I'm joining two tables and making a simple count, but I can't seem to rename the joined key variable into something more appropriate for the two tables, I keep getting the error '"CUSTOMER_NO" is not valid in the context where it is used.' I'm sure it's just a little syntax error, but I can't see it...
SELECT owner_no AS customer_no,
CASE
WHEN customer_no BETWEEN 5000 and 5999 THEN 'RENTER'
WHEN customer_no BETWEEN 6000 and 6999 THEN 'OWNER'
END AS customer_type
FROM owner_phone AS op
INNER JOIN renter_phone AS rp ON op.owner_no = rp.renter_no
GROUP BY customer_no
HAVING COUNT(*) > 1;
Use the actual column name in your CASE and GROUP BY, not the aliased column name.
CASE
WHEN owner_no BETWEEN 5000 and 5999 THEN 'RENTER'
WHEN owner_no BETWEEN 6000 and 6999 THEN 'OWNER'
END AS customer_type
FROM owner_phone AS op
INNER JOIN renter_phone AS rp ON op.owner_no = rp.renter_no
GROUP BY owner_no
HAVING Count(*) > 1;
You have to use OWNER_NO through the rest of your query but leave the AS CUSTOMER_NO to make that the column name.
SELECT owner_no AS customer_no,
CASE
WHEN owner_no BETWEEN 5000 and 5999 THEN 'RENTER'
WHEN owner_no BETWEEN 6000 and 6999 THEN 'OWNER'
END AS customer_type
FROM owner_phone AS op
INNER JOIN renter_phone AS rp ON op.owner_no = rp.renter_no
GROUP BY owner_no
HAVING COUNT(*) > 1;
Related
I am trying to achieve the following with Laravel Query builder.
I have a table called deals . Below is the basic schema
id
deal_id
merchant_id
status
deal_text
timestamps
I also have another table called merchants whose schema is
id
merchant_id
merchant_name
about
timestamps
Currently I am getting deals using the following query
$deals = DB::table('deals')
-> join ('merchants', 'deals.merchant_id', '=', 'merchants.merchant_id')
-> where ('merchant_url_text', $merchant_url_text)
-> get();
Since only 1 merchant is associated with a deal, I am getting deals and related merchant info with the query.
Now I have a 3rd table called tbl_deal_votes. Its schema looks like
id
deal_id
vote (1 if voted up, 0 if voted down)
timestamps
What I want to do is join this 3rd table (on deal_id) to my existing query and be able to also get the upvotes and down votes each deal has received.
To do this in a single query you'll probably need to use SQL subqueries, which doesn't seem to have good fluent query support in Laravel 4/5. Since you're not using Eloquent objects, the raw SQL is probably easiest to read. (Note the below example ignores your deals.deal_id and merchants.merchant_id columns, which can likely be dropped. Instead it just uses your deals.id and merchants.id fields by convention.)
$deals = DB::select(
DB::raw('
SELECT
deals.id AS deal_id,
deals.status,
deals.deal_text,
merchants.id AS merchant_id,
merchants.merchant_name,
merchants.about,
COALESCE(tbl_upvotes.upvotes_count, 0) AS upvotes_count,
COALESCE(tbl_downvotes.downvotes_count, 0) AS downvotes_count
FROM
deals
JOIN merchants ON (merchants.id = deals.merchant_id)
LEFT JOIN (
SELECT deal_id, count(*) AS upvotes_count
FROM tbl_deal_votes
WHERE vote = 1 && deal_id
GROUP BY deal_id
) tbl_upvotes ON (tbl_upvotes.deal_id = deals.id)
LEFT JOIN (
SELECT deal_id, count(*) AS downvotes_count
FROM tbl_deal_votes
WHERE vote = 0
GROUP BY deal_id
) tbl_downvotes ON (tbl_downvotes.deal_id = deals.id)
')
);
If you'd prefer to use fluent, this should work:
$upvotes_subquery = '
SELECT deal_id, count(*) AS upvotes_count
FROM tbl_deal_votes
WHERE vote = 1
GROUP BY deal_id';
$downvotes_subquery = '
SELECT deal_id, count(*) AS downvotes_count
FROM tbl_deal_votes
WHERE vote = 0
GROUP BY deal_id';
$deals = DB::table('deals')
->select([
DB::raw('deals.id AS deal_id'),
'deals.status',
'deals.deal_text',
DB::raw('merchants.id AS merchant_id'),
'merchants.merchant_name',
'merchants.about',
DB::raw('COALESCE(tbl_upvotes.upvotes_count, 0) AS upvotes_count'),
DB::raw('COALESCE(tbl_downvotes.downvotes_count, 0) AS downvotes_count')
])
->join('merchants', 'merchants.id', '=', 'deals.merchant_id')
->leftJoin(DB::raw('(' . $upvotes_subquery . ') tbl_upvotes'), function($join) {
$join->on('tbl_upvotes.deal_id', '=', 'deals.id');
})
->leftJoin(DB::raw('(' . $downvotes_subquery . ') tbl_downvotes'), function($join) {
$join->on('tbl_downvotes.deal_id', '=', 'deals.id');
})
->get();
A few notes about the fluent query:
Used the DB::raw() method to rename a few selected columns.
