Question

我目前正在通过twitter bootstrap将联系表单部署到网站，其中表单发布到发送电子邮件给管理员的php脚本。问题是如果文件js/plugins/validator/jquery.form-validator.min.js用于验证

，则表单不会提交

当我删除以下验证文件时，我可以让表单正常运行并成功提交电子邮件：

js/plugins/validator/jquery.form-validator.min.js

来自页面底部的脚本部分，但这会删除表单上的所有验证

<form  method="post" action="sendmail.php"  class="input-group input-group-lg" id="subscribeForm"   >
<input type="text class="form-control"  placeholder="Name"   id="name" name="name"  />
<input type="text" name="company" class="form-control" placeholder="company"   /><br />
<input type="text" name="phone"  class="form-control" placeholder="phone"  /><br />
<input type="email" class="form-control" placeholder="email" id="sEmail" name="email">
<span class="input-group-btn">
<input type="submit" button class="btn bg-primary">Invite Me!</button>
</span></form><!-- /input-group -->

sendmail.php的内容

<?php
$name = $_REQUEST['name'];
$company = $_REQUEST['company'];
$email = $_REQUEST['email'] ;
$phone = $_REQUEST['phone'];

$emess = "Name: ".$name."\n";
$emess.= "Company: ".$company."\n";
$emess.= "Email : ".$email."\n";
$emess.= "Phone number: ".$phone."\n";

mail( "send-to-email", "Information Request",
$emess, "From: $email" );
header( "Location: http://www.google.com" );?>

Answer 1

在您的代码中

<input type="text class="form-control"  placeholder="Name"   id="name" name="name"  />
                 ^ right there

您在文字后缺少引用"。扔进去，它应该工作。

在php方面需要注意的另一件事是，使用$name = $_POST['name'];代替，因为您通过帖子发送表单，默认情况下$_REQUEST包含$_GET的内容，{{1 }和$_POST。

编辑：正如Fred在输入按钮上指出的那样，它应该是

$_COOKIE

Answer 2

更改了输入提交标记

<form  method="post" action="sendmail.php"  class="input-group input-group-lg" id="subscribeForm"   >
<input type="text" class="form-control"  placeholder="Name"   id="name" name="name"  />
<input type="text" name="company" class="form-control" placeholder="company"   /><br />
<input type="text" name="phone"  class="form-control" placeholder="phone"  /><br />
<input type="email" class="form-control" placeholder="email" id="sEmail" name="email">
<span class="input-group-btn">
<input type="submit" value="Invite Me!" />
</span></form>

无法发布到PHP脚本

2 个答案: