5
나는 단일 주문에 관한 정보를 수집하기위한이 쿼리를 가지고 있으며 매우 복잡해졌습니다.MySQL InnerJoin에 제한이 있습니까?
데이터를 테스트 할 데이터가 없으므로 크고 작은 데이터 세트에서이 작업을 경험 한 사람이 하나의 쿼리에서 얼마나 많은 조인을 할 수 있는지 또는 제한해야하는지에 대한 제한이 있습니다. ? 대형 쿼리를 작은 부분으로 나누는 것이 좋을까요, 아니면 큰 차이를 만들지는 않습니까?
또한 각 INNER JOIN 다음에 WHERE 절을 포함해도됩니까?
귀하의 조언에 감사드립니다. 누군가가이 쿼리에 대한 스키마를 요청
# Order: Get Order
function getOrder($order_id) {
$sql = "SELECT (order.id, order.created, o_status.status,
/* payment info */
order.total, p_status.status,
/* ordered by */
cust_title.title, cust.forename, cust.surname,
customer.phone, customer.email,
cust.door_name, cust.street1,
cust.street2, cust.town,
cust.city, cust.postcode,
/* deliver to */
recip_title.title, recipient.forename, recipient.surname,
recipient.door_name, recipient.street1,
recipient.street2, recipient.town,
recipient.city, recipient.postcode,
/* deliver info */
shipping.name, order.memo,
/* meta data */
order.last_update)
FROM tbl_order AS order
INNER JOIN tbl_order_st AS o_status
ON order.order_status_id = o_status.id
INNER JOIN tbl_payment_st AS p_status
ON order.payment_status_id = p_status.id
INNER JOIN (SELECT (cust_title.title, cust.forename, cust.surname,
customer.phone, customer.email,
/* ordered by */ cust.door_name, cust.street1,
cust.street2, cust.town,
cust.city, cust.postcode)
FROM tbl_customer AS customer
INNER JOIN tbl_contact AS cust
ON customer.contact_id = cust.id
INNER JOIN tbl_contact_title AS cust_title
ON cust.contact_title_id = cust_title.id
WHERE order.customer_id = customer.id)
ON order.customer_id = customer.id
INNER JOIN (SELECT (recip_title.title, recipient.forename, recipient.surname,
/* deliver to */ recipient.door_name, recipient.street1,
recipient.street2, recipient.town,
recipient.city, recipient.postcode)
FROM tbl_contact AS recipient
INNER JOIN tbl_contact_title AS recip_title
ON recipient.contact_title_id = recip_title.id
WHERE order.contact_id = recipient.id)
ON order.contact_id = recipient.id
INNER JOIN tbl_shipping_opt AS shipping
ON order.shipping_option_id = shipping.id
WHERE order.id = '?';";
dbQuery($sql, array((int)$order_id));
$rows = dbRowsAffected();
if ($rows == 1)
return dbFetchAll();
else
return null;
}
때문에, 여기있다 :
# TBL_CONTACT_TITLE
DROP TABLE IF EXISTS tbl_contact_title;
CREATE TABLE tbl_contact_title(
id INT NOT NULL AUTO_INCREMENT,
PRIMARY KEY(id),
title CHAR(3)
) ENGINE = InnoDB;
INSERT INTO tbl_contact_title
(title)
VALUES ('MR'),
('MRS'),
('MS');
# TBL_CONTACT
DROP TABLE IF EXISTS tbl_contact;
CREATE TABLE tbl_contact(
id INT NOT NULL AUTO_INCREMENT,
PRIMARY KEY(id),
contact_title_id INT,
FOREIGN KEY(contact_title_id) REFERENCES tbl_contact_title(id) ON DELETE SET NULL,
forename VARCHAR(50),
surname VARCHAR(50),
door_name VARCHAR(25),
street1 VARCHAR(40),
street2 VARCHAR(40),
town VARCHAR(40),
city VARCHAR(40),
postcode VARCHAR(10),
currency_id INT,
FOREIGN KEY(currency_id) REFERENCES tbl_currency(id) ON DELETE SET NULL
) ENGINE = InnoDB;
# TBL_CUSTOMER
DROP TABLE IF EXISTS tbl_customer;
CREATE TABLE tbl_customer(
id INT NOT NULL AUTO_INCREMENT,
PRIMARY KEY(id),
contact_id INT,
FOREIGN KEY(contact_id) REFERENCES tbl_contact(id) ON DELETE SET NULL,
birthday DATE,
is_male TINYINT,
phone VARCHAR(20),
email VARCHAR(50) NOT NULL
) ENGINE = InnoDB, AUTO_INCREMENT = 1000;
# TBL_ORDER_ST
DROP TABLE IF EXISTS tbl_order_st;
CREATE TABLE tbl_order_st(
id INT NOT NULL AUTO_INCREMENT,
PRIMARY KEY(id),
status VARCHAR(25)
) ENGINE = InnoDB;
INSERT INTO tbl_order_st
(status)
VALUES
('NEW'),
('PROCESSING'),
('SHIPPED'),
('COMPLETED'),
('CANCELLED');
# TBL_SHIPPING_OPT
DROP TABLE IF EXISTS tbl_shipping_opt;
CREATE TABLE tbl_shipping_opt(
id INT NOT NULL AUTO_INCREMENT,
PRIMARY KEY(id),
name VARCHAR(50),
description VARCHAR(255),
cost DECIMAL(6,3)
) ENGINE = InnoDB;
INSERT INTO tbl_shipping_opt
(name, description, cost)
VALUES
('UK Premier', 'U.K. Mainland upto 30KG, Next Working Day', 8.00),
('Europe Standard', 'Most European Destinations* upto 30KG, 2 to 5 Working Days *please check before purchase', 15.00);
# TBL_PAYMENT_ST
DROP TABLE IF EXISTS tbl_payment_st;
CREATE TABLE tbl_payment_st(
id INT NOT NULL AUTO_INCREMENT,
PRIMARY KEY(id),
status VARCHAR(25)
) ENGINE = InnoDB;
INSERT INTO tbl_payment_st
(status)
VALUES
('UNPAID'),
('PAID');
# TBL_ORDER
DROP TABLE IF EXISTS tbl_order;
CREATE TABLE tbl_order(
id INT NOT NULL AUTO_INCREMENT,
PRIMARY KEY(id),
customer_id INT,
FOREIGN KEY(customer_id) REFERENCES tbl_customer(id) ON DELETE SET NULL,
contact_id INT,
FOREIGN KEY(contact_id) REFERENCES tbl_contact(id) ON DELETE SET NULL,
created DATETIME,
last_update TIMESTAMP,
memo VARCHAR(255),
order_status_id INT,
FOREIGN KEY(order_status_id) REFERENCES tbl_order_st(id),
shipping_option_id INT,
FOREIGN KEY(shipping_option_id) REFERENCES tbl_shipping_opt(id),
coupon_id INT,
FOREIGN KEY(coupon_id) REFERENCES tbl_coupon(id) ON DELETE SET NULL,
total DECIMAL(9,3),
payment_status_id INT,
FOREIGN KEY(payment_status_id) REFERENCES tbl_payment_st(id)
) ENGINE = InnoDB, AUTO_INCREMENT = 1000;
구문 오류나 "너무 많은 조인"이없는 한 계속하십시오. – hakre
@hakre 고마워 ... 그때, 쿼리의 크기는 중요하지 않습니다. – Ozzy