Otherwise, there would have been a conflict between deals.id
and merchants.id in the results.
Used COALESCE to default null votes to 0.
Split the subqueries into separate PHP strings to improve readability.
Used left joins for the subqueries so deals with no upvotes/downvotes still show up.
Thanks in advance for any help with this, it is highly appreciated.
So, basically, I have a Greenplum database and I am wanting to select the table size for the top 10 largest tables. This isn't a problem using the below:
select
sotaidschemaname schema_name
,sotaidtablename table_name
,pg_size_pretty(sotaidtablesize) table_size
from gp_toolkit.gp_size_of_table_and_indexes_disk
order by 3 desc
limit 10
;
However I have several partitioned tables in my database and these show up with the above sql as all their 'child tables' split up into small fragments (though I know they accumalate to make the largest 2 tables). Is there a way of making a script that selects tables (partitioned or otherwise) and their total size?
Note: I'd be happy to include some sort of join where I specify the partitoned table-name specifically as there are only 2 partitioned tables. However, I would still need to take the top 10 (where I cannot assume the partitioned table(s) are up there) and I cannot specify any other table names since there are near a thousand of them.
Thanks again,
Vinny.
Your friends would be pg_relation_size() function for getting relation size and you would select pg_class, pg_namespace and pg_partition joining them together like this:
select schemaname,
tablename,
sum(size_mb) as size_mb,
sum(num_partitions) as num_partitions
from (
select coalesce(p.schemaname, n.nspname) as schemaname,
coalesce(p.tablename, c.relname) as tablename,
1 as num_partitions,
pg_relation_size(n.nspname || '.' || c.relname)/1000000. as size_mb
from pg_class as c
inner join pg_namespace as n on c.relnamespace = n.oid
left join pg_partitions as p on c.relname = p.partitiontablename and n.nspname = p.partitionschemaname
) as q
group by 1, 2
order by 3 desc
limit 10;
select * from
(
select schemaname,tablename,
pg_relation_size(schemaname||'.'||tablename) as Size_In_Bytes
from pg_tables
where schemaname||'.'||tablename not in (select schemaname||'.'||partitiontablename from pg_partitions)
and schemaname||'.'||tablename not in (select distinct schemaname||'.'||tablename from pg_partitions )
union all
select schemaname,tablename,
sum(pg_relation_size(schemaname||'.'||partitiontablename)) as Size_In_Bytes
from pg_partitions
group by 1,2) as foo
where Size_In_Bytes >= '0' order by 3 desc;
The following code gets all the residences which have all the amenities which are listed in id_list. It works with out a problem with SQLite but raises an error with PostgreSQL:
id_list = [48, 49]
Residence.joins(:listed_amenities).
where(listed_amenities: {amenity_id: id_list}).
references(:listed_amenities).
group(:residence_id).
having("count(*) = ?", id_list.size)
The error on the PostgreSQL version:
What do I have to change to make it work with PostgreSQL?
