I am using MySQL and database server and I have two tables one is for customer and another for customer_contacts.
Here are the table structures:
customer(
id(pk, ai)
name
email
)
and
customer_contacts(
id(ai)
customer_id
first_name
last_name
)
Now my question is:
lets say I have one customer has many customer_contacts like this
customer
id name email
1 john john#example.com
and customer_contacts is like this (first row)
id customer_id first_name last_name
1 1 john doe
2 1 johnp pual
like this
So here I want to get all the contact details count for the id john. So can some one tell me how to get that?
Any help and suggestions will be really appreciable. Thanks
If the id is already known (as suggested by the question, it would be a plain
SELECT COUNT(*) from customer_contacts
WHERE customer_contacts.id = 1;
You just need to know the number of contactacs for the customer John, right?
SELECT COUNT(*)
FROM customer cus INNER JOIN customer_contacts con ON cus.id = con.customer_id
WHERE cus.name = 'john'
Nevertheless, It would be better if you know the id of John. Your query would be this:
SELECT COUNT(*)
FROM customer_contacts
WHERE customer_id = 1
Simply join both tables and use count() with group by c.id use where to filter records
select c.*,
count(*) total_contacts
from customer c
left join customer_contacts cc on(c.id = cc.customer_id)
group by c.id
Fiddle Demo
Related
I need to make one SQL command.
From table with comments i'll get comment id, then
with this ID I need to get count of reactions with the same comment ID and user's names.
So for example I have this 2 tables:
Comments:
ID
Comm_text
1
Example text
2
Another example
and Reactions:
ID
comm_id
usr
etc..
1
1
Peter
another
2
1
John
collon
3
1
Dog
cuz
4
2
Cat
why not
I need to get this:
ID
Comm_text
Reactions_Count
Users
1
Example text
3
Peter, John, Dog
2
Another example
1
Cat
I tried this:
SELECT k.id, k.comm, COUNT(r.id) as reactions, r.usr
FROM `comms` k
INNER JOIN `reactions` r ON r.id=k.id
It's just one row with one comment and count of all rows in reaction table.
Thanks.
Try this query that makes the same output:
select comments.id as ID , comments.Comm_text as Comm_text ,
(select count(id) from Reactions where comm_id = comments.id) as Reactions_Count ,
(select coalesce(GROUP_CONCAT(usr ORDER BY usr DESC) , '') from Reactions WHERE comm_id = comments.id) as Users
from comments group by comments.id
You should use group by to group the comments and have just one row then use query to count and gather the data, based on each row of the group.
The GROUP_CONCAT attach the output with , and the coalesce set the output to a given string if the output was empty.
Read more about:
GROUP BY
GROUP_CONCAT
COALESCE
subquery
According to the names that u set in the example, this will work. Just fix the table names for your database structure.
SELECT `Comments`.`ID`, `Comments`.`Comm_text`, count(`Reactions`.`comm_id`) as react, `Reactions`.`usr`
FROM `Comments`
INNER JOIN `Reactions`
ON `Comments`.`ID`=`Reactions`.`comm_id`
GROUP BY `Reactions`.`comm_id`
As I'm SQL beginner, I can't describe a problem in a simple way, so let me show you an example:
3 Tables:
PRODUCT
id
group_id
person_id
GROUP
id
name
PERSON
id
group_id
As you see, GROUP can have multiple PERSONs and PRODUCT can be connected with GROUP and PERSON.
From this point, I would like to count number of PERSONs having a PRODUCT within a GROUP
I don't really understand the background of IN or using another SELECT within FROM, so if that's the point, then I'm happy that I was one step before it lol.
