Php 使用多个表单添加记录以在数据库表上创建一条记录
我正在创建一个onlineshop，用户将在其中添加一个新记录，方法是键入标题、描述、价格、图像上载以及从下拉列表中选择，该列表包含要添加到特定类别中的产品的所有数据库表 我目前使用多个表单添加新产品，因此我想知道如何使用所有表单，在用户输入标题、说明、价格（第一个表单）、上传图片（第二个表单——使用php文件）和选择类别之后，点击提交按钮，收集上述三种表格中的所有信息 数据库的名称为“onlineshop” index.htmlPhp 使用多个表单添加记录以在数据库表上创建一条记录,php,jquery,mysql,forms,Php,Jquery,Mysql,Forms,我正在创建一个onlineshop，用户将在其中添加一个新记录，方法是键入标题、描述、价格、图像上载以及从下拉列表中选择，该列表包含要添加到特定类别中的产品的所有数据库表 我目前使用多个表单添加新产品，因此我想知道如何使用所有表单，在用户输入标题、说明、价格（第一个表单）、上传图片（第二个表单——使用php文件）和选择类别之后，点击提交按钮，收集上述三种表格中的所有信息 数据库的名称为“onlineshop” index.html 1st form(image uploading) &
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>
1表格（图像上传）
$（文档）.ready（函数（）{
$（'#photoimg'）.live（'change'，function（）{
$（“#预览”）.html（“”）；
$（“#预览”）.html（“”）；
$（“#imageform”）.ajaxForm({
目标：“#预览”
}).submit（）；
});
});
身体
{
字体系列：arial；
}
预览
{
宽度：160px；
边框：实心2px#dedede；
填充：10px；
}
#预演
{
颜色：#cc0000；
字体大小：10px
}
上传你的图片
标题：
描述：
价格：
insert.php
如果（！mysql_connect（'localhost'，'root'，''）{
echo“无法连接到mysql”；
出口
}
$sql=“显示来自$dbname的表”；
$result=mysql\u查询（$sql）；
如果（！$result）{
echo“数据库错误，无法列出表\n”；
echo'MySQL Error:'.MySQL_Error（）；
出口
}
$tables=''；
while（$row=mysql\u fetch\u row（$result））{
$tables.=“$row[0]”；
}
mysql_free_result（$result）；
?>
<?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(! $retval)
{
die('Could not create table: ' . mysql_error());
}
echo "Tables created successfully\n";
?>
我在所有这些方面都是新手，所以任何建议都将被欣然接受。请详细解释我该怎么做。@Allende我可以看到您编辑了我的源代码，但请添加一条关于如何编辑我的代码以修复的新注释。很抱歉，我是StackOverflow的新手，如果您足够友好，请帮助我@cmario该版本旨在提高可读性，使用4个空格标识，不打算修复它：如果你还没有读过，请注意：这可能会鼓励其他人阅读你的代码并回答。回到你的问题，你会看到这样的问题：如果你想一次提交所有信息，为什么不只使用一个表单？（至少您希望同时保留函数和/或使用AJAX）。如果您提供创建表的sql，也会对其他人有所帮助。我现在更新了问题，以显示用于创建表的sql语句（connect.php）。您能告诉我如何使用AJAX来完成我在上述问题中提出的任务吗？@cmario，您可以始终使用$\u SESSION全局变量来保存页面中的数据。
<?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);
?>
<?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";
?>