How to reduce SQL queries in only one in this case - mysql

I want to save the hassle of doing many querys for the following:
I have a table like this:
name, age
{
Mike, 7
Peter, 2
Mario, 1
Tony, 4
Mary, 2
Tom, 7
Jerry, 3
Nick, 2
Albert, 22
Steven, 7
}
And I want the following result:
Results(custom_text, num)
{
1 Year, 1
2 Year, 3
3 Year, 1
4 Year, 1
5 Year, 0
6 Year, 0
7 Year, 3
8 Year, 0
9 Year, 0
10 Year, 0
More than 10 Year, 1
}
I know how to do this but in 11 queries :( But how to simplify it?
EDIT:
Doing the following, I can obtain the non zero values, but I need the zeroes in the right places.
SELECT COUNT(*) AS AgeCount
FROM mytable
GROUP BY Age
How can I achieve this?
Thanks for reading.

you can use below query but it will not show the gaps if you want gaps then the use Linoff's answer:
select t.txt, count(t.age) from
(select
case
when age<11 then concat(age ,' year')
else 'more than 10'
end txt, age
from your_table)t
group by t.txt
order by 1
SQL FIDDLE DEMO

You can use left join and a subquery to get what you want:
select coalesce(concat(ages.n, ' year'), 'More than 10 year') as custom_text,
count(*)
from (select 1 as n union all select 2 union all select 3 union all select 4 union all
select 5 union all select 6 union all select 7 union all select 8 union all
select 9 union all select 10 union all select null
) ages left join
tabla t
on (t.age = ages.n or ages.n is null and t.age > 10)
group by ages.n;
EDIT:
I think the following is a better way to do this query:
select (case when least(age, 11) = 11 then 'More than 10 year'
else concat(age, ' year')
end) as agegroup, count(name)
from (select 1 as age, NULL as name union all
select 2, NULL union all
select 3, NULL union all
select 4, NULL union all
select 5, NULL union all
select 6, NULL union all
select 7, NULL union all
select 8, NULL union all
select 9, NULL union all
select 10, NULL union all
select 11, NULL
union all
select age, name
from tabla t
) t
group by least(age, 11);
Basically, the query need a full outer join and MySQL does not provide one. However, we can get the same result by adding in extra values for each age, so we know something is there. Then because name is NULL, the count(name) will return 0 for those rows.

Please try using this query for required output.
SQL FIDDLE link http://www.sqlfiddle.com/#!9/4e52a/6
select coalesce(concat(ages.n, ' year'), 'More than 10 year') as custom_text,
count(t.age) from (select 1 as n union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9 union all select 10 union all select null ) ages left join tabla t
on (case when ages.n<11 then t.age = ages.n else t.age > 10 end)
group by ages.n;

Related

Laravel Query Builder raw query with two FROM statements

my apologies if this question sounds a little boring for someone I really turn here because I have already tried to solve my problem in different ways and I have not been able to obtain good results.
I am trying to convert a MySQL query to Laravel Query Builder, I show you the query:
SELECT DAY(AAA.fecha_hora_entrada) as DAY, IFNULL(BBB.SALES, 0) SALES, IFNULL(BBB.NET, 0) NET
FROM (
SELECT fecha_hora_entrada
FROM (
SELECT MAKEDATE(YEAR(NOW()), 1) +
INTERVAL (MONTH(NOW()) - 1) MONTH +
INTERVAL daynum DAY fecha_hora_entrada
FROM (
SELECT t * 10 + u daynum
FROM (SELECT 0 t UNION SELECT 1 UNION SELECT 2 UNION SELECT 3) A,
(SELECT 0 u UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9) B
ORDER BY daynum
) AS AA
) AS AA
WHERE MONTH(fecha_hora_entrada) = MONTH(NOW())
) AS AAA
LEFT JOIN (SELECT (DATE(fecha_hora_entrada)) AS fecha_hora_entrada, (SUM(neto)) AS NET, (COUNT(neto)) AS SALES
FROM t_derivados
GROUP BY DATE(fecha_hora_entrada)) BBB
ON AAA.fecha_hora_entrada = BBB.fecha_hora_entrada;
What I have tried so far is the following. So far I have managed to pass part of the MySQL query to Query Builder
public function salesYear()
{
return DB::table(DB::table(DB::table(DB::table(DB::table(null, null)
->selectRaw('0 t UNION SELECT 2 UNION SELECT 3'),'A')
->orderBy('daynum')
->selectRaw('t * 10 + u daynum'), 'AA')
->selectRaw('MAKEDATE(YEAR(NOW()), 1) + INTERVAL (MONTH(NOW()) - 1) MONTH + INTERVAL daynum DAY fecha_hora_entrada'), 'AA')
->addSelect('fecha_hora_entrada')
->whereRaw('MONTH(fecha_hora_entrada) = MONTH(NOW())'), 'AAA')
->selectSub('DAY(AAA.fecha_hora_entrada)', 'DAY')
->selectSub('IFNULL(BBB.SALES, 0)', 'SALES')
->selectSub('IFNULL(BBB.NET, 0)', 'NET')
->leftJoinSub(DB::table('t_derivados')
->selectSub('DATE(fecha_hora_entrada)', 'fecha_hora_entrada')
->selectSub('SUM(neto)', 'NET')
->selectSub('COUNT(neto)', 'SALES')
->groupBy(DB::raw('DATE(fecha_hora_entrada)')), 'BBB', 'AAA.fecha_hora_entrada', 'BBB.fecha_hora_entrada')
->toSql();
}
I know it's a bit boring to read everything .. but let's focus on the next sub-sentence:
SELECT t * 10 + u daynum
FROM (SELECT 0 t UNION SELECT 1 UNION SELECT 2 UNION SELECT 3) A,
(SELECT 0 u UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9) B
ORDER BY daynum
...
The truth is I have not found a way to add the second FROM to Query Builder
SELECT 0 u UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9) B
...
Someone expert on the subject could help me please. Thank you very much in advance

Get number of monday in a rangedate mysql

Get number of monday in a rangedate MySQL, I run this code but it give me result 0:
select count(*) from tarif where weekday(`end_tarif`<= '2019-02-21'AND `start_tarif`>='2019-02-05') = 0;
my table:
CREATE TABLE `tarif` (
`tarif_id` int(11) NOT NULL AUTO_INCREMENT,
`start_tarif` date NOT NULL,
`end_tarif` date NOT NULL,
`day_tarif` varchar(50) NOT NULL,
PRIMARY KEY (`tarif_id`)
);
INSERT INTO `tarif` VALUES (1, '2019-02-01', '2019-02-10', '10'),
(2, '2019-02-11', '2019-02-20', '20'),
(3, '2019-02-21', '2019-02-28', '10'),
(4, '2019-03-01', '2019-02-10', '15');
You can use a solution using a calendar table. So you can use a solution like the following:
1. create a table with calendar data
-- create the table "calendar"
CREATE TABLE `calendar` (
`dateValue` DATE
);
-- insert the days to the table "calendar"
INSERT INTO calendar
SELECT adddate('1970-01-01',t4*10000 + t3*1000 + t2*100 + t1*10 + t0) gen_date from
(select 0 t0 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t0,
(select 0 t1 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t1,
(select 0 t2 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t2,
(select 0 t3 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t3,
(select 0 t4 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t4
HAVING gen_date BETWEEN '2019-01-01' AND '2019-12-31'
You can find the script to generate the calendar data on StackOverflow:
How to populate a table with a range of dates?
2. create the table with your data (with monday tarif)
-- create the table "tarif"
CREATE TABLE tarif (
tarif_id INT(11) NOT NULL AUTO_INCREMENT,
start_tarif DATE NOT NULL,
end_tarif DATE NOT NULL,
day_tarif VARCHAR(50) NOT NULL,
monday_tarif VARCHAR(50) NOT NULL,
PRIMARY KEY (tarif_id)
);
-- insert the tarif information
INSERT INTO tarif VALUES
(1, '2019-02-01', '2019-02-10', '10', '5'),
(2, '2019-02-11', '2019-02-20', '20', '5'),
(3, '2019-02-21', '2019-02-28', '10', '5'),
(4, '2019-03-01', '2019-02-10', '15', '5');
Note: To create a useful example I added the column monday_tarif and insert the value 5 on every date range.
3. get the result
Now you can get all days of your needed range (between 2019-02-05 and 2019-02-21) from the calendar table. With a LEFT JOIN you add your tarif table to all days of date range.
With a CASE WHEN and the condition DAYOFWEEK = 2 or DAYNAME = 'Monday' you can check if the current date is a Monday or not, to get the correct tarif value of the day.
SELECT SUM(CASE WHEN DAYOFWEEK(cal.dateValue) = 2 THEN tarif.monday_tarif ELSE tarif.day_tarif END) AS sumWithMondayTarif
FROM calendar cal
LEFT JOIN tarif ON cal.dateValue BETWEEN start_tarif AND end_tarif
WHERE cal.dateValue BETWEEN '2019-02-05' AND '2019-02-21';
You can also use a SELECT with a sub select of the calendar:
SELECT SUM(CASE WHEN DAYOFWEEK(cal.dateValue) = 2 THEN tarif.monday_tarif ELSE tarif.day_tarif END) AS sumWithMondayTarif FROM (
SELECT adddate('1970-01-01',t4*10000 + t3*1000 + t2*100 + t1*10 + t0) dateValue FROM
(select 0 t0 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t0,
(select 0 t1 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t1,
(select 0 t2 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t2,
(select 0 t3 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t3,
(select 0 t4 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t4
HAVING dateValue BETWEEN '2019-02-05' AND '2019-02-21'
) cal LEFT JOIN tarif ON cal.dateValue BETWEEN start_tarif AND end_tarif
demo on dbfiddle.uk
Below mentioned query is for sundays count. you can modify it as per your requirement
select ROUND((
(unix_timestamp(`end_tarif`) - unix_timestamp(`start_tarif`) )/(24*60*60)
-7+WEEKDAY(`start_tarif`)-WEEKDAY(`end_tarif`)
)/7)
+ if(WEEKDAY(`start_tarif`) <= 6, 1, 0)
+ if(WEEKDAY(`end_tarif`) >= 6, 1, 0) as Sunday
from tarif
where `end_tarif`<= '2019-02-21' AND `start_tarif`>='2019-02-05' ;

Insert where in join, code confused

This code returns me the most used words in a column in TEXT format called description and is on the table 'messages`.
However I can not stick this in the WHERE code:
messages.tag = 'HELLO'
I need it to do what it already does, but with this WHERE I tried this code and gave not sure:
SELECT message, count(message) as count
FROM (
SELECT
messages.id,
SUBSTRING_INDEX(SUBSTRING_INDEX(messages.message, ' ', count.n), ' ', -1) as message
FROM
(select (h*100+t*10+u+1) n from
(select 0 h union select 1 union select 2 union select 3 union select 4 union
select 5 union select 6 union select 7 union select 8 union select 9) A,
(select 0 t union select 1 union select 2 union select 3 union select 4 union
select 5 union select 6 union select 7 union select 8 union select 9) B,
(select 0 u union select 1 union select 2 union select 3 union select 4 union
select 5 union select 6 union select 7 union select 8 union select 9) C
) as count
INNER JOIN messages
ON CHAR_LENGTH(messages.message)-CHAR_LENGTH(REPLACE(messages.message, ' ',''))>=count.n-1
ORDER BY id, n
) x
WHERE LENGTH(message) >= 5
AND messages.tag = 'HELLO'
GROUP BY message
ORDER BY count DESC
LIMIT 10
You have to move that where clause into the subquery. The messages alias is not known in the outer query:
FROM (SELECT messages.id,
SUBSTRING_INDEX(SUBSTRING_INDEX(messages.message, ' ', count.n), ' ', -1) as message
FROM (select (h*100+t*10+u+1) n
from (select 0 h union select 1 union select 2 union select 3 union select 4 union
select 5 union select 6 union select 7 union select 8 union select 9) A,
(select 0 t union select 1 union select 2 union select 3 union select 4 union
select 5 union select 6 union select 7 union select 8 union select 9) B,
(select 0 u union select 1 union select 2 union select 3 union select 4 union
select 5 union select 6 union select 7 union select 8 union select 9) C
) count INNER JOIN
messages
ON CHAR_LENGTH(messages.message)-CHAR_LENGTH(REPLACE(messages.message, ' ','')) >= count.n-1
WHERE messages.tag = 'HELLO'
ORDER BY id, n
) x
The other condition stays in the outer query.
Try this,
I think messages.tag should be like, not equal to 'HELLO'
SELECT message, count(message) as count
FROM (
SELECT
messages.id,
SUBSTRING_INDEX(SUBSTRING_INDEX(messages.message, ' ', count.n), ' ', -1) as message
FROM
(
select (h*100+t*10+u+1) n
from
(
select 0 h union select 1 union select 2 union select 3 union select 4
union
select 5 union select 6 union select 7 union select 8 union select 9) A,
(select 0 t union select 1 union select 2 union select 3 union select 4
union
select 5 union select 6 union select 7 union select 8 union select 9) B,
(select 0 u union select 1 union select 2 union select 3 union select 4
union
select 5 union select 6 union select 7 union select 8 union select 9) C
) count
INNER JOIN messages
ON CHAR_LENGTH(messages.message)-CHAR_LENGTH(REPLACE(messages.message, ' ',''))>=count.n-1
WHERE messages.tag like '%HELLO%'
ORDER BY id, n
) x
WHERE LENGTH(message) >= 5
GROUP BY message
ORDER BY count DESC
LIMIT 10

Mysql Query to find distinct characters in a string

Is there a way to write recursive query in mysql. Equivalent of connect by (level or Prior) in oracle. I searched google as well as stackoverflow and there is no direct eqivalent. But is there any work around to get it.
I have a string, i have to iterate through individual characters in the string and print only the distinct characters of the string.
Input:
recursive
Output:
recusiv
I came up with this "simple" solution.
Functions used:
GROUP_CONCAT: http://dev.mysql.com/doc/refman/5.0/en/group-by-functions.html#function_group-concat
SUBSTRING http://dev.mysql.com/doc/refman/5.0/en/string-functions.html#function_substring
Live demo: http://www.sqlfiddle.com/#!2/d41d8/19186
SELECT
GROUP_CONCAT( chars.c SEPARATOR '') AS allchars
FROM (
SELECT DISTINCT
SUBSTRING( str.str, pos.pos, 1 ) AS c
FROM
( SELECT x1.x + x2.x*10 AS pos
FROM
( SELECT 0 AS x UNION ALL
SELECT 1 UNION ALL
SELECT 2 UNION ALL
SELECT 3 UNION ALL
SELECT 4 UNION ALL
SELECT 5 UNION ALL
SELECT 6 UNION ALL
SELECT 7 UNION ALL
SELECT 8 UNION ALL
SELECT 9
) AS x1
INNER JOIN
( SELECT 0 AS x UNION ALL
SELECT 1 UNION ALL
SELECT 2 UNION ALL
SELECT 3 UNION ALL
SELECT 4 UNION ALL
SELECT 5 UNION ALL
SELECT 6 UNION ALL
SELECT 7 UNION ALL
SELECT 8 UNION ALL
SELECT 9
) AS x2
) AS pos
INNER JOIN
( SELECT 'recursive' AS str UNION ALL
SELECT 'XYZ'
) AS str
) AS chars

SELECT de-normalized columns into separate records?

