SELECT alert,
(select created_at from alerts
WHERE alert = #ALERT ORDER BY created_at desc LIMIT 1)
AS latest FROM alerts GROUP BY alert;
I am having an issue with the above query where I would like to pass in each alert into the subquery so that I have a column called latest which displays the latest alert for each group of alerts. How should I do this?
This is called a correlated subquery. To make it work, you need table aliases:
SELECT a1.alert,
(select a2.created_at
from alerts a2
WHERE a2.alert = a1.alert
ORDER BY a2.created_at desc
LIMIT 1
) AS latest
FROM alerts a1
GROUP BY a1.alert;
Table aliases are a good habit to get into, because they often make the SQL more readable. It is also a good idea to use table aliases with column references, so you easily know where the column is coming from.
EDIT:
If you really want the latest, you can get it by simply doing:
select alert, max(created_at)
from alerts
group by alert;
If you are trying to get the latest created_at date for each group of alerts, there is a simpler way.
SELECT
alert,
max (created_at) AS latest
FROM
alerts
GROUP BY
alert;
I would do the following
SELECT
alert_group_name,
MAX(created_at) AS latest
FROM
alerts A
GROUP BY
alert_group_name;
For a correlated subquery, you need to reference an expression from the outer query.
The best way to do that is to assign an alias to the table on the outer query, and then reference that in the inner query. Best practice is to assign an alias to EVERY table reference, and qualify EVERY column reference.
All that needs to be done to "fix" your query is to replace the reference to "#ALERT" with a reference to the alert column from the table on the outer query.
In our shop, that statement would be formatted something like this:
SELECT a.alert
, (SELECT l.created_at
FROM alerts l
WHERE l.alert = a.alert
ORDER BY l.created_at DESC
LIMIT 1
) AS latest
FROM alerts a
GROUP
BY a.alert
Not because that's easier to write that way, but more importantly it's easier to read and understand what the statement is doing.
The correlated subquery approach can be efficient for a small number of rows returned (a very restrictive WHERE clause on the outermost query.) But in general, correlated subqueries in the SELECT list can make for a (what we refer to in our shop) an LDQ "light dimming query".
In our shop, if we needed the resultset returned by that query, that statement would likely be rewritten as:
SELECT a.alert
, MAX(a.created_at) AS latest
FROM alerts a
GROUP
BY a.alert
And we'd definitely have an index defined ON alerts(alert,created_at) (or an index with additional columns after those first two.)
size, we
(I don't anticipate any cases where this statement would return a different result.)
Related
I'm selecting a set of account records from a large table (millions of rows) with integer id values. As basic of a query as one gets, in a sense. What I'm doing us building a large comma separated list, and passing that into the query as an "in" clause. Right now the result is completely unordered. What I'd like to do is get the results back in the order of the values in the "in" clause.
I assume instead I'll have to build a temporary table and do a join instead, which I'd like to avoid, but may not be able to.
Thoughts? The size of the query right now is capped at about 60k each, as we're trying to limit the output size, but it could be arbitrarily large, which might rule out an "in" query anyway from a practical standpoint, if not a physical one.
Thanks in advance.
Actually, this is better:
SELECT * FROM your_table
WHERE id IN (5,2,6,8,12,1)
ORDER BY FIELD(id,5,2,6,8,12,1);
heres the FIELD documentation:
http://dev.mysql.com/doc/refman/5.0/en/string-functions.html#function_field
A bit of a trick....
SELECT * FROM your_table
WHERE id IN (5,2,6,8,12,1)
ORDER BY FIND_IN_SET(id,'5,2,6,8,12,1') DESC;
note that the list of ID's in the find_in_set is a string, so its quoted.
Also note that without DESC, they results are returned in REVERSE order to what the list specified.
If your query is 60K, that's a sign that you're doing it the wrong way.
There is no other way to order the result set than by using an ORDER BY clause. You could have a complicated CASE clause in your order by listing all the elements in your IN clause again, but then your query would probably be 120K.
I know you don't want to, but you should put the values in the IN clause in a table or a temporary table and join with it. You can also include a SortOrder column in the temporary table, and order by that. Databases like joins. Doing it this way will help your query to perform well.
This is what I get for mysql 8.0. It seems opposite to above answers.
sort in same order as list specified:
SELECT * FROM your_table
WHERE id IN (5,2,6,8,12,1)
ORDER BY FIND_IN_SET(id,'5,2,6,8,12,1');
sort in reverse order as list specified:
SELECT * FROM your_table
WHERE id IN (5,2,6,8,12,1)
ORDER BY FIND_IN_SET(id,'5,2,6,8,12,1') DESC;
You're first query surely uses an order by clause. So, you could just do a join, and use the same order by clause.
For example, if this was your first query
SELECT customer_id
FROM customer
WHERE customer_id BETWEEN 1 AND 100
ORDER
BY last_name
And this was your second query
SELECT inventory_id
FROM rental
WHERE customer_id in (...the ordered list...)
Combined would be
SELECT r.inventory_id
FROM rental r
INNER
JOIN customer c
ON r.customer_id = c.customer_id
WHERE c.customer_id BETWEEN 1 AND 100
ORDER
BY c.last_name
This is what worked for me
SELECT * FROM your_table
WHERE id IN ('5','2','6','8','12','1')
ORDER BY FIELD(id,'5','2','6','8','12','1');
I added the ids in quotes
My database is called: (training_session)
I try to print out some information from my data, but I do not want to have any duplicates. I do get it somehow, may someone tell me what I do wrong?
SELECT DISTINCT athlete_id AND duration FROM training_session
SELECT DISTINCT athlete_id, duration FROM training_session
It works perfectly if i use only one column, but when I add another. it does not work.
I think you misunderstood the use of DISTINCT.
There is big difference between using DISTINCT and GROUP BY.
Both have some sort of goal, but they have different purpose.
You use DISTINCT if you want to show a series of columns and never repeat. That means you dont care about calculations or group function aggregates. DISTINCT will show different RESULTS if you keep adding more columns in your SELECT (if the table has many columns)
You use GROUP BY if you want to show "distinctively" on a certain selected columns and you use group function to calculate the data related to it. Therefore you use GROUP BY if you want to use group functions.
Please check group functions you can use in this link.
https://dev.mysql.com/doc/refman/8.0/en/group-by-functions.html
EDIT 1:
It seems like you are trying to get the "latest" of a certain athlete, I'll assume the current scenario if there is no ID.
Here is my alternate solution:
SELECT a.athlete_id ,
( SELECT b.duration
FROM training_session as b
WHERE b.athlete_id = a.athlete_id -- connect
ORDER BY [latest column to sort] DESC
LIMIT 1
) last_duration
FROM training_session as a
GROUP BY a.athlete_id
ORDER BY a.athlete_id
This syntax is called IN-SELECT subquery. With the help of LIMIT 1, it shows the topmost record. In-select subquery must have 1 record to return or else it shows error.
MySQL's DISTINCT clause is used to filter out duplicate recordsets.
If your query was SELECT DISTINCT athlete_id FROM training_session then your output would be:
athlete_id
----------
1
2
3
4
5
6
As soon as you add another column to your query (in your example, the column called duration) then each record resulting from your query are unique, hence the results you're getting. In other words the query is working correctly.
I'm trying to get the highest version within a group. My query:
SELECT
rubric_id,
max(version) as version,
group_id
FROM
rubrics
WHERE
client_id = 1
GROUP BY
group_id
The Data:
The Results:
The rubric of ID 2 does not have a version of 2, why is this being mismatched? What do I need to do to correct this?
Edit, not a duplicate:
This is not a duplicate of SQL Select only rows with Max Value on a Column , which is a post I have read and referenced before writing this. My question is not how to find the max, my question is why is the version not matched to the correct ID
MySQL is confusing you by letting you get away with having a column in your select that isn't in your group by. To resolve the issue, make sure you don't select any field that isn't in the group by.
