2 is niet gelijk aan het getal dat u hebt ingevoerd.</p>';
    }
    } else {
        echo '<p>Alle velden met een * zijn verplicht in te vullen.</p>';
    }
}
    ?>

有人可以帮助我获得完整的工作表格吗？谢谢

Answer 1

你应该从php看到isset方法。

<?php
if(isset($_POST["name"]))
{
//code here

}
?>

问题在于，当您重新加载页面时，服务器端代码会查找索引name并且找不到它，因为name还没有任何值。提交页面后会收到该值。这就是错误存在的原因。

<强>更新

错误是当你提交表单时，有些字段可能会被发送为空，所以一旦post值变为send.php，就会有$ _POST预期值。如果他们中的一些人没有回复确实是值，它会告诉你：“嘿，我没有价值观，所以我无效”。这就是为什么你必须检查字段是否已填满或在服务器端进行验证的原因。

<?php
if(isset($_POST["name"],$_POST["last_name"],etc,etc))
{
//if everything ok, ill go on.
}
else{
//if there are empty field, go back
}

?>

更新2

if (isset($_POST['name'],$_POST['email'],$_POST['message'])){

    $name = $_POST['name'];
    $email = $_POST['email'];
    $message = $_POST['message'];
    $from = 'From: ' . $_POST['email']; 
    $to  = 'myemail@mail.com' . ', '; 
    $to .= $_POST['email'];
    $subject = 'Uw vraag op www.aledvertising.be';
    $human = $_POST['human'];

    $body = "From: $name\n E-Mail: $email\nMessage:\n $message";
    if ($name != '' && $email != '' && $message != '') {
        if ($human == '4') {                 
            if (mail ($to, $subject, $body, $from)) { 
            echo '<p class="green">Uw bericht is succesvol verzonden.</p>';
        } else { 
            echo '<p>Er iets misgelopen, probeer opnieuw aub.</p>'; 
        } 
    } else if ($_POST['submit'] && $human != '4') {
        echo '<p>2+2 is niet gelijk aan het getal dat u hebt ingevoerd.</p>';
    }
    } else {
        echo '<p>Alle velden met een * zijn verplicht in te vullen.</p>';
    }
}

Answer 2

<?php
if (isset($_POST['submit'])) {
    $name = $_POST['name'];
    $email = $_POST['email'];
    $message = $_POST['message'];
    $from = 'From: ' . $_POST['email']; 
    $to  = 'myemail@mail.com' . ', '; 
    $to .= $_POST['email'];
    $subject = 'Uw vraag op www.aledvertising.be';
    $human = $_POST['human'];

    $body = "From: $name\n E-Mail: $email\nMessage:\n $message";
    if ($name != '' && $email != '' && $message != '') {
        if ($human == '4') {                 
            if (mail ($to, $subject, $body, $from)) { 
            echo '<p class="green">Uw bericht is succesvol verzonden.</p>';
        } else { 
            echo '<p>Er iets misgelopen, probeer opnieuw aub.</p>'; 
        } 
    } else if ($_POST['submit'] && $human != '4') {
        echo '<p>2+2 is niet gelijk aan het getal dat u hebt ingevoerd.</p>';
    }
    } else {
        echo '<p>Alle velden met een * zijn verplicht in te vullen.</p>';
    }
}
?>

设置isset（）函数，然后声明变量适用于我，在我的服务器中测试它，没有给出任何错误，表单已成功提交...

请检查一次......

Php联系表单错误

2 个答案: