无法使用php提交表单内部javascript popover

Question

I am trying to submit a bootstrap popover form with php, however it is not populating the table. any suggestions?

javascript:

<script type="text/javascript">
$(document).ready(function() {
$("[rel='android']").popover({
html: 'true', 
content : '<div id="popOver-input"><p id="coming-soon">Coming March 2014</p><p>Be the first to know when buddyTruk goes live!</p><form class="form-horizontal" action="temp.php"  method="post"><input type="text" class="form-control input-lg" name="emailFuture"  placeholder="Email"/><button class="btn btn-success btn-lg" name="comingSoon" type="submit"   value="comingSoon">Submit</button></form></div>'
});
});

php:

<?php

if (isset($_POST['comingSoon']) && strlen($_POST['email'])>0)
{
$coming_soon=$_POST['emailFuture']; 
mysql_connect("*****", "****", "*****") or die(mysql_error()); 
mysql_select_db("buddyTruk") or die(mysql_error());
mysql_query("ALTER TABLE coming_soon ADD PRIMARY KEY (email)");
mysql_query("REPLACE INTO `coming_soon` VALUES ('$coming_soon')"); 

 } 


 ?> 

Answer 1

Because there is no input with name email but you trying : $_POST['email'], So, change the name:

<script type="text/javascript">
$(document).ready(function() {
$("[rel='android']").popover({
html: 'true', 
content : '<div id="popOver-input"><p id="coming-soon">Coming March 2014</p><p>Be the first to know when buddyTruk goes live!</p><form class="form-horizontal" action="temp.php"  method="post"><input type="text" class="form-control input-lg" name="email"  placeholder="Email"/><button class="btn btn-success btn-lg" name="comingSoon" type="submit"   value="comingSoon">Submit</button></form></div>'
});
});