i have this query and i want to concat the names of the student. Where should I put the concat statement?
"concat(text,LPAD(id,4,'0'))"
The text and id here is from students table. this is the query:
"SELECT p.*, s.* FROM students s, payments p
where s.id=p.id and level='Grade 3' and amount>='1500'"
Table
-students table-
create table students(
text char(5)NOT NULL,
id int(11)NOT NULL AUTO_INCREMENT,
name varchar(250),
address varchar(250)
PRIMARY KEY(id)
)
-payments-
create table payments(
p_id int(11)NOT NULL AUTO_INCREMENT,
amount varchar(250),
id int,
PRIMARY KEY(p_id)
FOREIGN KEY(id) REFERENCES students(id);
)
Thank you!
Try this :
SELECT p.*, s.*, concat(s.text,LPAD(s.id,4,'0')) as student_names
FROM students s, payments p
where s.id=p.id and level='Grade 3' and amount>='1500'
Related
How can I combine these queries into a single query with where clause from another parent table? Please consider my SQL code and suggest a better method to work with
//look my code
CREATE TABLE IF NOT EXISTS first (
fid int(11) NOT NULL PRIMARY KEY AUTO_INCREMENT,
p_name varchar(60) NOT NULL
);
CREATE TABLE IF NOT EXISTS second (
sed int(11) NOT NULL PRIMARY KEY AUTO_INCREMENT,
firstname varchar(20) NOT NULL,
fid int(11) NOT NULL,
FOREIGN KEY (fid) REFERENCES first(fid)
);
CREATE TABLE IF NOT EXISTS third (
thid int(11) NOT NULL PRIMARY KEY AUTO_INCREMENT,
start_date date NOT NULL,
end_date date NOT NULL,
sed int(11) NOT NULL,
FOREIGN KEY (sed) REFERENCES second(sed),
fid int(11) NOT NULL,
FOREIGN KEY (fid) REFERENCES first(fid)
);
CREATE TABLE IF NOT EXISTS fourth (
fid int(11) NOT NULL PRIMARY KEY AUTO_INCREMENT,
start_date date NOT NULL,
end_date date NOT NULL,
sed int(11) NOT NULL,
FOREIGN KEY (sed) REFERENCES second(sed),
fid int(11) NOT NULL,
FOREIGN KEY (fid) REFERENCES first(fid)
);
CREATE TABLE IF NOT EXISTS fifth (
fiid int(11) NOT NULL PRIMARY KEY AUTO_INCREMENT,
start_date date NOT NULL,
end_date date NOT NULL,
sed int(11) NOT NULL,
FOREIGN KEY (sed) REFERENCES second(sed),
fid int(11) NOT NULL,
FOREIGN KEY (fid) REFERENCES first(fid)
);
CREATE TABLE IF NOT EXISTS sixth (
sid int(11) NOT NULL PRIMARY KEY AUTO_INCREMENT,
start_date date NOT NULL,
end_date date NOT NULL,
sed int(11) NOT NULL,
FOREIGN KEY (sed) REFERENCES second(sed),
fid int(11) NOT NULL,
FOREIGN KEY (fid) REFERENCES first(fid)
);
//As you can see above, I want to create a single query to query all data at the samee time i.e
//All table from third table depend on first and second table, but the second table have column firstname and the first table has the p_name column
//I want
SELECT second.*, third.* FROM second INNER JOIN third ON third.sed = second.sed
SELECT second.*, fourth.* FROM second INNER JOIN fourth ON fourth.sed = second.sed
SELECT second.*, fifth.* FROM second INNER JOIN fifth ON fifth.sed = second.sed
SELECT second.*, sixth.* FROM second INNER JOIN sixth ON sixth.sed = second.sed
....WHERE fid = 1;
I want to combine these queries into a single query ie, $newqueries = '.....';
The concept
The second table is used to carry all details, ie student details, but the third to sixth tables are tables with few different details but they took all other details from the second table, ie a student can be a chairman, secretary and vice secretary but not all students so that I classified them in third to sixth table. The first table used to keep few info about i.e classes so I want to differentiate chairman etc base on class tables but all of them are students
In short
A chairman, secretary and vice secretary are students but not all students have these role in a class but we have more than one classes, how to differentiate these leaders based on class
in a single query
You can use left join
SELECT second.*, third.*,fourth.*,fifth.*,sixth.* FROM second
LEFT JOIN third ON third.sed = second.sed
LEFT JOIN fourth ON fourth.sed = second.sed
LEFT JOIN fifth ON fifth.sed = second.sed
LEFT JOIN sixth ON sixth.sed = second.sed
WHERE second.fid = 1;
