I'm currently using 1-inf request in MySQL and I have a little problem right now.
Is it possible to join, concat, regroup some results who has the same column result in common ?
This are my ouptut
You have lessons (column one) who can be teach by teachers (column two) and the class (column three).
Is there a way to join them and have a result like
cours | professeurs1 professeurs2 professeurs3 | classe
Here is my query:
SELECT cours.cours, professeurs.nom, classe.classe
FROM cours
JOIN classe
ON cours.id_classe = classe.id
JOIN cours_professeurs
ON cours_professeurs.id_cours = cours.id
JOIN professeurs
ON cours_professeurs.id_prof = professeurs.id
Thanks in advance.
You can use GROUP_CONCAT and you will have to use group by with it too.
SELECT
cours.cours,
GROUP_CONCAT(professeurs.nom) AS teachers,
classe.classe
FROM cours
JOIN classe
ON cours.id_classe = classe.id
JOIN cours_professeurs
ON cours_professeurs.id_cours = cours.id
JOIN professeurs
ON cours_professeurs.id_prof = professeurs.id
GROUP BY cours.cours,classe.classe
Yes, you can do this using the GROUP BY() MySQL statement.
Related
I’ve a BD with the next tables.
TABLE detalle_contrato
TABLE detalle_tradicional
There is a relation with ID_CONTRATO and i need to view the table with the next data.
SELECT
ID_CONTRATO,
TRADICIONAL,
NOM_VARIEDAD,
SUM(CANTIDAD)
FROM detalle_contrato
WHERE ID_CONTRATO = '$ID' AND TIPO_VARIEDAD = 'TRADICIONAL';
SELECT
SUM(CANTIDAD_D)
FROM detalle_tradicional
WHERE ID_CONTRATO = '$ID'
GROUP BY NOM_VARIEDAD ";
There are a filter different in this two select and I need this in a table but i don't know together.
The idea is this:
ID_CONTRATO,
NOM_VARIEDAD,
CANTIDAD
( THIS IS THE SUM THE ALL CANTIDAD DUKE AND
LEGACY IN GROUP THE TABLE DETALLE_CONTRATO) ,
CANTIDAD_D
(TABLE DETALLE_TRADICIONAL THIS IS SUM
THE ALL DUKE AND LEGACY SEPARATE THE CANTIDAD_D
I need exactly this using the data the photos
You can use LEFT JOIN. Left join your second table with id_contrato and detalle_contrato id_contrato.
SELECT
dc.ID_CONTRATO,
dc.TRADICIONAL,
dc.NOM_VARIEDAD,
dc.IFNULL(SUM(CANTIDAD),0) AS CANTIDAD,
dc.IFNULL(SUM(CANTIDAD_D),0) AS CANTIDAD_D
FROM
detalle_contrato dc
LEFT JOIN TABLE_NAME t2 ON t2.ID_CONTRATO = dc.ID_CONTRATO
WHERE dc.ID_CONTRATO = '$ID' AND t2.TIPO_VARIEDAD = 'TRADICIONAL'
I got the bellow piece of select statement that got level 2 child records, having problems to got deeper, can anyone help out?
SELECT
id_mobile AS ID_PROJETO,
UM.qtd_UC,
AM.qtd_AMBIENTE
FROM projetos_mobile AS PM
LEFT JOIN (
SELECT
COUNT(id) AS qtd_UC,
projeto,
data_hora_importacao,
id_uc_mobile
FROM ucs_mobile
WHERE data_hora_importacao = '2015-05-15 17:21:02'
GROUP BY projeto) AS UM
ON PM.id_mobile = UM.projeto
LEFT JOIN (
SELECT
COUNT(id_uc_mobile) AS qtd_AMBIENTE,
id_uc_mobile
FROM ucs_mobile
LEFT JOIN (
SELECT
uc
FROM ambientes_mobile AS s
WHERE data_hora_importacao = '2015-05-15 17:21:02') AS G
ON G.uc = ucs_mobile.id_uc_mobile
WHERE data_hora_importacao = '2015-05-15 17:21:02') AS AM
ON UM.id_uc_mobile = AM.id_uc_mobile
WHERE PM.data_hora_importacao = '2015-05-15 17:21:02'
http://sqlfiddle.com/#!9/2eecf
here is a sqlfiddle if anyone want to try a solution. I have the specific hierarchy: projeto>uc>ambiente>secao>medicoes
ucs_mobile.projeto refers to projetos_mobile.id_mobile
ambientes_mobile.uc refers to ucs_mobile.id_uc_mobile
secoes_iluminacao_mobile.ambiente refers to ambientes_mobile.id_ambiente_mobile
I need a count of each child for the parent I pass, I will have 5 functions that
return the count of each child for a given parent, for example, for a projeto parent I should have count(ucs),count(ambientes),count(secoes),count(medicoes)
So, hope you guys can help me. The database is terrible ugly but that's is what I got. Appreciate any help.
When you have really large queries like this, it can often be helpful to break them down individually, starting from the ground up and patching them together.
I started by just getting the count of each ucs_mobile row for each projetos_mobile value. You can do that by joining the two tables on the related row, and using COUNT(DISTINCT um.id) to get the number of rows. There are other ways to do it, but this particular method will scale better for the rest of your query:
SELECT pm.id, COALESCE(COUNT(DISTINCT um.id), 0) AS qty_uc
FROM projetos_mobile pm
LEFT JOIN ucs_mobile um ON um.data_hora_importacao = '2015-05-15 17:21:02' AND um.projeto = pm.id_mobile
GROUP BY pm.id;
The COALESCE function will be used to fill 0 counts. As long as you remember to use the DISTINCT keyword, and group by the proper id, you can just add in the child rows like so:
SELECT
pm.id,
COALESCE(COUNT(DISTINCT um.id), 0) AS qty_uc,
COALESCE(COUNT(DISTINCT am.id), 0) AS qty_am,
COALESCE(COUNT(DISTINCT sim.id), 0) AS qty_sim
FROM projetos_mobile pm
LEFT JOIN ucs_mobile um ON um.data_hora_importacao = '2015-05-15 17:21:02' AND um.projeto = pm.id_mobile
LEFT JOIN ambientes_mobile am ON am.data_hora_importacao = um.data_hora_importacao AND am.uc = um.id_uc_mobile
LEFT JOIN secoes_iluminacao_mobile sim ON sim.data_hora_importacao = am.data_hora_importacao AND sim.ambiente = am.id_ambiente_mobile
GROUP BY pm.id;
Here is an SQL Fiddle example. NOTE I changed your sample data slightly to ensure my query was working as expected.
Also, a side note. I noticed as you went along that you kept using the same date in your WHERE clauses, so I just joined each table on the date as well, and made sure that in my very first join I looked for the date specified, which in turn will carry its way over to the other tables.
$sql = 'SELECT *
FROM table_all ta
INNER JOIN table11 ta11
ON ta11.column1 = ta.column1
INNER JOIN table12 ta12
ON ta12.column2 = ta.column2
INNER JOIN table13 ta13
ON ta13.column3 = ta.column3';
This is a sql statement of a PHP-programm that I am currently working on. It works fine actually, but I am unsatisfied with the outcome of this statement. It would be nice to have
only the content of the table_all and then
only the three columns of those other three tables (table11, table12, table13).
The asterisk creates some problems in this regard.
The length of table_all is not known. It is able to change in size.
I tried Union and multiple selects, but only received errors. How would one deal with such a problem?
You can try specifying which columns do you want to get.
As you can see asterisk returns all columns.
