I've encountered a problem with PostgreSQL in Grafana when im searching for name in my DB with accents.
PostgreSQL doesn't search for those names and they don't apear in my dashboard, does anyone knows a way to work arround this problem?
Select vgtuser, count(*)
From (
SELECT DISTINCT(a.ticket_id), max(a.updated_on), b.name as vgtuser
FROM ticket_messages a
INNER JOIN admins b ON a.admin_id=b.admin_id
INNER JOIN ticket_status_history c ON a.ticket_id=c.ticket_id
WHERE a.updated_by in('Ana Monteiro', 'Nuno Gonçalves', 'Henrique Espinha', 'Ricardo Sousa', 'João Fernandes', 'Pedro Pereira', 'Luis Moreno', 'Gonçalo Rodrigues', 'Nuno Coelho') AND c.status_id IN(50, 60, 65)
GROUP BY a.ticket_id, b.name
ORDER BY 1 ) as media
group by 1
ORDER BY 2 desc
To be more precise the names that dosent appear on grafana display are "Nuno Gonçalves" , "João Fernandes", "Gonçalo Rodrigues" all the other names are appearing great, i tried working around the problem using the LIKE clause but the results were not correct and i couldn`t get it to work properly.
Related
This is a current postgres query I have:
sql = """
SELECT
vms.campaign_id,
avg(vms.open_rate_uplift) as open_rate_average,
avg(vms.click_rate_uplift) as click_rate_average,
avg(vms.conversion_rate_uplift) as conversion_rate_average,
avg(cms.incremental_opens),
avg(cms.incremental_clicks),
avg(cms.incremental_conversions)
FROM
experiments.variant_metric_snapshot vms
INNER JOIN experiments.campaign_metric_snapshot cms ON vms.campaign_id = cms.campaign_id
WHERE
vms.campaign_id IN %(campaign_ids)s
GROUP BY
vms.campaign_id
"""
whereby I get the average incremental_opens, incremental_clicks, and incremental_conversions per campaign group from the cms table. However, instead of the average, I want the most recent values for the 3 fields. See the cms table screenshot below - I want the values from the record with the greatest (i.e. most recent) event_id (instead of an average for all records) for a given group).
How can I do this? Thanks
It sounds like you want a lateral join.
FROM
experiments.variant_metric_snapshot vms
CROSS JOIN LATERAL (select * from experiments.campaign_metric_snapshot cms where vms.campaign_id = cms.campaign_id order by event_id desc LIMIT 1) cms
WHERE...
If you are after a quick and dirty solution you can use array_agg function with minimal change to your query.
SELECT
vms.campaign_id,
avg(vms.open_rate_uplift) as open_rate_average,
avg(vms.click_rate_uplift) as click_rate_average,
avg(vms.conversion_rate_uplift) as conversion_rate_average,
(array_agg(cms.incremental_opens ORDER BY cms.event_id DESC))[1] AS incremental_opens,
..
FROM
experiments.variant_metric_snapshot vms
INNER JOIN experiments.campaign_metric_snapshot cms ON vms.campaign_id = cms.campaign_id
WHERE
vms.campaign_id IN %(campaign_ids)s
GROUP BY
vms.campaign_id;
Running out of spool space wondering if the query can be optimized.
I've tried running a DISTINCT and UNION ALL, Group By doesn't make sense.
SELECT DISTINCT T1.EMAIL, T2.BILLG_STATE_CD, T2.BILLG_ZIP_CD
FROM
(SELECT EMAIL
FROM CAT
UNION ALL
SELECT EMAIL
FROM DOG
UNION ALL
SELECT email As EMAIL
FROM MOUSE) As T1
LEFT JOIN HAMSTER As T2 ON T1.EMAIL =T2.EMAIL_ADDR;
I will need to do this same type of data pull often, looking for a viable solution other than doing three separate joins.
I need to union multiple tables (T1) and join columns from another table (T2) on (T1).
WHERE T2.ord_creatd_dt > DATE '2019-01-01' and T2.ord_creatd_dt < DATE '2019-11-08'
I have two postgres tables where one column listing a city name matches. I'm trying to create a view of some records which I'm displaying on a map via WMS on my GeoServer.
I need to select only records from table1 of 100k records that has a city name that matches those cities listed in table2 of 20 records.
To list everything I've tried would be a waste of your time. I've tried every join tutorial and example but, am perplexed why I can't get any success. I would really appreciate some direction.
Here's a latest query but, if this is the wrong approach just ignore since I have about 50 similar attempts.
SELECT t1.id,
t1.dba,
t1.prem_city,
t1.geom
t2.city_label
FROM schema1.table1 AS t1
LEFT JOIN schema2.table2 AS t2
ON t2.city_label = t1.prem_city;
Thanks for any help!
Your query seems correct, just a minor change - LEFT JOIN keeps all the records from the left table and only the matching record from the right one. If you want only those that appear in both - an INNER JOIN is required .
SELECT t1.id,
t1.dba,
t1.prem_city,
t1.geom,
t2.city_label
FROM schema1.table1 t1
JOIN schema2.table2 t2
ON t2.city_label = t1.prem_city;
really struggling with this... I have written the following code that seems to work and identifies the row ID of 40,000 addresses that match where FrontDoorColour is RED.
