dqpfl2508589 2015-09-19 03:02
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php表单不发布电子邮件和textarea数据

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>
  • 写回答

1条回答 默认 最新

  • dongwei7913 2015-09-19 03:06
    关注

    replace

    id="SharingEmail name="SharingEmail"
    

    with

    id="SharingEmail" name="SharingEmail"
    

    and

    id="SharingShared_items name="SharingShared_Items"
    

    with

    id="SharingShared_items" name="SharingShared_Items"
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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