A few things:
references should only be used with includes; it tells ActiveRecord to perform a join, so it's redundant when using an explicit joins.
You need to fully qualify the argument to group, i.e. group('residences.id').
For example,
id_list = [48, 49]
Residence.joins(:listed_amenities).
where(listed_amenities: { amenity_id: id_list }).
group('residences.id').
having('COUNT(*) = ?", id_list.size)
The query the Ruby (?) code is expanded to is selecting all fields from the residences table:
SELECT "residences".*
FROM "residences"
INNER JOIN "listed_amenities"
ON "listed_amentities"."residence_id" = "residences"."id"
WHERE "listed_amenities"."amenity_id" IN (48,49)
GROUP BY "residence_id"
HAVING count(*) = 2
ORDER BY "residences"."id" ASC
LIMIT 1;
From the Postgres manual, When GROUP BY is present, it is not valid for the SELECT list expressions to refer to ungrouped columns except within aggregate functions or if the ungrouped column is functionally dependent on the grouped columns, since there would otherwise be more than one possible value to return for an ungrouped column.
You'll need to either group by all fields that aggregate functions aren't applied to, or do this differently. From the query, it looks like you only need to scan the amentities table to get the residence ID you're looking for:
SELECT "residence_id"
FROM "listed_amenities"
WHERE "listed_amenities"."amenity_id" IN (48,49)
GROUP BY "residence_id"
HAVING count(*) = 2
ORDER BY "residences"."id" ASC
LIMIT 1
And then fetch your residence data with that ID. Or, in one query:
SELECT "residences".*
FROM "residences"
WHERE "id" IN (SELECT "residence_id"
FROM "listed_amenities"
WHERE "listed_amenities"."amenity_id" IN (48,49)
GROUP BY "residence_id"
HAVING count(*) = 2
ORDER BY "residences"."id" ASC
LIMIT 1
);
I have two tables in hive:
Table1: uid,txid,amt,vendor Table2: uid,txid
Now I need to join the tables on txid which basically confirms a transaction is finally recorded. There will be some transactions which will be present only in Table1 and not in Table2.
I need to find out number of avg of transaction matches found per user(uid) per vendor. Then I need to find the avg of these averages by adding all the averages and divide them by the number of unique users per vendor.
Let's say I have the data:
Table1:
u1,120,44,vend1
u1,199,33,vend1
u1,100,23,vend1
u1,101,24,vend1
u2,200,34,vend1
u2,202,32,vend2
Table2:
u1,100
u1,101
u2,200
u2,202
Example For vendor vend1:
u1-> Avg transaction find rate = 2(matches found in both Tables,Table1 and Table2)/4(total occurrence in Table1) =0.5
u2 -> Avg transaction find rate = 1/1 = 1
Avg of avgs = 0.5+1(sum of avgs)/2(total unique users) = 0.75
Required output:
vend1,0.75
vend2,1
I can't seem to find count of both matches and occurrence in just Table1 in one hive query per user per vendor. I have reached to this query and can't find how to change it further.
SELECT A.vendor,A.uid,count(*) as totalmatchesperuser FROM Table1 A JOIN Table2 B ON A.uid = B.uid AND B.txid =A.txid group by vendor,A.uid
Any help would be great.
I think you are running into trouble with your JOIN. When you JOIN by txid and uid, you are losing the total number of uid's per group. If I were you I would assign a column of 1's to table2 and name the column something like success or transaction and do a LEFT OUTER JOIN. Then in your new table you will have a column with the number 1 in it if there was a completed transaction and NULL otherwise. You can then do a case statement to convert these NULLs to 0
Query:
select vendor
,(SUM(avg_uid) / COUNT(uid)) as avg_of_avgs
from (
select vendor
,uid
,AVG(complete) as avg_uid
from (
select uid
,txid
,amt
,vendor
,case when success is null then 0
else success
end as complete
from (
select A.*
,B.success
from table1 as A
LEFT OUTER JOIN table2 as B
ON B.txid = A.txid
) x
) y
group by vendor, uid
) z
group by vendor
Output:
vend1 0.75
vend2 1.0
B.success in line 17 is the column of 1's that I put int table2 before the JOIN. If you are curious about case statements in Hive you can find them here
Amazing and precise answer by GoBrewers14!! Thank you so much. I was looking at it from a wrong perspective.
