mysql-display different columns from multiple tables - mysql

Consider the following table and values..i need to display all the column and its values from table "TAG" along with single column value from another table named as "ITEM"
TABLE NAME "TAG"
TAGID NAME RATE
1 A 100
2 B 200
3 C 300
4 D 8099
5 E 400
6 F 500
7 G 600
8 H 700
9 I 200
10 J 400
TABLE NAME "ITEM"
ITEMID Name ITEM_QTY
1 A 11
2 B 22
3 I 3
4 A 8
5 B 42
6 C 5
7 I 61
8 B 74
9 C 99
10 A 48
EXPECTED OUTPUT
TAGID NAME RATE ITEM_QTY
1 A 100 48 - (VALUE OF LAST INSERTED RECORD)
2 B 200 74
3 C 300 99
4 D 8099 0
5 E 400 0
6 F 500 0
7 G 600 0
8 H 700 0
9 I 200 61
10 J 400 0
i have tried some sort of quires but its all vien..
this is my query,
select *
from tag t1
join ( select item_qty
from item t2
join tag t1
on t1.name=t2.name;
can anyone pls help me out of this issue...
thanks in advance...

try like this
select tag.id,tag.name,rate,qty from tag join
(select qty,name from
ITEM where ITEM.id in
(select max(id) from ITEM group by name)) as tt on tt.name=tag.name

Assuming, that the last inserted record has the highest ID in your item table, you first need to get the highest ID per item name.
SELECT
Name, MAX(ITEMID) AS max_id
FROM
item
GROUP BY Name;
With this query (executed in the following query as subquery) we get the corresponding row for each max_id.
SELECT i.*
FROM item i
JOIN (
SELECT
MAX(ITEMID) AS max_id
FROM
item
GROUP BY Name
) si ON i.ITEMID = si.max_id;
Here are more examples of how to get The Rows Holding the Group-wise Maximum of a Certain Column.
The final query is
SELECT TAGID, tag.NAME, RATE, COALESCE(ITEM_QTY, 0) AS ITEM_QTY
FROM
tag
LEFT JOIN (
SELECT i.*
FROM item i
JOIN (
SELECT
MAX(ITEMID) AS max_id
FROM
item
GROUP BY Name
) si ON i.ITEMID = si.max_id
) items ON tag.Name = items.Name
see it working live in an sqlfiddle

You may try as
select
t.tagid,
t.name,
t.rate,
i.item_qty
from tag t
join
(
select name,max(item_qty) as item_qty
from item group by name
)i
on i.name = t.name
DEMO
If you need all the data from item table as you have shown in the output you may use left join instead of inner join
select
t.tagid,
t.name,
t.rate,
coalesce(i.item_qty,0) as item_qty
from tag t
left join
(
select name,max(item_qty) as item_qty
from item group by name
)i
on i.name = t.name
DEMO

Related

mysql count children of selected rows having selected rows as parent?

Table structure(representative)
ID NAME PARENT
--------------------
1 cat1 0
2 cat1 1
3 cat2 1
4 cat1 2
5 cat2 2
6 cat3 2
7 cat1 3
8 cat2 3
9 cat3 3
10 cat1 1
FOREIGN TABLE data for foreign_sub_category_count
id_parent name
-----------------------
2 a
2 b
2 c
3 a
3 b
3 c
categories may have sub categories.
SELECT t.name,t.id
FROM TABLE_NAME AS t
WHERE t.parent = SOME_ID
SOME_ID = 1
gives me the name,id of all categories with SOME_ID parent id
what i want is to get a count of all sub categories of each row in above result set besides the name
WHERE t.id is parent of sub categories and get count of categories from another table which has the same t.id as parent
EXPECTED RESULT
t.id t.name sub_category_count foreign_sub_category_count
2 cat1 3 3
3 cat2 3 3
10 cat1 0 0
I suspect that you are looking for a recursive query - available in MySQL 8.0:
with recursive cte as (
select id root_id, id from mytable
union all
select c.root_id, t.id from cte c inner join mytable t on t.parent = c.id
)
select
t.*,
(select count(*) - 1 from cte c where c.root_id = t.id) no_children
from mytable t
This adds one column to your original table, which contains the number of direct and indirect descendants of the current row.
Try this:
select
tab1.id,
tab1.name,
coalesce(tab2.counts,0) as sub_category_count,
coalesce(tab3.counts,0) as foreign_sub_category_count
from
(select id,name,parent from representative) tab1
left join
(select t1.parent tab_id, count(*) as counts from representative t1 inner join representative t2 on t1.parent=t2.id group by t1.parent) tab2
on tab1.id=tab2.tab_id
left join
(select parent_id,count(*) as counts from foreign_table group by parent_id) tab3
on tab1.id=tab3.parent_id
where tab1.parent=1 --SOME_ID
you can change the parent_id in where tab1.parent=1 of you choice
Example on DB-FIDDLE

