PHP - 插入 Xampp/Mysql
我正在用这样的插入 php 制作产品数据库
$save_sale = mysql_query("INSERT INTO sale(no_receipt,no_confirm,time,date,day,total) values
('$_POST[no_receipt]',
'$_POST[no_confirm]',
'$hour_now',
'$date_now',
'$day_now',
'$_POST[total]')")
or die(mysql_error());
$save_sale_detail = mysql_query("INSERT INTO sale_detail(no_receipt,id_product,quantity,price,subtotal) values
('$_POST[no_receipt]',
'$_POST[id_product]',
'$_POST[quantity]',
'$_POST[price]',
'$_POST[subtotal]')")
or die(mysql_error());
但每次我提交不同数量、价格和小计的多个产品订单时,它只会保存一笔交易。我的意思是在销售储蓄中创建 no_receipt,然后sale_detail在销售之后创建 no_receipt 。示例:销售
No. Receipt = 001
No. Confirm = 205850 <-- it just confirmation order
hour = 17.00
date = 31/08/2013
day = saturday
total = $770 <-- subtotal calculation
sale_detail
No. Receipt = 001
id_product = 1 <-- let say it's a hat
quantity = 3
price = $150 <-- price by each product
subtotal = $450 <-- price * quantity
----- and come other product order ---
sale_detail
No. Receipt = 001
id_product = 4 <-- let's say it's a glasses
quantity = 4
price = $80 <-- price by each product
subtotal = $320 <-- price * quantity
----- and come other.. ---------------
但我卡住了,只能保存一笔交易,如果销售_detail成功保存,销售仍然是空的。
任何帮助将不胜感激..