我想要的是当用户单击一个链接时，它应该一次自动创建两个文本框，我们可以从中单击并创建无限数量的文本框，当提交时它应该将所有动态创建的文本框保存在一个连续的两个文本框中。
含义 textboxA textboxB
以这种方式......
我在网上找到了一个代码，它的工作方式与我想要的非常相似......但是当点击链接时，它一次只创建一个文本框，而不是两个文本框，首先我会给你完整的原始代码......
1) 索引.php
<?php
//Include the database class
require("classes/db.class.php");
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<title>jQuery</title>
<script type="text/javascript" src="js/jquery.js"></script>
<link rel="stylesheet" type="text/css" href="css/css.css" />
<script type="text/javascript">
var count = 0;
$(function(){
$('p#add_field').click(function(){
count += 1;
$('#container').append(
'<strong>Link #' + count + '</strong><br />'
+ '<input id="field_' + count + '" name="fields[]' + '" type="text" /><br />' );
});
});
</script>
<body>
<?php
//If form was submitted
if (isset($_POST['btnSubmit'])) {
//create instance of database class
$db = new mysqldb();
$db->select_db();
//Insert static values into users table
$sql_user = sprintf("INSERT INTO users (Username, Password) VALUES ('%s','%s')",
mysql_real_escape_string($_POST['name']),
mysql_real_escape_string($_POST['password']) );
$result_user = $db->query($sql_user);
//Check if user has actually added additional fields to prevent a php error
if ($_POST['fields']) {
//get last inserted userid
$inserted_user_id = $db->last_insert_id();
//Loop through added fields
foreach ( $_POST['fields'] as $key=>$value ) {
//Insert into websites table
$sql_website = sprintf("INSERT INTO websites (Website_URL) VALUES ('%s')",
mysql_real_escape_string($value) );
$result_website = $db->query($sql_website);
$inserted_website_id = $db->last_insert_id();
//Insert into users_websites_link table
$sql_users_website = sprintf("INSERT INTO users_websites_link (UserID, WebsiteID) VALUES ('%s','%s')",
mysql_real_escape_string($inserted_user_id),
mysql_real_escape_string($inserted_website_id) );
$result_users_website = $db->query($sql_users_website);
}
} else {
//No additional fields added by user
}
echo "<h1>User Added, <strong>" . count($_POST['fields']) . "</strong> website(s) added for this user!</h1>";
//disconnect mysql connection
$db->kill();
}
?>
<?php if (!isset($_POST['btnSubmit'])) { ?>
<h1>New User Signup</h1>
<form name="test" method="post" action="">
<label for="name">Username:</label>
<input type="text" name="name" id="name" />
<div class="spacer"></div>
<label for="name">Password:</label>
<input type="text" name="password" id="password" />
<div class="spacer"></div>
<div id="container">
<p id="add_field"><a href="#"><span>» Add your favourite links.....</span></a></p>
</div>
<div class="spacer"></div>
<input id="go" name="btnSubmit" type="submit" value="Signup" class="btn" />
</form>
<?php } ?>
</body>
</html>
2) db.class.php
<?php
class mysqldb {
/*
FILL IN YOUR DATABASE DETAILS BEFORE RUNNING THE EXAMPLE
*/
var $hostname = "localhost";
var $username = "root";
var $password = "mypassword";
var $database = "unlimited";
function db_connect() {
$result = mysql_connect($this->hostname,$this->username,$this->password);
if (!$result) {
echo 'Connection to database server at: '.$this->hostname.' failed.';
return false;
}
return $result;
}
function select_db() {
$this->db_connect();
if (!mysql_select_db($this->database)) {
echo 'Selection of database: '.$this->database.' failed.';
return false;
}
}
function query($query) {
$result = mysql_query($query) or die("Query failed: $query<br><br>" . mysql_error());
return $result;
mysql_free_result($result);
}
function fetch_array($result) {
return mysql_fetch_array($result);
}
function num_rows($result) {
return mysql_num_rows($result);
}
function last_insert_id() {
return mysql_insert_id();
}
function kill() {
mysql_close();
}
}
?>
3）css.css
html, input {font-family: Verdana, Arial, Helvetica, sans-serif; font-size: 0.8em;}
body { width: 500px; margin: 50px auto 0 auto; display: block;}
h1 { font-size: 1.5em; color: #333; }
input { font-size: 0.9em; padding: 5px; border: 1px solid #ccc; margin: 0; display: block;}
a { text-decoration: none; color: #666; font-weight: bold; }
a:hover { color: #ff0000; }
#divTxt { width:400px; padding: 5px; }
p a img { border: none; vertical-align: middle; }
.spacer {clear: both; height: 10px; }
.btn { width: 90px; font-weight: bold; }
#container { border: 1px solid #ccc; padding: 2px; }
.clear {overflow: hidden;width: 100%;
}
4) JQUERY.js
使用此代码，我只允许在单击链接时动态创建一个文本框，正如我之前所说的，所以为了让它供我使用，因为我想要两个文本框，我在index.php页面中编辑了 jquery 部分，如下所示......
<script type="text/javascript">
var count = 0;
$(function(){
$('p#add_field').click(function(){
count += 1;
$('#container').append(
'<strong>Link #' + count + '</strong><br />'
+ '<label for="fields[]' + '">Colour</label><input id="field_' + count + '" name="fields[]' + '" type="text" /><label for="fields2[]' + '">Quantity</label><input id="field2_' + count + '" name="fields2[]' + '" type="text" /><br />');
});
});
</script>
直到这里我成功了......但主要问题是我无法将它们两个文本框都保存在 mysql 表中的一行中..
如果您得到任何答案，请查看此代码并回复我......
我一定会单击绿色箭头作为工作答案作为接受的答案..
请帮助伙计们......