You should use something like:
$sql = 'SELECT ta11.column1, taa12.column2, ta13.column13
FROM table_all ta
INNER JOIN table11 ta11
ON ta11.column1 = ta.column1
INNER JOIN table12 ta12
ON ta12.column2 = ta.column2
INNER JOIN table13 ta13
ON ta13.column3 = ta.column3';
I don't see a point in getting ta11.column1, taa12.column2, ta13.column13 because these fields are already in table_all (you use these columns in Join). But you can select any columns you want that way.
I keep falling back into questions with MySQL joining.
And I would like to request a very simple example I could use to continue my journey of understanding learning the MySQL syntax.
Let's say I got the following table's
test_testtable
testtable_id
testtable_name
testtable_user
testtable_option
testtable_textfield
test_testlink
testlink_id
testlink_link
testlink_address
test_address
address_id
address_name
address_phone
address_email
address_street
address_city
address_zip
I would like to make a selection like :
SELECT * (lets say I would define the fields) FROM `test_testable`
JOIN `test_testtable`.`testtable_id` = `test_testlink`.`testlink_link`
AND
JOIN `test_testlink`.`testlink_addres` = `test_address`.`address_id`
WHERE `user_id` = 5
Hence the linking structure is like:
test_testtable.testtable_id = leading
table test_testlink is a table to link the table test_testtable and test_address
And linking table test_testlink uses the field testlink_link to link to the table test_testtable, and uses the field testlink_address to link to the table test_address
This does not work. FOR ME.. Since I continuously seem to fail of catching the correct syntax logic.
So I hope that someone could give me a small example of how to correctly implement such a simple yet critical query!
TIAD!!
A general approach :
SELECT table1.* FROM table1
JOIN table2 ON table2.id_table1 = table1.id
JOIN table3 ON table3.id_table2 = table2.id
WHERE table1.id = 10
For your purpose :
SELECT * (lets say I would define the fields) FROM `test_testable`
JOIN `test_testlink` ON `test_testtable`.`testtable_id` = `test_testlink`.`testlink_link`
JOIN `test_address` ON `test_testlink`.`testlink_addres` = `test_address`.`address_id`
WHERE `user_id` = 5
Please read the reference
You are using wrong syntax. You should mention which tables to join first then based on which fields.
SELECT * (lets say I would define the fields) FROM `test_testable`
INNER JOIN test_testlink
ON `test_testtable`.`testtable_id` = `test_testlink`.`testlink_link`
INNER JOIN `test_address`
ON `test_testlink`.`testlink_addres` = `test_address`.`address_id`
AND `test_testtable`.`user_id` = 5
select * from testlink JOIN testtable ON testlink.tableid = testtable.ID
JOIN testaddress ON testlink.addressid = testaddress.ID
WHERE testtable.ID = 5
I have two tables and joined them to one different table
1 table named 'rec_dept'
id_dept
id_divisi
nama_dept
2 table named 'rec_divisi'
id_divisi
nama_div
3 joined table named 'rec_divdep'
id_divdep
id_divisi
id_dept
How to get nama_dept where in the same id_divisi?
Maybe you're looking for this:
SELECT `nama_dept` FROM `rec_dept` WHERE `id_divisi` IN (SELECT `id_divisi` FROM `rec_divdep`);
Hope that helps
you can do a SELECT query with a LEFT JOIN function to get data
SELECT a.`nama_dept` FROM `rec_dept` a
LEFT JOIN `rec_divisi` b
ON a.`id_divisi` = b.`id_divisi`
ORDER BY a.`id_divisi` ASC
SELECT documentation
LEFT JOIN documentation
select a.id_dept, a.id_divisi, a.nama_dept, b.id_divisi, b.nama_div, c.id_divdep, c.id_divisi from rec_divdep as c left join rec_divisi as b on (c.id_divisi = b.id_divisi) left join rec_dept as a on (c.id_divisi = a.id_divisi)
what database do you use. I code mine as mysql, basically I condition the three tables that has same id_divisi. I did not test it but I am pretty sure of the logic based on what I understand.