以尽可能最佳的规范化方式将可变数量的数据输入数据库
好的,所以我有一个包含两个表、产品和供应商的数据库。
所有供应商都填写一个表格,然后他们的数据存储在供应商表中,而产品表包含所有产品的列表,因此当供应商填写表格时,他可以像我一样选择任意数量的产品使用 jQuery JSON 和 AJAX 获取所有产品的列表,然后填充包含所有产品的下拉列表,然后可以根据需要多次克隆该列表。
我现在面临的问题是,如何将供应商选择的所有不同产品插入供应商表中,或者我应该将他选择的所有产品与一个供应商相关联,以便更好地规范化,因为所有产品都是已经在那里了?
我将使用 jQuery $.ajax 将 JSON 格式的表单数据 POST 到等待的 PHP 文件,然后该文件将解析它并将数据插入数据库。
所以基本上,我需要弄清楚如何关联数据库中的数据以实现尽可能最佳的标准化,并且我需要找出一种将可变数量的产品插入供应商表中的方法,或者找到一种方法来关联许多产品他选择的产品交给了一家供应商。
我对关系数据库非常陌生,所以任何有关如何继续的建议都会有很大的帮助,你们可能有任何其他建议!
我用来填充克隆并发布供应商选择的产品的 jQuery 代码:
$(document).ready(function() {
var count = 0;
//when clicked it will remove the closest div with a class of 'container'
$("span.remove").live('click', function(){
$(this).closest("div.container").fadeOut(400, function(){
$(this).remove();
$('#button').attr('disabled','');
});
});
//initialize the button
$('#button').attr('disabled','');
$('#button').click(function(){
var count = $("#systems_wrapper > .container").size();
var lastID = $("#systems_wrapper > .container:last").attr('id');
var exploded = lastID.split("_");
var increment = Number(exploded[1])+1;
//if the user has selected 5 products, disable the 'add' button
if(count >= 5){
$('#button').attr('disabled','disabled');
}else {
$('#button').attr('disabled','');
}
//clone the first drop down and give it a different ID, as well as it's child elements
var test = $('#systems_0.container').clone().attr('id', 'system_' + increment).appendTo('#systems_wrapper');
test.children(':nth-child(2)').append('<span class="remove"></span>');
test.children(':nth-child(2)').children(':first').attr('id', 'mail_' + increment).attr('class','dropDowns').attr('onchange','test();');
});
//get the products JSON object returned from test_post.php and run the necessary functions on the returned data
$.getJSON("test_post.php", function(data){
//clean out the select list
$('#box').html('');
//run the loop to populate the drop down list
$.each(data, function(i, products) {
$('#box').append(
$('<option></option>').html(products.products)
);
});
});
});
//this gets all of the products chosen and then gets each ones value and ID, and then posts it to the qwer.php file
function test(){
var sections = $('#systems_wrapper').find('.dropDowns');
var newArray = new Array();
sections.each(function(){
var id = $(this).attr('id');
var val = $(this).val();
var o = { 'id': id, 'value': val };
newArray.push(o);
});
alert(newArray);
$.ajax({
type: "POST",
url: "qwer.php",
dataType: 'json',
data: { json: JSON.stringify(newArray) }
});
}
提前感谢!
ok, so I have a database comprising of two tables, products and suppliers.
All suppliers fill in a form and their data is then stored in the suppliers table, and the products table contains a list of all of the products, so when the supplier fills in the form, he can choose as many products as he wishes as I use jQuery JSON and AJAX to get the list of all of the products and then populate a drop down list with all of them in it, which can then be cloned as many times as is needed.
The problem I am sitting with now is, how do I insert all of the different products the supplier chooses into the supplier table, or should I rather just relate all of the products he chooses to the one supplier for better normalization since all the products are already there?
I will be using jQuery $.ajax to POST the form data in JSON format to a waiting PHP file, which will then parse it and insert the data into the database.
So basically, I need to figure out how to relate the data in the database to achieve the best normalization possible, and I need to figure out a way of inserting a variable amount of products into the suppliers table or find a way to relate the many products he chooses to the one supplier.
I am very new to relational databases, so any advice on how to proceed would be a great help, so would any other advice you guys may have!
The jQuery code I use to populate clone and POST the products the supplier chooses:
$(document).ready(function() {
var count = 0;
//when clicked it will remove the closest div with a class of 'container'
$("span.remove").live('click', function(){
$(this).closest("div.container").fadeOut(400, function(){
$(this).remove();
$('#button').attr('disabled','');
});
});
//initialize the button
$('#button').attr('disabled','');
$('#button').click(function(){
var count = $("#systems_wrapper > .container").size();
var lastID = $("#systems_wrapper > .container:last").attr('id');
var exploded = lastID.split("_");
var increment = Number(exploded[1])+1;
//if the user has selected 5 products, disable the 'add' button
if(count >= 5){
$('#button').attr('disabled','disabled');
}else {
$('#button').attr('disabled','');
}
//clone the first drop down and give it a different ID, as well as it's child elements
var test = $('#systems_0.container').clone().attr('id', 'system_' + increment).appendTo('#systems_wrapper');
test.children(':nth-child(2)').append('<span class="remove"></span>');
test.children(':nth-child(2)').children(':first').attr('id', 'mail_' + increment).attr('class','dropDowns').attr('onchange','test();');
});
//get the products JSON object returned from test_post.php and run the necessary functions on the returned data
$.getJSON("test_post.php", function(data){
//clean out the select list
$('#box').html('');
//run the loop to populate the drop down list
$.each(data, function(i, products) {
$('#box').append(
$('<option></option>').html(products.products)
);
});
});
});
//this gets all of the products chosen and then gets each ones value and ID, and then posts it to the qwer.php file
function test(){
var sections = $('#systems_wrapper').find('.dropDowns');
var newArray = new Array();
sections.each(function(){
var id = $(this).attr('id');
var val = $(this).val();
var o = { 'id': id, 'value': val };
newArray.push(o);
});
alert(newArray);
$.ajax({
type: "POST",
url: "qwer.php",
dataType: 'json',
data: { json: JSON.stringify(newArray) }
});
}
Thanx in advance!
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如果我从数据库级别正确理解了问题,您是否应该使用名为 ProductSupplier 的中间表,其中包含 Product_ID 和 Supply_ID 列。
然后,当供应商选择产品时,将供应商和产品 ID 添加到此表中的新列中。
这将允许多个供应商挑选相同的产品,并且允许同一供应商挑选多种产品。
编辑:我的意思是“将供应商和产品 ID 添加到此表中的新行”
If i understand the problem correctly from a database level, should you be using an intermediate table called something like ProductSupplier containing a Product_ID and Supplier_ID column.
Then when a supplier selects a product, add both the supplier and product id to a new column in this table.
This will allow multiple suppliers to pick the same product and multiple products to be picked by the same supplier.
EDIT: I meant to say "add both the supplier and product id to a new ROW in this table"