我遇到的問題是我的值沒有通過我的PHP代碼。我將php文件路徑名放在下面表單的action部分,並將驗證的Javascript代碼附加到submit按鈕。我在php中訪問的唯一值是複選框和廣播值。訪問收音機和複選框值以外的任何值
** * ** * ** * *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>
** * ** * ** * ** * ** * **個PHP代碼* ** * ** * ** * ** * ** * ** * *
<?= '<' . '?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>
Dooh!坦克星期一:) – Mike 2011-04-30 13:22:39