我遇到的问题是我的值没有通过我的PHP代码。我将php文件路径名放在下面表单的操作部分,并将验证Javascript代码附加到提交按钮。我在php中唯一可以访问的值是复选框和单选值。
* * * * * * * * * * HTML * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
<div id="right-cont">
<form name = "contact_Us" action="http://nova.umuc.edu/cgi-bin/cgiwrap/ct386a28/eContact.php" method = "post">
<div id="style2">
<p>Please enter contact information below:</p>
</div>
<div class="style7">
<label>First Name: </label>
<br /><input type="text" id="firstName" tabindex="1"
style="width: 176px" />
</div>
<div class="style7">
<label>Middle Name: </label>
<br /><input type="text" id ="middleName" tabindex="2"
style="width: 176px" />
</div>
<div class="style7">
<label>Last Name: </label>
<br /><input type="text" id ="lastName" tabindex="3"
style="width: 176px" />
</div>
<div class =" buttons">
<input type="submit" value="SUBMIT" onclick = "return validate()"/><input type="reset" value="CLEAR"/> <br />
</div>
</form>
</div>
* * * * * * * * * * * * * * * * * PHP代码 * * * * * * * * * * * * * * * * * * * *
<?= '<' . '?xml version="1.0" encoding="utf-8"?' . '>' ?>
<?php
$fName = $_POST["firstName"];
$lName = $_POST["lastName"];
$mName = $_POST["middleName"];
$email = $_POST["email"];
$phone = $_POST["phone_Num"];
$comment = $_POST["comment"];
$phone_Type = $_POST["phone"];
$specialty_Type = $_POST["specType"];
?>
<div id="right-cont">
<div style="style8">
<h2>The Below Information has been sent to Pierre Law, LLC:</h2>
</div>
<div class="style7">
<label>First Name: </label>
<?php echo $fName; ?>
</div>
<div class="style7">
<label>Middle Name: </label>
<?php echo $mName;?>
</div>
<div class="style7">
<label>Last Name: </label>
<?php echo $lName;?>
</div>
</div>
您的输入标签缺少name属性。请求数据以"NAME=VALUE"的形式发送。您只放置了元素的id。您可以在输入元素中使用与name相同的id属性值,这些值将在PHP代码中被接收
应该是这样的:
<div class="style7">
<label for="firstName">First Name: </label>
<input type="text" name="firstName" id="firstName" tabindex="1" />
</div>
输入宽度应该来自.style7 input{}
css规则和请,停止使用<br />
和
进行格式化,这就是css的作用。
注:CSS类的名称应该描述标签的内容("article"、"important"等)。form是一个字段列表