mysql query to get all reviews for every company - php

I have a database with tables like below:
Reviews
id | review | companyid
companies
id | name
Now i want to get the data back so that i can show each company name with the total number of reviews for the company. Like seen below:
company 1 (company name) | 345
company 2 (company name) | 28
company 3 (company name) | 794
From here i will make a table using php to display the results
How can i achieve this with MYSQl?

Try this way:
SELECT Count(`r`.`review`) AS `total_reviews`,
`c`.`company`
FROM `reviews` AS `r`
JOIN `companies` AS `c`
ON `c`.`id` = `r`.`companyid`

Use COUNT and GROUP BY to count the reviews per company and use JOIN to get the company name from the other table.
Query
select t2.name as companyName,coalesce(t1.`count`,0) as `count` from
(
select companyid,count(companyid) as `count`
from reviews
group by companyid
)t1
right join companies t2
on t1.companyid= t2.id;
Sample Output
Table - reviews
+----+--------+-----------+
| id | review | companyid |
+----+--------+-----------+
| 1 | r1 | 1 |
| 2 | r2 | 2 |
| 3 | r3 | 1 |
+----+--------+-----------+
Table - companies
+----+------+
| id | name |
+----+------+
| 1 | C1 |
| 2 | C2 |
| 3 | C3 |
+----+------+
Output
+------+-------+
| name | count |
+------+-------+
| C1 | 2 |
| C2 | 1 |
| C3 | 0 |
+------+-------+
SQL Fiddle

You need to use a GROUP BY
SELECT r.companyid, c.name, count(r.id) as nb_review
FROM reviews r
INNER JOIN companies c ON (r.companyid = c.id)
GROUP BY r.companyid, c.name;
If you also want to see the companies with no reviews, do :
SELECT c.id, c.name, count(r.id) as nb_review
FROM companies c
LEFT JOIN reviews r ON (r.companyid = c.id)
GROUP BY c.id, c.name;

Related

INNERJOIN WITH COUNT DISTINCT [duplicate]

I have a tagging system for my events system I would like to create a 'tag cloud'.
I have Events, which can have multiple 'categories'.
Here's the table structure:
**Event_Categories** (Stores Tags / Categories)
| id | name |
+-----------------+
+ 1 | sport |
+ 2 | charity |
+ 3 | other_tag |
**Events_Categories** (Linking Table)
| event_id | event_category_id |
+-------------------------------+
+ 1 | 1 |
+ 2 | 2 |
+ 3 | 1 |
+ 3 | 2 |
Summary:
Event ID 1 -> Sport
Event ID 2 -> Charity
Event ID 3 -> Sport, Charity
I'd like to return the following:
| tag_name | occurrences |
+-----------+-------------+
| sport | 2 |
| society | 2 |
other_tag - Not actually returned, as it has 0 occurrences
Thanks! :)
this will work:
SELECT c.name AS tag_name, COUNT(ec.event_id) AS occurrences
FROM Event_Categories c
INNER JOIN Events_Categories ec ON c.id = ec.event_category_id
GROUP BY c.id
change the INNER JOIN to LEFT JOIN if you want to include categories with 0 occurances
How about something like
SELECT e.name,
COUNT(1)
FROM Event_Categories e INNER JOIN
Events_Categories_Linking ec ON e.id = ec.event_category_id
GROUP BY e.name
SQL Fiddle DEMO

How do I select distinct rows and cross check in another table?

I have 3 tables, users and tasks and completed_tasks. So basically I want to select all tasks where the user_id = 2 AND also check that the task does not exist in another table` the So here is my tables:
users table:
+----+-------+
| id | name |
+----+-------+
| 1 | John |
| 2 | Sally |
+----+-------+
tasks table:
+----+-----------+---------+
| id | task_name | user_id |
+----+-----------+---------+
| 1 | mop floor | 2 |
| 2 | dishes | 1 |
| 3 | laundry | 2 |
| 4 | cook | 2 |
+----+-----------+---------+
completed_tasks table:
+----+---------+---------+
| id | task_id | user_id |
+----+---------+---------+
| 1 | 1 | 2 |
+----+---------+---------+
Here is my current SELECT code for my MySQL database:
$db = "SELECT DISTINCT tasks.task_name, users.name FROM tasks LEFT JOIN ON users.id = tasks.user_id WHERE tasks.user_id = 2";
THe problem I'm having is: I want it to search in completed_tasks table and if the task exists, then don't select that task.
I tried to do that by adding the following but it did not work:
LEFT JOIN completed_tasks ON completed_tasks.user_id = 2
That did not work because if I had multiple completed tasks, it would just ignore it all together.
I want the end result should return the user's name and task name of task 3 and 4.
Also, performance is critical in my application. I could use PHP and loop through the arrays and do SELECT for each of them but that would not be good for performance.
You have a few ways to do this.
You can use a LEFT JOIN and then check for NULL in the optional table.
SELECT a.name, b.task_name
FROM users a
JOIN tasks b ON a.id = b.user_id
LEFT JOIN completed_tasks c ON c.task_id = b.id AND c.user_id = b.user_id
WHERE c.id IS NULL
;
You can do a NOT EXISTS sub-query
SELECT a.name, b.task_name
FROM users a
JOIN tasks b ON a.id = b.user_id
WHERE NOT EXISTS (
SELECT 1
FROM completed_tasks c
WHERE c.task_id = b.id AND c.user_id = b.user_id)
;

PDO Query : Count related and repeated values then inner join

I work with PHP and PDO.
So I have 2 tables like,
Table 1
| id | name | age |
| 1 | John | 25 |
| 2 | Tom | 32 |
| 3 | James| 45 |
Table 2
| id | Comment | Link |
| 1 | some text | 3 |
| 2 | some text | 3 |
| 3 | some text | 1 |
So, Link column numbers represent id's in table1. For example Link = 3s in table 2 represent James in table 1. I need a query which brings all table1's data and also a number of repeated value for related Link column which comes from table2.
For example, the query should give me (let's choose James),
| id | name | age | Value |
| 3 | James | 45 | 2 |
value=2, because there are two 3s in link column which related to James
I tried somethings but got lots of errors.
I think you just need the GROUP BY
SELECT a.id,
a.name,
a.age,
count(*) as value
FROM table1 a
JOIN table2 b ON a.id = b.link
GROUP BY a.id, a.name, a.age
If you really want just one row then add WHERE
SELECT a.id,
a.name,
a.age,
count(*) as value
FROM table1 a
JOIN table2 b ON a.id = b.link
WHERE a.name = 'James'
GROUP BY a.id, a.name, a.age
or use subquery
SELECT a.id,
a.name,
a.age,
(SELECT count(*) FROM table2 b WHERE a.id = b.link) as value
FROM table1 a
WHERE a.name = 'James'

Insert `COUNT(*)` based on separate table

In MySQL is it possible to select columns from one table while also creating a column for COUNT(*) based on other tables? That way a summary of the results from all tables can be returned. This might be a bit confusing to explain in words so I made some sample tables instead:
events_tbl
----------------------------
id | eventname
1 | Anime Festival
2 | Food Festival
----------------------------
booths_tbl
-------------------------
id | boothname
1 | Walmart
2 | Pizza Hut
3 | Nike
4 | North Face
-------------------------
participants_tbl
-----------------------------
id | participantname
1 | John
2 | Mike
3 | Rambo
4 | Minnie
-----------------------------
event_booths_tbl
--------------------------------
event_id | booth_id
1 | 1
1 | 2
1 | 5
2 | 3
2 | 4
--------------------------------
event_participants_tbl
-------------------------------------
event_id | booth_id
1 | 1
1 | 2
1 | 3
1 | 4
-------------------------------------
Is there a way to get results like this in MySQL:
summary_tbl
------------------------------------------------------------------------
id | eventname | booth_count | participant_count
1 | Anime Festival | 3 | 4
2 | Food Festival | 2 | 0
------------------------------------------------------------------------
The event_participants_tbl should contain participant_id instead of booth_id.
Its irrelevant otherwise.
Your MySQL query would be like this :
select
et.id,
et.eventname,
count(distinct ebt.booth_id) as booth_count,
count(distinct ept.participant_id) as participant_count
from
event_booths_tbl ebt
left join events_tbl et on et.id=ebt.event_id
left join event_participants_tbl ept on ept.event_id=ebt.event_id
group by et.event_id;
Join with subqueries that count in each table:
SELECT e.id, e.event_name,
IFNULL(b.booth_count, 0) AS booth_count,
IFNULL(p.participant_count, 0) AS participant_count
FROM events_table AS e
LEFT JOIN (SELECT event_id, COUNT(*) AS booth_count
FROM event_booths_table
GROUP BY event_id) AS b ON e.id = b.event_id
LEFT JOIN (SELECT event_id, COUNT(*) AS participant_count
FROM event_participants_table
GROUP BY event_id) AS p ON e.id = p.event_id
Try this :
select event.id,
event.name,
count(distinct eventBooth.booth_id),
count(distinct eventParitcipant.booth_id)
from events_tbl event
LEFT JOIN event_booths_tbl eventBooth on eventBooth.event_id=event.id
LEFT JOIN event_participants_tbl eventParitcipant
on eventParitcipant.event_id=event.id
group by event.id

How to SELECT records from One table If Matching Record In Not Found In Other Table

Halo i am trying for a query to select record from one table say 'deal_asking' only if a matching record is not found in the second table 'deal_unpluged'. this what i does (below) selecting record using LEFT JOIN and then filter the record in PHP side.
What i am looking for an Mysql query way of solution for this problem please help..
thanks in advance...
SELECT DA.das_id, DU.das_id_fk
FROM deal_asking DA
LEFT JOIN deal_unpluged DU ON DA.das_id= DU.das_id_fk
WHERE department='8'
ORDER BY das_id ASC LIMIT 10 OFFSET 0
Just add this to your WHERE clause:
AND DU.das_id_fk IS NULL
Say I have the following two tables:
+-------------------------+ +-------------------------+
| Person | | Pet |
+----------+--------------+ +-------------------------+
| PersonID | INT(11) | | PetID | INT(11) |
| Name | VARCHAR(255) | | PersonID | INT(11) |
+----------+--------------+ | Name | VARCHAR(255) |
+----------+--------------+
And my tables contain the following data:
+------------------------+ +---------------------------+
| Person | | Pet |
+----------+-------------+ +-------+----------+--------+
| PersonID | Name | | PetID | PersonID | Name |
+----------+-------------+ +-------+----------+--------+
| 1 | Sean | | 5 | 1 | Lucy |
| 2 | Javier | | 6 | 1 | Cooper |
| 3 | tradebel123 | | 7 | 2 | Fluffy |
+----------+-------------+ +-------+----------+--------+
Now, if I want a list of all Persons:
SELECT pr.PersonID, pr.Name
FROM
Person pr
If I want a list of Persons that have pets (including their pet's names):
SELECT pr.PersonID, pr.Name, pt.Name AS PetName
FROM
Person pr
INNER JOIN Pet pt ON pr.PersonID = pt.PersonID
If I want a list of Persons that have no pets:
SELECT pr.PersonID, pr.`Name`
FROM
Person pr
LEFT JOIN Pet pt ON pr.PersonID = pt.PersonID
WHERE
pt.`PetID` IS NULL
If I want a list of all Persons and their pets (even if they don't have pets):
SELECT
pr.PersonID,
pr.Name,
COALESCE(pt.Name, '<No Pet>') AS PetName
FROM
Person pr
LEFT JOIN Pet pt ON pr.PersonID = pt.PersonID
If I want a list of Persons and a count of how many pets they have:
SELECT pr.PersonID, pr.Name, COUNT(pt.PetID) AS NumPets
FROM
Person pr
LEFT JOIN Pet pt ON pr.PersonID = pt.PersonID
GROUP BY
pr.PersonID, pr.Name
Same as above, but don't show Persons with 0 pets:
SELECT pr.PersonID, pr.Name, COUNT(pt.PetID) AS NumPets
FROM
Person pr
LEFT JOIN Pet pt ON pr.PersonID = pt.PersonID
GROUP BY
pr.PersonID, pr.Name
HAVING COUNT(pt.PetID) > 0

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