SELECT DISTINCT ID
FROM Database.table1
WHERE table1.address = table2.address
AND table1.FrontDoorColour = 'RED'
The problem I have is when I want to reverse this and identify the 10,000 addresses where FrontDoorColour is RED but where the address does NOT match.
I run the same query but swap
WHERE table1.address = table2.address
for
WHERE table1.address <> table2.addres
Instead of generating the 10,000 NON-matching rows, I get a spool space error (2646)
Any suggestions would be greatly appreciated!
Thanks
An EXPLAIN output of the second query should yield PRODUCT JOIN and is likely the reason for the spool error you received. The first query may also yield a product join but it may process within your spool allocation. The following SQL should help you find address ids from Table1 where the address is not found in Table2 and the door in Table1 is RED for the address id.
SELECT DISTINCT t1.id
FROM Database.Table1 t1
WHERE NOT EXISTS (SELECT 1
FROM Database.Table2 t2
WHERE t1.address = t2.address)
AND t1.FrontDoorColour = 'RED';
I need to show all clients entered into the system for a date range.
All clients are assigned to a group, but not necessarily to a staff.
When I run the query as such:
SELECT
clients.name_lastfirst_cs,
to_char (clients.date_intake,'MM/DD/YY')AS Date_Created,
clients.client_id,
clients.display_intake,
staff.staff_name_cs,
groups.name
FROM
public.clients,
public.groups,
public.staff,
public.link_group
WHERE
clients.zrud_staff = staff.zzud_staff AND
clients.zzud_client = link_group.zrud_client AND
groups.zzud_group = link_group.zrud_group AND
clients.date_intake BETWEEN (now() - '8 days'::interval)::timestamp AND now()
ORDER BY
groups.name ASC,
clients.client_id ASC,
staff.staff_name_cs ASC
I get 121 entries
if I comment out:
SELECT
clients.name_lastfirst_cs,
to_char (clients.date_intake,'MM/DD/YY')AS Date_Created,
clients.client_id,
clients.display_intake,
-- staff.staff_name_cs, -- Line Commented out
groups.name
FROM
public.clients,
public.groups,
public.staff,
public.link_group
WHERE
-- clients.zrud_staff = staff.zzud_staff AND --Line commented out
clients.zzud_client = link_group.zrud_client AND
groups.zzud_group = link_group.zrud_group AND
clients.date_intake BETWEEN (now() - '8 days'::interval)::timestamp AND now()
ORDER BY
groups.name ASC,
clients.client_id ASC,
staff.staff_name_cs ASC
I get 173 entries
I know I need to do an outer join to capture all clients regardless of if there
is a staff assigned, but each attempt has failed. I have done outer joins with
two tables, but adding a third has twisted my brain.
Thanks for any suggestions
I have no way of testing this (or of knowing that it is right) but what I read in your query is that you want something similar to this:
SELECT --I just used short aliases. I choose something other than the table name so I know it is an alias "c" for client etc...
c.name_lastfirst_cs,
to_char (c.date_intake,'MM/DD/YY')AS Date_Created,
c.client_id,
c.display_intake,
s.staff_name_cs,
g.name,
l.zrud_client AS "link_client",--I'm selecting some data here so that I can debug later, you can just filter this out with another select if you need to
l.zzud_group AS "link_group" --Again, so I can see these relationships
FROM
public.clients c
LEFT OUTER JOIN staff s ON --is staff required? If it isn't then outer join (optional)
s.zzud_staff = c.zrud_staff --so we linked staff to clients here
LEFT OUTER JOIN public.link_group l ON --this looks like a lookup table to me so we select the lookup record
l.zrud_client = c.zzud_client -- this is how I define the lookup, a client id
LEFT OUTER JOIN public.groups g ON --then we use that to lookup a group
g.zzup_group = l.zrud_group --which is defined by this data here
WHERE -- the following must be true
c.date_intake BETWEEN (now() - '8 days'::interval)::timestamp AND now()
Now for the why: I've basically moved your where clause to JOIN x ON y=z syntax. In my experience this is a better way to write an maintain queries as it allows you to specify relationships between tables rather than doing a big-ol'-join and trying to filter that data with the where clause. Keep in mind each condition is REQUIRED not optional so when you say you want records with the following conditions you're going to get them (and if I read this right--I probably don't as I don't have a schema in-front of me) if a record is missing a link-table record OR a staff member you're going to filter it out.
Alternatively (possibly significantly slower) You can SELECT anything so you can chain it like:
SELECT
*
FROM
(
SELECT
*
FROM
public.clients
WHERE
x condition
)
WHERE
y condition
OR
SELECT * FROM x WHERE x.condition IN (SELECT * FROM y)
In your case this tactic probably won't be easier than a standard join syntax.
^And some serious opinion here: I recommend you use the join syntax I outlined above here. It is functionally the same as joining and specifying a where clause, but as you noted, if you don't understand the relationships it can cause a Cartesian join. http://www.tutorialspoint.com/sql/sql-cartesian-joins.htm . Lastly, I tend to specify what type of join I want. I write INNER JOIN and OUTER JOIN a lot in my queries because it helps the next person (usually me) figure out what the heck I meant. If it is optional use an outer join, if it is required use an inner join (default).
Good luck! There are much better SQL developers out there and there's probably another way to do it.