sql group by although there is not value

I have a table where exists 4 entries like this.
class_type
id type
1 A
2 B
3 M
4 T
and another table where these values are foreign key.
id number id_class_type
1 10 1
2 11 1
3 12 2
4 13 1
5 14 2
6 15 3
7 16 1
8 17 3
So what i want is count(*) and group by id_class_type but with all class_type (1,2,3,4) although there is not present the id 4.
if you want only the class tha match you can use inner join
select a.class_type, count(*)
from class_type a
inner join table2 b on a.id = b.id_class_type
group by a.class_type
otherwise you can use left join
select a.class_type, count(*)
from class_type a
left join table2 b on a.id = b.id_class_type
group by a.class_type

MYSQL filter multiply row by two lists of id

I have results:
item_id subitem_id
----------------------
1 35
1 25
1 8
2 10
2 25
3 60
4 35
4 25
4 44
5 1
5 23
5 15
5 13
5 9
and I have two lists of subitem
(25,44,1)
(8,9)
how do I set the where clause in order to filter the result and return this
item_id subitem_id
----------------------
1 35
1 25 <-- first set
1 8 <-- second set
-----------------
5 1 <-- first set
5 23
5 15
5 13
5 9 <-- second set
because this item_id contain both subitem_id from two lists
SELECT
`item_id`
FROM table
WHERE `subitem_id` in (25,44,1)
AND `subitem_id` in (8,9)
Did not work, because in one time subitem_id have one id (not all list)
P.S.
This is a simple example, in reality we have more than 100k records with some join construction
http://sqlfiddle.com/#!9/71c28e5/3
SELECT t1.*
FROM (
SELECT DISTINCT(t1.item_id)
FROM t1
INNER JOIN t1 t2
ON t1.item_id = t2.item_id
AND t2.subitem_id in (8,9)
WHERE t1.subitem_id in (25,44,1)
) t
LEFT JOIN t1
ON t.item_id = t1.item_id
Another approach to avoid big number of executed records for mysql:
http://sqlfiddle.com/#!9/71c28e5/10
SELECT t1.*
FROM t1
WHERE item_id in (
SELECT DISTINCT(t1.item_id)
FROM t1
INNER JOIN t1 t2
ON t1.item_id = t2.item_id
AND t2.subitem_id in (25,44,1)
WHERE t1.subitem_id in (8,9)
)
SQL Fiddle
I think you're trying to make sure a item_ID has subcategories in 2 differen sets..
Select * from table A
where exists (Select 1 from table B where A.Item_Id = B.Item_ID and subitem_ID in (25,44,1))
and exists (Select 1 from table C where A.Item_Id = C.Item_ID and subitem_ID in (8,9))

MySQL - Display null column from child table if all values are not distinct

I have the following tables, for example:
invoices
ID Name
1 A
2 B
3 C
4 D
5 E
transactions
ID Invoice_ID User_ID
1 1 10
2 1 10
3 1 10
4 2 30
5 3 20
6 3 40
7 2 30
8 2 30
9 4 40
10 3 50
Now I want to make a select that will pull the invoices and the user_id from the related transactions, but of course if I do that I won't get all the ids, since they may be distinct but there will be only one column for that. What I want to do is that if there are distinct User_ids, I will display a pre-defined text in the column instead of the actual result.
select invoices.id, invoices.name, transactions.user_id(if there are distinct user_ids -> return null)
from invoices
left join transactions on invoices.id = transactions.invoice_id
and then this would be the result
ID Name User_ID
1 A 10
2 B 30
3 C null
4 D 40
5 E null
Is this possible?
You can do the following :
select
invoices.id,
invoices.name,
IF (
(SELECT COUNT(DISTINCT user_id) FROM transactions WHERE transactions.invoice_id = invoices.id) = 1,
(SELECT MAX(user_id) FROM transactions WHERE transactions.invoice_id = invoices.id),
null
) AS user_id
from invoices
Or, alternatively, you can use the GROUP_CONCAT function to output a comma-separated list of users for each invoice. It is not exactly what you asked, but maybe in fact it will be more useful :
select
invoices.id,
invoices.name,
GROUP_CONCAT(DISTINCT transactions.user_id SEPARATOR ',') AS user_ids
from invoices
left join transactions on invoices.id = transactions.invoice_id
group by invoices.id
Try somethingh like:
select i.id, i.name, t.user_id
from invoices i left join
(
select invoice_ID, User_ID
from transactions
group by invoice_ID
having count(invoice_ID)=1
) t on i.id=t.invoice_id
SQL fiddle
You could list all the transactions that have multiple user ids, like this:
select invoices.id, invoices.name, null
from invoices
left join transactions on invoices.id = transactions.invoice_id having count(distinct transactions.user_id) > 1
Also, I think this CASE might suit your needs here:
select invoices.id, invoices.name,
case when count(distinct transactions.user_id) > 1 then null else transactions.user_id end
from invoices
left join transactions on invoices.id = transactions.invoice_id
group by invoices.id
although, I'm not sure this is syntactically correct

SQL - How to calculate column value and join with another table

As I am not good with MySQL query's so I wish someone help me for creating this kind of sql query.
I having two MySQL tables which is describe bellow:
Table Name: rating
-------------------
property_id user_id area_rate_count safety_rate_count friendly_rate_count walkability_rate_count
4 28 1 1 1 2
5 38 2 3 4 1
5 40 2 2 3 1
6 40 2 3 1 4
10 43 2 2 3 1
Table Name: listing
-------------------
property_id title
4 Sample 1
5 Sample 2
6 Sample 3
10 Sample 4
11 Sample 5
12 Sample 6
Now first I want to sum each column and divide. (area_rate_count, safety_rate_count, friendly_rate_count, walkability_rate_count). For example In property_id:5 having two times so first calculate column sum and divide by 2.
After calculation we will get this output:
Table Name: rating (After Calculation)
--------------------------------------
property_id rate
4 5
5 9 (Divided by 2 because this property_id is two times in table)
6 10
10 8
And Finally I want join this result to my listing table and result looks something like this:
Table Name: listing
-------------------
property_id title rate
4 Sample 1 5
5 Sample 2 9 (Divided by 2 becouse property_id is two times in table)
6 Sample 3 10
10 Sample 4 8
11 Sample 5 0
12 Sample 6 0
Thanks.
I think you want the avg() aggregation function along with a join:
select l.property_id, l.title,
coalesce(avg(area_rate_count + safety_rate_count + friendly_rate_count + walkability_rate_count
), 0) as rate
from listing l left outer join
property_id p
on l.property_id = p.property_id
group by l.property_id, l.title ;
If I understood it right I think you need this:
select l.property_id, l.title, coalesce(r.ssum/if(r.ct=0,1,r.ct), 0) as rate
from listing l LEFT JOIN
(select property_id,
sum(area_rate_count+safety_rate_count
+friendly_rate_count+walkability_rate_count) ssum,
count(*) ct
from rating
group by property_id ) r
ON l.property_id = r.property_id
order by l.property_id
See it here on fiddle: http://sqlfiddle.com/#!2/589d6/5
Edit
As OP asked on the comments that he wants all columns from listing here is what he want:
select l.*, coalesce(r.ssum/if(r.ct=0,1,r.ct), 0) as rate
from listing l LEFT JOIN
(select property_id,
sum(area_rate_count+safety_rate_count
+friendly_rate_count+walkability_rate_count) ssum,
count(*) ct
from rating
group by property_id ) r
ON l.property_id = r.property_id
order by l.property_id
CREATE TEMPORARY TABLE IF NOT EXISTS
temp_table ( INDEX(col_2) )
ENGINE=MyISAM
AS (
SELECT
property_id,
AVG(area_rate_count) as area_rate_count,
AVG(safety_rate_count) as safety_rate_count,
AVG(friendly_rate_count) as friendly_rate_count,
AVG(walkability_rate_count) as walkability_rate_count
FROM rating
GROUP BY property_id
)
SELECT * FROM listing L
JOIN temp_table T
ON L.property_id = T.property_id
Use the below statement to get distinct property_id with its own rate
select property_id, sum(separaterating)/count(property_id) from (
select property_id,sum(area_rate_count , safety_rate_count , friendly_rate_count , walkability_rate_count) as separaterating from rating group by property_id AS temp ) group by
property_id
you can then join with the other table to get the final result as below
select * from ( select property_id, sum(separaterating)/count(property_id) from (
select property_id,sum(area_rate_count , safety_rate_count , friendly_rate_count , walkability_rate_count) as separaterating from rating group by property_id AS temp ) group by
property_id) AS A inner join listing AS B on A.property_id = B.property_id
try this:
select a.prop_id as property_id, l.title, a.allratings / b.numberofreviews as rate
from
(
select property_id as prop_id, SUM(coalesce(area_rate_count,0) + coalesce(safety_rate_count,0) + coalesce(friendly_rate_count,0) + coalesce(walkability_rate_count,0)) as allratings
from rating
group by property_id
) a inner join
(
select property_id, count(distinct user_id) as numberofreviews
from rating
group by property_id
) b on a.property_id = b.property_id
inner join listing l on a.property_id = l.property_id
Try This Query
select ls.property_id,ls.title,inr.rate from listing as ls
left join
(select r.property_id as pid,r.rate/r.cnt as rate from
(select property_id,user_id,(area_rate_count+safefty_rate_count+friendly_rate_count+walkability_rate_count) as rate,count(*) as cnt from rating group by property_id) as r) as inr on inr.pid=ls.property_id