Question

我有这个查询并且收到错误显示＆＃34;您的SQL语法中有错误;查看与您的MySQL服务器版本相对应的手册，以便在＆＃39; FROM Sales JOIN上使用正确的语法。销售ON Inventory.SaleID = Sales.ID JOIN Sales ON Sales.custo＆＃39;在第1行

这是我的查询。我知道我没有使用别名，所以它看起来很丑，但语法似乎是正确的，我传入的变量不是null我已经回复了它们。

 $sql = "UPDATE Sales SET salequantity='$takeCount' 
       FROM `Sales` 
       JOIN Sales ON Inventory.saleID=Sales.ID 
       JOIN Customers ON Sales.customerID=Customers.ID 
       JOIN Products ON Inventory.productID = Products.ID 
      JOIN Categories ON Products.categoryID=Categories.ID 
      WHERE Customers.ID='$customersID' AND Products.ID='$id'";

这是我的架构

CREATE TABLE Categories
(
ID int NOT NULL auto_increment,
type varchar(255),
description varchar(255),
PRIMARY KEY (ID)
);

CREATE TABLE Products
(
ID int NOT NULL auto_increment,
categoryID int,
customerID int,
name varchar(255),
description varchar(255),
price float(10),
quantity int(10),
PRIMARY KEY (ID),
FOREIGN KEY (categoryID) REFERENCES Categories(ID)
);
CREATE TABLE Customers
(
ID int NOT NULL auto_increment,
productID int,
name varchar(255),
street varchar(255),
city varchar(255),
state varchar(255),
zip int(8),
username varchar(255),
password varchar(255),
confirm_password varchar(255),
phone varchar(255),
PRIMARY KEY (ID),
);

CREATE TABLE Sales
(
ID int NOT NULL auto_increment,
customerID int,
amount float(15),
salequantity int(10),
PRIMARY KEY (ID),
FOREIGN KEY (customerID) REFERENCES Customers(ID)
);

CREATE TABLE Inventory
(
productID int,
saleID int,
FOREIGN KEY (productID) REFERENCES Products(ID),
FOREIGN KEY (saleID) REFERENCES Sales(ID)
);

这是最终查询。只想要这个简单的工作。

UPDATE Sales  INNER JOIN Sales ON Inventory.saleID=Sales.ID SET salequantity=2 where Sales.customerID=1 AND Inventory.productID=1;

从第一条评论中查询并失败。

UPDATE Sales SET salequantity=5 FROM Sales JOIN Inventory ON Inventory.saleID=Sales.ID JOIN Customers ON Sales.customerID=Customers.ID JOIN Products ON Inventory.productID = Products.ID JOIN Categories ON Products.categoryID=Categories.ID where Customers.ID=1 AND Products.ID=1;

解聚合函数完美无缺

$customerProducts = mysql_query("select Products.ID, Products.name, Products.description, Products.quantity,Products.price, Categories.type, Sales.ID, Sales.customerID, Sales.amount, Sales.salequantity, Inventory.productID, Inventory.saleID, Customers.ID, Customers.productID, COUNT(Products.name) AS productsCount from(((( Inventory INNER JOIN Sales ON Inventory.saleID=Sales.ID) JOIN Customers ON Sales.customerID=Customers.ID) JOIN Products ON Inventory.productID = Products.ID) JOIN Categories ON Products.categoryID=Categories.ID) where Customers.ID ='$customersID' GROUP BY Products.name");

Answer 1

<强>模式

create table sales (id varchar(10), customerid varchar(10), salequantity varchar(10))
create table Inventory (id varchar(10), productID varchar(10), saleID varchar(10))
create table Customers (id varchar(10))
create table Products (id varchar(10), categoryID varchar(10))
create table Categories (id varchar(10))

更正了查询

UPDATE Sales SET salequantity='$takeCount' 
  FROM Sales JOIN Inventory ON Inventory.saleID=Sales.ID
             JOIN Customers ON Sales.customerID=Customers.ID
             JOIN Products ON Inventory.productID = Products.ID
             JOIN Categories ON Products.categoryID=Categories.ID
  where Customers.ID='$customersID' AND Products.ID='$id'

<强>输出

（0行（s）受影响）

更改为查询

在第一次加入时，实际查询类似于“FROM Sales JOIN Sales ON Inventory.saleID = Sales.ID”，将其更正为“FROM Sales JOIN 库存 ON Inventory.saleID = Sales.ID“。目前尚不清楚您的所需输出到底是什么。但上面的查询将解决编译错误。

Answer 2

$customerProducts = mysql_query("select Products.ID, Products.name, Products.description, Products.quantity,Products.price, Categories.type, Sales.ID, Sales.customerID, Sales.amount, Sales.salequantity, Inventory.productID, Inventory.saleID, Customers.ID, Customers.productID, COUNT(Products.name) AS productsCount from(((( Inventory INNER JOIN Sales ON Inventory.saleID=Sales.ID) JOIN Customers ON Sales.customerID=Customers.ID) JOIN Products ON Inventory.productID = Products.ID) JOIN Categories ON Products.categoryID=Categories.ID) where Customers.ID ='$customersID' GROUP BY Products.name");

这完全符合我的要求。我试图让它复杂化并忘记聚合。

带连接的SQL更新失败

2 个答案: