Laravel: Adding multiple count results of sub queries - sql

I wanted to add the count result of more than one queries that belong to different tables.
I am using the below problem as a reference to my actual problem because this problem has already a solution (How do I add two count(*) results together on two different tables?) but I am facing problem in implementing the solution in laravel.
I have two tables: Toys and Games.
+--------------------+------------------+
| Field | Type |
+--------------------+------------------+
| toy_id | int(10) unsigned |
| little_kid_id | int(10) unsigned |
+--------------------+------------------+
+--------------------+------------------+
| Field | Type |
+--------------------+------------------+
| game_id | int(10) unsigned |
| little_kid_id | int(10) unsigned |
+--------------------+------------------+
A little kid can have multiple toys. A little kid can be participating in multiple games at once.
I want a query that will give me the total number of toys + games that a little_kid is involved with.
Basically, I want the sum of these two queries:
SELECT COUNT(*) FROM Toys WHERE little_kid_id = 900;
SELECT COUNT(*) from Games WHERE little_kid_id = 900
The above problem has the following accepted answer
SELECT
(SELECT COUNT(*) FROM Toys WHERE little_kid_id = 900)+
(SELECT COUNT(*) from Games WHERE little_kid_id = 900)
AS SumCount
I wanted to implement the above solution in Laravel.
I have tried the following method but to no avail. It gives syntax error.
$sub=DB::tabel('toys')->select(DB::raw('count(*) as total'))
->where('little_kid_id',$id);
$sub1=DB::tabel('games')->select(DB::raw('count(*) as total'))
->where('little_kid_id',$id);
$result = DB::table( DB::raw("( ({$sub->toSql()})+({$sub1->toSql()}) ) as
total_count") )
->mergeBindings($sub)
->mergeBindings($sub1)
->select('total_count.total')
->get();
I have also tried this method. It works but gives collection but I need an integer value of total count
$result=$sub->unionAll($sub1);
dd($result->get());
In short I wanted to perform the accepted solution of that problem (How do I add two count(*) results together on two different tables?) in laravel.

You can use those codes :
$q = DB::tabel('toys')->select('toy_id','little_kid_id')->where('little_kid_id',900);
$q1 = DB::tabel('games')->select('game_id','little_kid_id')->where('little_kid_id',900)
$data = $q->union($q1)->count();
Don't forget Union require the tables must have the same columns so that I select the columns if your columns will not match each other then don't touch the select statement otherwise feel free to remove the codes
The second way is :
DB::table(DB::raw('(SELECT COUNT(*) as c FROM Toys WHERE little_kid_id = 900) t,(SELECT COUNT(*) as c1 from Games WHERE
little_kid_id = 900) t2'))->selectRaw('t.c+t2.c1 as
SumCount')->toSql(); //change toSql() to get() if you want to get
datas instead of sql code
You can try this. for more information look at https://laravel.com/docs/5.8/queries#raw-expressions. if you want to get request parameters by any user then be confirmed to prevent SQL Injection

Related

Get total count and first 3 columns

I have the following SQL query:
SELECT TOP 3 accounts.username
,COUNT(accounts.username) AS count
FROM relationships
JOIN accounts ON relationships.account = accounts.id
WHERE relationships.following = 4
AND relationships.account IN (
SELECT relationships.following
FROM relationships
WHERE relationships.account = 8
);
I want to return the total count of accounts.username and the first 3 accounts.username (in no particular order). Unfortunately accounts.username and COUNT(accounts.username) cannot coexist. The query works fine removing one of the them. I don't want to send the request twice with different select bodies. The count column could span to 1000+ so I would prefer to calculate it in SQL rather in code.
The current query returns the error Column 'accounts.username' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause. which has not led me anywhere and this is different to other questions as I do not want to use the 'group by' clause. Is there a way to do this with FOR JSON AUTO?
The desired output could be:
+-------+----------+
| count | username |
+-------+----------+
| 1551 | simon1 |
| 1551 | simon2 |
| 1551 | simon3 |
+-------+----------+
or
+----------------------------------------------------------------+
| JSON_F52E2B61-18A1-11d1-B105-00805F49916B |
+----------------------------------------------------------------+
| [{"count": 1551, "usernames": ["simon1", "simon2", "simon3"]}] |
+----------------------------------------------------------------+
If you want to display the total count of rows that satisfy the filter conditions (and where username is not null) in an additional column in your resultset, then you could use window functions:
SELECT TOP 3
a.username,
COUNT(a.username) OVER() AS cnt
FROM relationships r
JOIN accounts a ON r.account = a.id
WHERE
r.following = 4
AND EXISTS (
SELECT 1 FROM relationships t1 WHERE r1.account = 8 AND r1.following = r.account
)
;
Side notes:
if username is not nullable, use COUNT(*) rather than COUNT(a.username): this is more efficient since it does not require the database to check every value for nullity
table aliases make the query easier to write, read and maintain
I usually prefer EXISTS over IN (but here this is mostly a matter of taste, as both techniques should work fine for your use case)

How to create two JOIN-tables so that I can compare attributes within?

I take a Database course in which we have listings of AirBnBs and need to be able to do some SQL queries in the Relationship-Model we made from the data, but I struggle with one in particular :
I have two tables that we are interested in, Billing and Amenities. The first one have the id and price of listings, the second have id and wifi (let's say, to simplify, that it equals 1 if there is Wifi, 0 otherwise). Both have other attributes that we don't really care about here.
So the query is, "What is the difference in the average price of listings with and without Wifi ?"
My idea was to build to JOIN-tables, one with listings that have wifi, the other without, and compare them easily :
SELECT avg(B.price - A.price) as averagePrice
FROM (
SELECT Billing.price, Billing.id
FROM Billing
INNER JOIN Amenities
ON Billing.id = Amenities.id
WHERE Amenities.wifi = 0
) A, (
SELECT Billing.price, Billing.id
FROM Billing
INNER JOIN Amenities
ON Billing.id = Amenities.id
WHERE Amenities.wifi = 1) B
WHERE A.id = B.id;
Obviously this doesn't work... I am pretty sure that there is a far easier solution to it tho, what do I miss ?
(And by the way, is there a way to compute the absolute between the difference of price ?)
I hope that I was clear enough, thank you for your time !
Edit : As mentionned in the comments, forgot to say that, but both tables have idas their primary key, so that there is one row per listing.
Just use conditional aggregation:
SELECT AVG(CASE WHEN a.wifi = 0 THEN b.price END) as avg_no_wifi,
AVG(CASE WHEN a.wifi = 1 THEN b.price END) as avg_wifi
FROM Billing b JOIN
Amenities a
ON b.id = a.id
WHERE a.wifi IN (0, 1);
You can use a - if you want the difference instead of the specific values.
Let's assume we're working with data like the following (problems with your data model are noted below):
Billing
+------------+---------+
| listing_id | price |
+------------+---------+
| 1 | 1500.00 |
| 2 | 1700.00 |
| 3 | 1800.00 |
| 4 | 1900.00 |
+------------+---------+
Amenities
+------------+------+
| listing_id | wifi |
+------------+------+
| 1 | 1 |
| 2 | 1 |
| 3 | 0 |
+------------+------+
Notice that I changed "id" to "listing_id" to make it clear what it was (using "id" as an attribute name is problematic anyways). Also, note that one listing doesn't have an entry in the Amenities table. Depending on your data, that may or may not be a concern (again, refer to the bottom for a discussion of your data model).
Based on this data, your averages should be as follows:
Listings with wifi average $1600 (Listings 1 and 2)
Listings without wifi (just 3) average 1800).
So the difference would be $200.
To achieve this result in SQL, it may be helpful to first get the average cost per amenity (whether wifi is offered). This would be obtained with the following query:
SELECT
Amenities.wifi AS has_wifi,
AVG(Billing.price) AS avg_cost
FROM Billing
INNER JOIN Amenities ON
Amenities.listing_id = Billing.listing_id
GROUP BY Amenities.wifi
which gives you the following results:
+----------+-----------------------+
| has_wifi | avg_cost |
+----------+-----------------------+
| 0 | 1800.0000000000000000 |
| 1 | 1600.0000000000000000 |
+----------+-----------------------+
So far so good. So now we need to calculate the difference between these 2 rows. There are a number of different ways to do this, but one is to use a CASE expression to make one of the values negative, and then simply take the SUM of the result (note that I'm using a CTE, but you can also use a sub-query):
WITH
avg_by_wifi(has_wifi, avg_cost) AS
(
SELECT Amenities.wifi, AVG(Billing.price)
FROM Billing
INNER JOIN Amenities ON
Amenities.listing_id = Billing.listing_id
GROUP BY Amenities.wifi
)
SELECT
ABS(SUM
(
CASE
WHEN has_wifi = 1 THEN avg_cost
ELSE -1 * avg_cost
END
))
FROM avg_by_wifi
which gives us the expected value of 200.
Now regarding your data model:
If both your Billing and Amenities table only have 1 row for each listing, it makes sense to combine them into 1 table. For example: Listings(listing_id, price, wifi)
However, this is still problematic, because you probably have a bunch of other amenities you want to model (pool, sauna, etc.) So you might want to model a many-to-many relationship between listings and amenities using an intermediate table:
Listings(listing_id, price)
Amenities(amenity_id, amenity_name)
ListingsAmenities(listing_id, amenity_id)
This way, you could list multiple amenities for a given listing without having to add additional columns. It also becomes easy to store additional information about an amenity: What's the wifi password? How deep is the pool? etc.
Of course, using this model makes your original query (difference in average cost of listings by wifi) a bit tricker, but definitely still doable.

How to do an exact match followed by ORDER BY in PostgreSQL

I'm trying to write a query that puts some results (in my case a single result) at the top, and then sorts the rest. I have yet to find a PostgreSQL solution.
Say I have a table called airports like so.
id | code | display_name
----+------+----------------------------
1 | SDF | International
2 | INT | International Airport
3 | TES | Test
4 | APP | Airport Place International
In short, I have a query in a controller method that gets called asynchronously when a user text searches for an airport either by code or display_name. However, when a user types in an input that matches a code exactly (airport code is unique), I want that result to appear first, and all airports that also have int in their display_name to be displayed afterwards in ascending order. If there is no exact match, it should return any wildcard matches sorted by display_name ascending. So if a user types in INT, The row (2, INT, International Airport) should be returned first followed by the others:
Results:
1. INT | International Airport
2. APP | Airport Place International
3. SDF | International
Here's the kind of query I was tinkering with that is slightly simplified to make sense outside the context of my application but same concept nonetheless.
SELECT * FROM airports
WHERE display_name LIKE 'somesearchtext%'
ORDER BY (CASE WHEN a.code = 'somesearchtext` THEN a.code ELSE a.display_name END)
Right now the results if I type INT I'm getting
Results:
1. APP | Airport Place International
2. INT | International Airport
3. SDF | International
My ORDER BY must be incorrect but I can't seem to get it
Any help would be greatly appreciated :)
If you want an exact match on code to return first, then I think this does the trick:
SELECT a.*
FROM airports a
WHERE a.display_name LIKE 'somesearchtext%'
ORDER BY (CASE WHEN a.code = 'somesearchtext' THEN 1 ELSE 2 END),
a.display_name
You could also write this as:
ORDER BY (a.code = 'somesearchtext') DESC, a.display_name
This isn't standard SQL, but it is quite readable.
I think you can achieve your goal by using a UNION.
First get an exact match and then add that result to rest of the data as you which.
e.g.. (you will need to work in this a bit)
SELECT * FROM airports
WHERE code == 'somesearchtext'
ORDER BY display_name
UNION
SELECT * FROM airports
WHERE code != 'somesearchtext' AND display_name LIKE 'somesearchtext%'
ORDER BY display_name

PostgreSQL finding the 3 most popular articles in a news database

I'm currently trying to find the 3 most popular articles in a database. I want to print out the title and amount of views for each. I know I'll have to join two of the tables together (articles & log) in order to do so.
The articles table has a column of the titles, and one with a slug for the title.
The log table has a column of the paths in the format of /article/'slug'.
How would I join these two tables, filter out the path to compare to the slug column of the articles table, and use count to display the number of times it was viewed?
The correct query used was:
SELECT title, count(*) as views
FROM articles a, log l
WHERE a.slug=substring(l.path, 10)
GROUP BY title
ORDER BY views DESC
LIMIT 3;
If I understood you correctly you just need to join two tables based on one column using aggregation. The catch is that you can't compare them directly but have to use some string functions before.
Assuming a schema like this:
article
| title | slug |
-------------------
| title1 | myslug |
| title2 | myslug |
log
| path |
--------------------------
| /article/'myslug' |
| /article/'unmentioned' |
Try out something like the following:
select title, count(*) from article a join log l where concat('''', a.slug, '''') = substring(l.path, 10) group by title;
For more complex queries it can be helpful to at first write smaller queries which help you to figure out the whole query later. For example just check if the string functions return what you expect:
select substring(l.path, 10) from log l;
select concat('''', a.slug, '''') from article a;

SQL: SUM of MAX values WHERE date1 <= date2 returns "wrong" results

Hi stackoverflow users
I'm having a bit of a problem trying to combine SUM, MAX and WHERE in one query and after an intense Google search (my search engine skills usually don't fail me) you are my last hope to understand and fix the following issue.
My goal is to count people in a certain period of time and because a person can visit more than once in said period, I'm using MAX. Due to the fact that I'm defining people as male (m) or female (f) using a string (for statistic purposes), CHAR_LENGTH returns the numbers I'm in need of.
SELECT SUM(max_pers) AS "People"
FROM (
SELECT "guests"."id", MAX(CHAR_LENGTH("guests"."gender")) AS "max_pers"
FROM "guests"
GROUP BY "guests"."id")
So far, so good. But now, as stated before, I'd like to only count the guests which visited in a certain time interval (for statistic purposes as well).
SELECT "statistic"."id", SUM(max_pers) AS "People"
FROM (
SELECT "guests"."id", MAX(CHAR_LENGTH("guests"."gender")) AS "max_pers"
FROM "guests"
GROUP BY "guests"."id"),
"statistic", "guests"
WHERE ( "guests"."arrival" <= "statistic"."from" AND "guests"."departure" >= "statistic"."to")
GROUP BY "statistic"."id"
This query returns the following, x = desired result:
x * (x+1)
So if the result should be 3, it's 12. If it should be 5, it's 30 etc.
I probably could solve this algebraic but I'd rather understand what I'm doing wrong and learn from it.
Thanks in advance and I'm certainly going to answer all further questions.
PS: I'm using LibreOffice Base.
EDIT: An example
guests table:
ID | arrival | departure | gender |
10 | 1.1.14 | 10.1.14 | mf |
10 | 15.1.14 | 17.1.14 | m |
11 | 5.1.14 | 6.1.14 | m |
12 | 10.2.14 | 24.2.14 | f |
13 | 27.2.14 | 28.2.14 | mmmmmf |
statistic table:
ID | from | to | name |
1 | 1.1.14 | 31.1.14 |January | expected result: 3
2 | 1.2.14 | 28.2.14 |February| expected result: 7
MAX(...) is the wrong function: You want COUNT(DISTINCT ...).
Add proper join syntax, simplify (and remove unnecessary quotes) and this should work:
SELECT s.id, COUNT(DISTINCT g.id) AS People
FROM statistic s
LEFT JOIN guests g ON g.arrival <= s."from" AND g.departure >= s."too"
GROUP BY s.id
Note: Using LEFT join means you'll get a result of zero for statistics ids that have no guests. If you would rather no row at all, remove the LEFT keyword.
You have a very strange data structure. In any case, I think you want:
SELECT s.id, sum(numpersons) AS People
FROM (select g.id, max(char_length(g.gender)) as numpersons
from guests g join
statistic s
on g.arrival <= s."from" AND g.departure >= s."too"
group by g.id
) g join
GROUP BY s.id;
Thanks for all your inputs. I wasn't familiar with JOIN but it was necessary to solve my problem.
Since my databank is designed in german, I made quite the big mistake while translating it and I'm sorry if this caused confusion.
Selecting guests.id and later on grouping by guests.id wouldn't make any sense since the id is unique. What I actually wanted to do is select and group the guests.adr_id which links a visiting guest to an adress databank.
The correct solution to my problem is the following code:
SELECT statname, SUM (numpers) FROM (
SELECT statistic.name AS statname, guests.adr_id, MAX( CHAR_LENGTH( guests.gender ) ) AS numpers
FROM guests
JOIN statistics ON (guests.arrival <= statistics.too AND guests.departure >= statistics.from )
GROUP BY guests.adr_id, statistic.name )
GROUP BY statname
I also noted that my database structure is a mess but I created it learning by doing and haven't found any time to rewrite it yet. Next time posting, I'll try better.