I am playing around with SQL a little just so I am not completely ignorant about it if I am ever asked in a job interview. My friend was recently asked the following question at an interview and he couldn't get it and I asked somebody at work who knows SQL decently and he didn't know. Can you guys answer this problem for me and then explain how it works? Please?
*The problem*
Database normalization (or lack of normalization) often presents a challenge for developers.
Consider a database table of employees that contains three fields:
EmployeeID
EmployeeName
EmailAddresses
Every employee, identified by a unique EmployeeID, may have one or more comma-separated, #rockauto.com email address(es) in the EmailAddresses field.
The database table is defined below:
CREATE TABLE Employees
(
EmployeeID int UNSIGNED NOT NULL PRIMARY KEY,
EmployeeName varchar(50) NOT NULL,
EmailAddresses varchar(40) NOT NULL ,
PRIMARY KEY(EmployeeID)
);
For testing purposes, here is some sample data:
INSERT INTO Employees (EmployeeID, EmployeeName, EmailAddresses) VALUES
('1', 'Bill', 'bill#companyx.com'),
('2', 'Fred', 'fred#companyx.com,freddie#companyx.com'),
('3', 'Fred', 'fredsmith#companyx.com'),
('4', 'Joe', 'joe#companyx.com,joe_smith#companyx.com');
Your task is to write a single MySQL SELECT query that will show the following output for the sample data above:
Employee EmailAddress
Bill bill#companyx.com
Fred (2) fred#companyx.com
Fred (2) freddie#companyx.com
Fred (3) fredsmith#companyx.com
Joe joe#companyx.com
Joe joe_smith#companyx.com
Please take note that because there is more than one person with the same name (in this case, "Fred"), the EmployeeID is included in parenthesis.
Your query is required to written in MySQL version 5.1.41 compatible syntax. You should assume that the ordering is accomplished using standard database ascending ordering: "ORDER BY EmployeeID ASC"
For this problem, you need to submit a single SQL SELECT query. Your query should be able to process a table of 1000 records in a reasonable amount of time.
only if you have less than 10000 emails.... is that acceptable?
select
if(t1.c > 1, concat(e.employeename, ' (', e.employeeid, ')'), e.employeename) as Employee,
replace(substring(substring_index(e.EmailAddresses, ',', n.row), length(substring_index(e.EmailAddresses, ',', n.row - 1)) + 1), ',', '') EmailAddress
from
(select employeename, count(*) as c from Employees group by employeename) as t1,
(select EmployeeID, length(EmailAddresses) - length(replace(EmailAddresses,',','')) + 1 as emails from Employees) as t2,
(SELECT #row := #row + 1 as row FROM
(select 0 union all select 1 union all select 3 union all select 4 union all select 5 union all select 6 union all select 6 union all select 7 union all select 8 union all select 9) x,
(select 0 union all select 1 union all select 3 union all select 4 union all select 5 union all select 6 union all select 6 union all select 7 union all select 8 union all select 9) x2,
(select 0 union all select 1 union all select 3 union all select 4 union all select 5 union all select 6 union all select 6 union all select 7 union all select 8 union all select 9) x3,
(select 0 union all select 1 union all select 3 union all select 4 union all select 5 union all select 6 union all select 6 union all select 7 union all select 8 union all select 9) x4,
(SELECT #row:=0) as ff) as n,
Employees e
where
e.employeename = t1.employeename and
e.employeeid = t2.employeeid and
n.row <= t2.emails
order by e.employeeid;
EDIT:
With less useless numbers generated:
select
if(t1.c > 1, concat(e.EmployeeName, ' (', e.EmployeeID, ')'), e.EmployeeName) as Employee,
replace(substring(substring_index(e.EmailAddresses, ',', n.row), length(substring_index(e.EmailAddresses, ',', n.row - 1)) + 1), ',', '') as EmailAddress
from
(select EmployeeName, count(*) as c from Employees group by EmployeeName) as t1,
(select EmployeeID, length(EmailAddresses) - length(replace(EmailAddresses,',','')) + 1 as emails from Employees) as t2,
(select `1` as row from (select 1 union all select 2 union all select 3 union all select 4) x) as n,
Employees e
where
e.EmployeeName = t1.EmployeeName and
e.EmployeeID = t2.EmployeeID and
n.row <= t2.emails
order by e.EmployeeID;
And what did we learn? Poor database design results awful queries. And you can do stuff with SQL, that are probably supported only because people do poor database designs... :)