php表单不发布电子邮件和textarea数据

Question

The form below is collecting the following information from users: - Name - Email Address - Type of item being shared - The number of items - List of items being shared

The problem is that the email address and list of items being shared are not getting stored to the database. All other fields are being stored properly.

<form name="form1" method="post" action="feedback.php" class="pure-form">
    <fieldset class="pure-group">
        <input type="text" class="pure-input-1-2" placeholder="Sharetype" value="Nativity" id="SharingSharetype" name="SharingSharetype" readonly>
        <input type="text" class="pure-input-1-2" placeholder="Name" id="SharingName" name="SharingName">
        <input type="email" class="pure-input-1-2" placeholder="Email" id="SharingEmail name="SharingEmail">
    </fieldset><br>
    <fieldset class="pure-group">
        <input type="text" pattern="\d*" class="pure-input-1-2" placeholder="Number of Items" id="SharingNum" name="SharingNum">
        <textarea class="pure-input-1-2" placeholder="List items you want to share" id="SharingShared_items name="SharingShared_Items"></textarea>
    </fieldset>
    <button type="submit" class="pure-button pure-input-1-2 pure-button-primary">Submit</button>
</form>

<?php

$host="localhost"; // Host name 
$username="root"; // Mysql username 
$password=""; // Mysql password 
$db_name="db"; // Database name 
$tbl_name="table"; // Table name 

// Connect to server and select database.
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");

// Get values from form 
    $Name = mysql_escape_string($_POST['SharingName']);
    $Email = mysql_escape_string($_POST['SharingEmail']);
    $Sharetype = mysql_escape_string($_POST['SharingSharetype']);
    $Shared_Items = mysql_escape_string($_POST['SharingItems']);
    $Num = $_POST[SharingNum];

// Insert data into mysql 
$sql="INSERT INTO $tbl_name(Name, Email, Sharetype, Shareditems, Num, ShareDate)VALUES('$Name', '$Email', '$Sharetype', '$Shared_Items', '$Num', Now())";
$result=mysql_query($sql);

// if successfully insert data into database, displays message "Successful". 
if($result){
mysql_close();
header( 'Location: http://www.googe.com/' );
exit;
}
else {
mysql_close();
header( 'Location: http://www.google.com/' );
exit;
}
?> 

<?php 
// close connection 
mysql_close();
?>

</div>

Answer 1

replace

id="SharingEmail name="SharingEmail"

with

id="SharingEmail" name="SharingEmail"

and

id="SharingShared_items name="SharingShared_Items"

with

id="SharingShared_items" name="SharingShared_Items"