I assume that if student is chairman then there will be an entry of that student in third table. Above query will return null if the student is normal student. You can use CASE Statement if you want role as well. For example,
CASE WHEN third.startdate IS NULL THEN '' ELSE 'Chairman' END
these are the tables
and it results
I have two values on the first row of the table it_courses field course_branch. When i join branch table with it_courses table result is like this.it doesnt shows the second value in the 1st row that was entered in the it_coursestable
Query :
SELECT branch.branch_name
FROM `branch`
JOIN `it_courses` ON it_courses.course_branch = branch.branch_id
You can try this database structure and query :
Table:
tbl_it_courses (id - PK, cource_name, branch_id - FK)
tbl_branch (id - PK, branch_id, branch_name)
Table : tbl_it_courses
CREATE TABLE tbl_it_courses (
id INT NOT NULL AUTO_INCREMENT,
cource_name VARCHAR(70) NOT NULL,
branch_id VARCHAR(70) DEFAULT NULL,
PRIMARY KEY(id)
);
INSERT INTO `tbl_it_courses` (`Id`, `cource_name`, `branch_id`) VALUES (NULL, 'PHP', '1,4'),(NULL, '.NET', '3'), (NULL, 'CCNA', '3');
Table : tbl_branch
CREATE TABLE tbl_branch (
id INT NOT NULL AUTO_INCREMENT,
branch_id VARCHAR(70) DEFAULT NULL,
branch_name VARCHAR(70) NOT NULL,
PRIMARY KEY(id)
);
INSERT INTO `tbl_branch` (`Id`, `branch_id`, `branch_name`) VALUES (NULL, 'Ern', 'brn92b224'),(NULL, 'Klm', 'brnaf3650'), (NULL, 'Tvm', 'brn272493');
Query :
SELECT tbl_it_courses.cource_name, (SELECT GROUP_CONCAT(tbl_branch.branch_name) FROM tbl_branch WHERE FIND_IN_SET(tbl_branch.id,tbl_it_courses.branch_id) > 0) as branch_name
FROM tbl_it_courses
I Hope It will Help you.
Try this.
SELECT branch.branch_id,it_courses.course_name
FROM `branch`
left JOIN `it_courses` ON LOCATE(branch.branch_name,it_courses.course_branch) > 0
I have two tables employees and departments.
Departments:
create table dept (
id INT NOT NULL AUTO_INCREMENT,
dept_name VARCHAR(50) NOT NULL,
PRIMARY KEY (id)
);
Employees:
create table department_master (
id INT NOT NULL AUTO_INCREMENT,
dept_id INT NOT NULL,
emp_name VARCHAR(100) NOT NULL,
PRIMARY KEY (id)
);
I want to prevent departments being deleted from the UI if they are assigned to one of the employees in employee table. Left join is giving me duplicate values.
How do I see if the departments are being used in the employees table.
If you want to prevent a department from being deleted, you can simply add a foreign constraint to the table department_master for dept_id column.
create table department_master (
id INT NOT NULL AUTO_INCREMENT,
dept_id INT NOT NULL,
emp_name VARCHAR(100) NOT NULL,
PRIMARY KEY (id),
constraint con_dm foreign key dept_id references dept( id )
);
It's default behavior is ON DELETE RESTRICT which means that if there is atleast one row present in the department_master for a given dept_id, it can't be deleted from dept table.
If you want to fetch, the department that don't have any employee record, you can use NOT EXISTS:
select *
from dept d
where not exists (
select 1
from department_master m
where d.id = m.dept_id
);
I believe you want a count of the number of employees grouped by the department, like so:
SELECT count(*) as employees, d.dept_name FROM dept AS d LEFT JOIN department_master AS e ON e.dept_id = d.id group by e.dept_id
I have a table that contains millions of sales records and looks like this:
CREATE TABLE `sales` (
`dollar_amount` INT NULL,
`transaction_date` DATE NULL,
`company_name` VARCHAR(45) NULL,
`company_id` INT NULL);
The first three columns are populated with data. I would like to insert data into the company_id column that will identify each company with an auto_incremented integer. I plan to use the company_id field as a foreign key referencing another table that will contain each company's details. Many companies have multiple transactions, so the code needs to assign the same company_id to each row in the sales table with a matching company_name.