SELECT
group.name as GROUP_name,
COUNT(DISTINCT person_id) AS PERSON_having_min_one_PRODUCT
FROM products
LEFT JOIN groups ON groups.id = products.group_id
LEFT JOIN persons ON persons.id = products.person_id;
With this data:
GROUP
ExampleGroupName1 has 3 PERSONs, but 2 of them has >0 PRODUCTS
ExampleGroupName2 has 3 PERSONs and all of them has >0 PRODUCTS
ExampleGroupName3 has 2 PERSONs, but none of them has the PRODUCT
ExampleGroupName4 has 2 PERSONs, but only 1 has >0 PRODUCT
I would like to have an output like this:
GROUP_name | PERSON_having_min_one_PRODUCT
ExampleGroupName1 | 2
ExampleGroupName2 | 3
ExampleGroupName4 | 1
I would like to count number of PERSONs having a PRODUCT within a GROUP
Note: I will assume the table product does not have the column group_id, since it is redundant and can lead to a lot of errors.
The following query will show you the result you want by joining the tables person and product:
select
count(distinct x.id)
from person x
join product p on p.person_id = x.id
where x.group_id = 123 -- choosing a specific group
and p.id = 456 -- choosing a specific product
This would rather be simple like below meaning all the groups with some group_id with count(persons) and those count who has some product via id used in having clause
Select group_id,
count( distinct id ) AS "PERSON_WITH_PRODUCT"
from
person group by group_id having id
in (Select id from product);
Students Table:
ID | Family_ID | Student_name | F_name
Families Table:
ID | F_name | Contact_No
i want to get all records from students where family_id is repeating.(basically i want to know students Brothers/Sisters records if there is any in student table.
i tried it this way but got wrong output;
SELECT students.*
FROM students
INNER JOIN families
ON families.id = students.family_id
ORDER BY families.id ASC
my query result in image: as you can see some ids are showing once others are more then once, but i think all should appear more then once.
If you want to see only relevant people you don't need to link it to the families table. You can group the student with family_id. Here is your query :
SELECT *
FROM Student
WHERE family_id IN (SELECT family_id
FROM students
GROUP BY family_id
HAVING COUNT(1)>1)
ORDER BY family_id
You could try using a join on subquery for the family_id that have more that one rows in students
SELECT students.*
FROM students
inner join (
select students.family_id
FROM students
group by students.family_id
having count(*)>1
) t on t. family_id = students.family_id
I have the following tables:
Table 1 : Contacts
Fields : id first_name
Values :
1 Reeta
2 Rohan
3 John
Table 2 : email (it contains contact_id of contacts table)
Fields : id contact_id email_address
Values :
1 1 r#gmail.com
2 2 r#gmail.com
3 3 j#gmail.com
I want to display all duplicates by email. Like this:
cont_id first_name email_address
1 Reeta r#gmail.com
2 Rohan r#gmail.com
Here is my query :
select contact_id
from contacts
where email_address IN (
SELECT S.email_address
FROM contacts R
INNER JOIN email
ON R.id = S.contact_id
Group By email_address
Having Count(S.id) > 1
);
The query takes long time to execute with large number of records. However the inner query works faster but not the outer one. Please Help.
I would move your INNER JOIN outside of your subquery.
SELECT
c.contact_id,
c.first_name,
e.email_address
FROM contacts c
INNER JOIN email e ON c.id = e.contact_id
WHERE e.email_address IN (
SELECT email_address
FROM contacts
GROUP BY email_address
HAVING COUNT(id) > 1
);
You could also implement MySQL's EXPLAIN to get a better idea what's choking your query.
Another way to do it, which should be quicker, is something like this:
select email.email_address, group_concat(contacts.contact_id)
from contacts inner join email on contacts.contact_id=email.contact_id
group by email.email_address
having count(contacts.contact_id) > 1;
Now, you're getting exactly what you want; the only thing is, the contact IDs will be concatenated as a comma-separated string. But you'll know which email addresses are non-unique.
You should also have indexes on contact_id in all tables (because you join on these fields), and probably email_address as well (since you search on it).
ok i've been trying for 2 days to figure this out. I have 2 tables
Organization_dep
id
orgid
depname
organization_dep_users
id
orgid
depid
userid
I want to list the count of user ids from the departments of the organizations.
Can you try this:
SELECT
b.depname, count(a.id)
FROM organization_dep_users a
INNER JOIN Organization_dep b ON a.depid = b.id
GROUP BY b.depname;
Edit:
Thanks for Barmar for additional details.