我正在创建一个onlineshop，用户将在其中添加新记录，方法是在下拉列表中键入标题，说明，价格，图片上传和选择，其中包含要添加的产品的所有数据库表在特定的类别。
我目前使用多个表单添加新产品，所以我想知道如何使用所有表单并在用户输入标题，描述，价格（第一个表单），上传图片（第二个表单 - 使用通过单击提交按钮并收集上面列出的所有3种表格中的所有信息，选择类别，然后选择类别。
数据库的名称是'onlineshop'
的index.html
1st form(image uploading)
<script type="text/javascript" src="../cms/scripts/jquery.min.js"></script>
<script type="text/javascript" src="../cms/scripts/jquery.form.js"></script>
<script type="text/javascript" >
$(document).ready(function() {
$('#photoimg').live('change', function(){
$("#preview").html('');
$("#preview").html('<img src="loader.gif" alt="Uploading"/>');
$("#imageform").ajaxForm({
target: '#preview'
}).submit();
});
});
</script>
<style>
body
{
font-family:arial;
}
.preview
{
width:160px;
border:solid 2px #dedede;
padding:10px;
}
#preview
{
color:#cc0000;
font-size:10px
}
</style>
<body bgcolor="#ffdd55">
<font face=Arial size=3 color="#880088">
<form id="imageform" method="post" action='ajaximage.php' >
<small>Upload your image <input type="file" name="photoimg" id="photoimg" /></small>
</form>
<div id='preview'>
</div>
第二种形式（文本形式）
<form id="form1" action="insert.php" method="post" enctype="multipart/form-data">
<div>
<label for="title">Title: </label><input type="text" name="title"/>
</div>
<div>
<label for="description">Desc: </label><input type="text" name="description"/>
</div>
<div>
<label for="price">Price: </label><input type="text" name="price" />
</div>
<input type="submit" value="Submit">
</form>
insert.php
<?php session_start();
$mysqli = new mysqli('localhost', 'root', '',"onlineshop");
// Check connection
if ($mysqli->connect_error)
{
echo "Failed to connect to MySQL: " . $mysqli->error;
}
$validTableNames = array('my_table_1', 'my_table_2', 'another_table_3');
$tablename = isset($_POST['tablename']) ? $mysqli->real_escape_string($_POST['tablename']) : '';
$tableNameOk = in_array($tablename, $validTableNames);
if (!$tableNameOk) {
die('Error: Invalid table name:' . $tablename);
}
$title = !empty($_POST['title']) ? $mysqli->real_escape_string($_POST['title']) : null;
$description = !empty($_POST['description']) ? $mysqli->real_escape_string($_POST['description']) : null;
$price = !empty($_POST['price']) ? $mysqli->real_escape_string($_POST['price']) : null;
$sql = "INSERT INTO {$tablename} (title, description, price) VALUES (?, ?, ?)";
$stmt = $mysqli->prepare($sql);
$stmt->bind_param("sss", $title, $description, $price);
$allOk = $stmt->execute();
if (!$allOk)
{
die('Error: ' . $mysqli->error);
}
echo "1 record added";
$mysqli->close();
3rd form(dropdown list)
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST">
<select id = "form3" name="Tables" id="ddTables">
<?php
echo $tables;
?>
</select>
</form>
dropdown.php
<?php
$dbname = 'onlineshop';
?>
if (!mysql_connect('localhost', 'root', '')) {
echo 'Could not connect to mysql';
exit;
}
$sql = "SHOW TABLES FROM $dbname";
$result = mysql_query($sql);
if (!$result) {
echo "DB Error, could not list tables\n";
echo 'MySQL Error: ' . mysql_error();
exit;
}
$tables = '';
while ($row = mysql_fetch_row($result)) {
$tables .="<option value='$row[0]'>$row[0]</option>";
}
mysql_free_result($result);
?>
connect.php
<?php
// Try to connect to MySQL
$connect = mysql_connect('localhost','root', '') or die('Sorry could not connect to database');
// Check connect and return error if failed
$use_db = mysql_select_db('onlineshop');
$create_db = "CREATE DATABASE onlineshop";
if(!$use_db) {
echo mysql_error();
mysql_query($create_db);
mysql_select_db('onlineshop');
}
$con=mysqli_connect('localhost','root', '');
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
// Create database
$sql="CREATE DATABASE onlineshop";
if (mysqli_query($con,$sql))
{
echo "Database my_db created successfully";
}
else
{
echo "Error creating database: " . mysqli_error($con);
}
//main table
$sql = 'CREATE TABLE mens( '.
'id INT NOT NULL AUTO_INCREMENT, '.
'title VARCHAR(20) NOT NULL, '.
'description VARCHAR(45) NOT NULL, '.
'price FLOAT NOT NULL, '.
'image varchar(200),'.
'image_small varchar(200),'.
'primary key ( id ))';
//copy attributes of the main table
$sql2= 'CREATE TABLE women AS ( SELECT * FROM mens where 1=2)';
$sql3= 'CREATE TABLE kids AS ( SELECT * FROM mens where 1=2)';
$sql4= 'CREATE TABLE infants AS ( SELECT * FROM mens where 1=2)';
$sql5= 'CREATE TABLE baby_books AS ( SELECT * FROM mens where 1=2)';
$sql6= 'CREATE TABLE garden AS ( SELECT * FROM mens where 1=2)';
$sql7= 'CREATE TABLE comics AS ( SELECT * FROM mens where 1=2)';
$sql8= 'CREATE TABLE cooking AS ( SELECT * FROM mens where 1=2)';
$sql9= 'CREATE TABLE desktop AS ( SELECT * FROM mens where 1=2)';
$sql10= 'CREATE TABLE laptop AS ( SELECT * FROM mens where 1=2)';
$sql11= 'CREATE TABLE mobile AS ( SELECT * FROM mens where 1=2)';
$sql12= 'CREATE TABLE misc AS ( SELECT * FROM mens where 1=2)';
$sql13= 'CREATE TABLE moviestv AS ( SELECT * FROM mens where 1=2)';
$sql14= 'CREATE TABLE music AS ( SELECT * FROM mens where 1=2)';
$sql15= 'CREATE TABLE games AS ( SELECT * FROM mens where 1=2)';
$retval = mysql_query( $sql, $connect );
$retval2 = mysql_query($sql2, $connect);
$retval3 = mysql_query($sql3, $connect);
$retval4 = mysql_query($sql4, $connect);
$retval5 = mysql_query($sql5, $connect);
$retval6 = mysql_query($sql6, $connect);
$retval7 = mysql_query($sql7, $connect);
$retval8 = mysql_query($sql8, $connect);
$retval9 = mysql_query($sql9, $connect);
$retval10 = mysql_query($sql10, $connect);
$retval11 = mysql_query($sql11, $connect);
$retval12 = mysql_query($sql12, $connect);
$retval13 = mysql_query($sql13, $connect);
$retval14 = mysql_query($sql14, $connect);
$retval15 = mysql_query($sql15, $connect);
//this checks only for table1, check for all of them
if(! $retval)
{
die('Could not create table: ' . mysql_error());
}
echo "Tables created successfully\n";
?>
我是所有这些的新人，所以任何建议都会被接受。请详细解释我该怎么做。