Is there a way to do this using only MySQL?
First, I'd recommend creating the company table:
CREATE TABLE company (
company_id INT NOT NULL AUTO_INCREMENT,
company_name VARCHAR(45),
PRIMARY KEY(company_id));
Then insert the companies from your sales data:
INSERT INTO company (company_name)
SELECT distinct company_name
FROM sales;
Finally, update your sales table with a join to get the company_id:
UPDATE sales s
JOIN company c ON s.company_name = c.company_name
SET s.company_id = c.company_id;
SQL Fiddle Demo
You should also remove the company_name field from the sales table since this is now stored in the company table.
To define an auto incremented integer, you just use the AUTO_INCREMENT keyword. However, if you define any columns as auto_increment, you must also make that column your primary key. Which, in this case, would make sense in order for it to be a foreign key elsewhere.
Try this:
CREATE TABLE `sales` (
`dollar_amount` INT NULL,
`transaction_date` DATE NULL,
`company_name` VARCHAR(45) NULL,
`company_id` INT NOT NULL AUTO_INCREMENT,
PRIMARY KEY(company_id);
SQL Fiddle
I have three tables "user" , "bidding" and "item".
I need to find the query in order to get the completed item auctions for a buyer. The way how to find this in my database is the following, item.received=1 AND u.userid=X (this X will be filled in from my PHP which gives the userID of the highest bid). (note that received=1 implies that the deadline is over so this check is not necessary anymore).
Short explanation of the system: it is an auction website, where a user places bids on items and on the users personal account page I want to show the amount of auctions which he bought (and are processed, thus completed).
The 3 tables look like this:
CREATE TABLE user (
userid INT NOT NULL AUTO_INCREMENT,
username CHAR(30) NOT NULL UNIQUE,
password CHAR(32) NOT NULL,
firstname CHAR(30) NOT NULL,
lastname CHAR(30) NOT NULL,
gender CHAR(1) NOT NULL,
email CHAR(50) NOT NULL UNIQUE,
birthdate DATE NOT NULL,
addressid INT NOT NULL,
picture CHAR(50),
lastlogin TIMESTAMP NOT NULL,
role CHAR(30),
paymentid INT NOT NULL,
PRIMARY KEY (userid),
FOREIGN KEY (addressid)
REFERENCES address(addressid),
FOREIGN KEY (paymentid)
REFERENCES payment(paymentid)
);
CREATE TABLE item (
itemid INT NOT NULL AUTO_INCREMENT,
name CHAR(40) NOT NULL,
description CHAR(255) NOT NULL,
originalpurchasedate DATE,
deadline TIMESTAMP NOT NULL,
minprice DOUBLE,
received BOOLEAN NOT NULL,
dateadded TIMESTAMP NOT NULL,
openbidding BOOLEAN NOT NULL,
categoryid INT NOT NULL,
ownerid INT NOT NULL,
PRIMARY KEY (itemid),
FOREIGN KEY (categoryid)
REFERENCES category(categoryid),
FOREIGN KEY (ownerid)
REFERENCES user(userid)
);
CREATE TABLE bidding (
userid INT NOT NULL,
itemid INT NOT NULL,
amount DOUBLE,
bidtime TIMESTAMP NOT NULL,
FOREIGN KEY (userid)
REFERENCES user(userid),
FOREIGN KEY (itemid)
REFERENCES item(itemid)
);
The malfunctioning solution I have already is: the result is 3 rows and results being: 3 , 1 , 5. The solution I expect to get only has to be 1 row, containing the number of distinct items.
SELECT DISTINCT COUNT(u.userid) FROM `item` i
INNER JOIN `bidding` b ON i.itemid = b.itemid
INNER JOIN `user` u ON b.userid = u.userid
WHERE i.received=1 AND u.userid=2
GROUP BY i.itemid
You need to change your query to group on userid instead of item id, and count different items instead of different users.
SELECT DISTINCT COUNT(i.itemid) FROM `item` i
INNER JOIN `bidding` b ON i.itemid = b.itemid
INNER JOIN `user` u ON b.userid = u.userid
WHERE i.received=1 AND u.userid=2
GROUP BY u.userid