我想要的是当用户点击一个链接，它应该自动创建两个文本框的时间，从我们可以点击和创建无限数量的文本框，当提交它应该保存所有动态创建的文本框两个文本框在一行。
意思是textboxA textboxB
以这种方式......
我在网上找到了一个代码，它的工作原理与我想要的非常相似…但是当点击链接时，它每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) index . 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();
}
}
?>
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
有了这段代码，我只允许动态创建一个文本框，当点击链接，我之前说过，所以为了使它为我使用，因为我想有两个文本框的我已经编辑了jquery部分在index.php页面如下…
<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…但主要的问题是我不能保存这两个文本框的一行在mysql表..
请查看此代码并回复我，如果你得到任何答案.....
我肯定会点击绿色箭头，将工作答案作为接受的答案..
请帮助大家......
试试这个
jQuery<script type="text/javascript">
var count = 0;
$(function(){
$('p#add_field').click(function(){
count += 1;
$('#container').append(
'<strong>Link #' + count + '</strong><br />'
+ '<label for="field_'+count+'_1">Name</label><input id="field_'+count+'_1" name="fields[]['name']" type="text" /><label for="field2_'+count+'_2">URL</label><input id="field2_'+count+'_2" name="fields[]['url']" type="text" /><br />');
});
});
</script>
//Insert into websites table
$sql_website = sprintf("INSERT INTO websites (Website_Name,Website_URL) VALUES ('%s','%s')",
mysql_real_escape_string($value['name']),
mysql_real_escape_string($value['url']) );
$result_website = $db->query($sql_website);
$inserted_website_id = $db->last_insert_id();
我假设第一列是Website_Name第二列是Website_URL
注::您已经说过它创建了两个文本框，这意味着应该有两个表字段，您希望在其中添加这两个值。但是在你的MySQL查询中，只有一个列插入。
"INSERT INTO websites (Website_URL) VALUES ('%s')"
为正确回答问题指定第二列名称