I made little changes in the query to get things finally done.
I didn't need to add a "success" colummn to table2. I checked B.txid in the above query instead of B.success. B.txid will be null in case a match is not found and be some value if a match is found. That checks the success & failure conditions itself without adding a new column. And then I set NULL as 0 and !NULL as 1 in the part above it. Also I changed some variable names as hive was finding it ambiguous.
The final query looks like :
select vendr
,(SUM(avg_uid) / COUNT(usrid)) as avg_of_avgs
from (
select vendr
,usrid
,AVG(complete) as avg_uid
from (
select usrid
,txnid
,amnt
,vendr
,case when success is null then 0
else 1
end as complete
from (
select A.uid as usrid,A.vendor as vendr,A.amt as amnt,A.txid as txnid
,B.txid as success
from Table1 as A
LEFT OUTER JOIN Table2 as B
ON B.txid = A.txid
) x
) y
group by vendr, usrid
) z
group by vendr;
I’m working in SQL Server with the following sample problem. Brandon prefers PC’s and Macs, Sue prefers PC’s only, and Alan Prefers Macs. The data would be represented something like this. I've had to make some compromises here but hopefully you get the picture:
TABLE 1: User
uID (INT PK), uName (VARCHAR)
1 'Brandon'
2 'Sue'
3 'Alan'
TABLE 2: Computer
cID (INT PK), cName (VARCHAR)
1 'Mac'
2 'PC'
TABLE 3: UCPref --Stores the computer preferences for each user
uID (INT FK), cID (INT FK)
1 1
1 2
2 1
3 2
Now, if I want to select everyone who likes PC’s OR Macs that would be quite easy. There's a dozen ways to do it, but if I'm having a list of items fed in, then the IN clause is quite straight-forward:
SELECT u.uName
FROM User u
INNER JOIN UCPref p ON u.uID = p.uID
WHERE cID IN (1,2)
The problem I have is, what happens when I ONLY want to select people who like BOTH PC’s AND Mac’s? I can do it in multiple sub queries, however that isn’t very scalable.
SELECT u.uName
FROM User u
INNER JOIN UCPref p ON u.uID = p.uID
WHERE u.uID IN (SELECT uID FROM UCPref WHERE cID = 1)
AND u.uID IN (SELECT uID FROM UCPref WHERE cID = 2)
How does one write this query such that you can return the users who prefer multiple computers taking into consideration that there may be hundreds, maybe thousands of different kinds of computers (meaning no sub queries)? If only you could modify the IN clause to have a key word like 'ALL' to indicate that you want to match only those records that have all of the items in the parenthesis?
SELECT u.uName
FROM User u
INNER JOIN UCPref p ON u.uID = p.uID
WHERE cID IN *ALL* (1,2)
Using JOINs:
SELECT u.uname
FROM USERS u
JOIN UCPREF ucp ON ucp.uid = u.uid
JOIN COMPUTER mac ON mac.cid = ucp.cid
AND mac.cname = 'Mac'
JOIN COMPUTER pc ON pc.cid = ucp.cid
AND pc.cname = 'PC'
I'm using table aliases, because I'm JOINing onto the same table twice.
Using EXISTS:
SELECT u.uname
FROM USERS u
JOIN UCPREF ucp ON ucp.uid = u.uid
WHERE EXISTS (SELECT NULL
FROM COMPUTER c
WHERE c.cid = ucp.cid
AND c.cid IN (1, 2)
GROUP BY c.cid
HAVING COUNT(*) = 2)
If you're going to use the IN clause, you have to use GROUP BY/HAVING but there is a risk in the COUNT(). Some db's don't allow more than the *, while MySQL allows DISTINCT .... The problem is that if you can't use DISTINCT in the count, you could get two instances of the value 2, and it would valid to SQL - giving you a false positive.