Instead of trying to get everything in one statement, you will need to use a subquery to find the max_version_id and then join to it.
SELECT T.*
FROM rubrics T
JOIN
(
SELECT
group_id,
max(version) as max_version
FROM
rubrics
GROUP BY
group_id
) dedupe
on T.group_id = dedupe.group_id
and T.version_id = dedupe.max_version_id
WHERE
T.client_id = 1
Edit: So MySQL allows it, but I don't think it's a good practise to use it.
You are trying to query non-aggregated data from an aggregated query. You should not do that.
A GROUP BY takes the field it should make group of rows with (in your case, what you say with your GROUP BY is: give me a result per different group_id) and gives a result (the aggregated data) based on the grouping.
Here, you try to access non aggregated data (rubric_id in your case). For some reason, the query does not crash and picks a "random" id in your aggregated data.
Is it possible to have count in the select clause with a group by which is suppressed in the count? I need the count to ignore the group by clause
I got this query which is counting the total entries. The query is generic generated and therefore I can't make any comprehensive changes like subqueries etc.
In some specific cases a group by is needed to retrieve the correct rows and because of this the group by can't be removed
SELECT count(dv.id) num
FROM `data_voucher` dv
LEFT JOIN `data_voucher_enclosure` de ON de.data_voucher_id=dv.id
WHERE IF(de.id IS NULL,0,1)=0
GROUP BY dv.id
Is it possible to have count in the select clause with a group by which is suppressed in the count? I need the count to ignore the group by clause
well, the answer to your question is simply you can't have an aggregate that works on all the results, while having a group by statement. That's the whole purpose of the group by to create groups that change the behaviour of aggregates:
The GROUP BY clause causes aggregations to occur in groups (naturally) for the columns you name.
cf this blog post which is only the first result I found on google on this topic.
You'd need to redesign your query, the easiest way being to create a subquery, or a hell of a jointure. But without the schema and a little context on what you want this query to do, I can't give you an alternative that works.
I just can tell you that you're trying to use a hammer to tighten a screw...
Have found an alternative where COUNT DISTINCT is used
SELECT count(distinct dv.id) num
FROM `data_voucher` dv
LEFT JOIN `data_voucher_enclosure` de ON de.data_voucher_id=dv.id
WHERE IF(de.id IS NULL,0,1)=0
Is it possible to order the GROUP BY chosen results of a MySQL query w/out using a subquery? I'm finding that, with my large dataset, the subquery adds a significant amount of load time to my query.
Here is a similar situation: how to sort order of LEFT JOIN in SQL query?
This is my code that works, but it takes way too long to load:
SELECT tags.contact_id, n.last
FROM tags
LEFT JOIN ( SELECT * FROM names ORDER BY timestamp DESC ) n
ON (n.contact_id=tags.contact_id)
WHERE tags.tag='$tag'
GROUP BY tags.contact_id
ORDER BY n.last ASC;
I can get a fast result doing a simple join w/ a table name, but the "group by" command gives me the first row of the joined table, not the last row.
I'm not really sure what you're trying to do. Here are some of the problems with your query:
selecting n.last, although it is neither in the group by clause, nor an aggregate value. Although MySQL allows this, it's really not a good idea to take advantage of.
needlessly sorting a table before joining, instead of just joining
the subquery isn't really doing anything
I would suggest carefully writing down the desired query results, i.e. "I want the contact id and latest date for each tag" or something similar. It's possible that will lead to a natural, easy-to-write and semantically correct query that is also more efficient than what you showed in the OP.
To answer the question "is it possible to order a GROUP BY query": yes, it's quite easy, and here's an example:
select a, b, sum(c) as `c sum`
from <table_name>
group by a,b
order by `c sum`
You are doing a LEFT JOIN on contact ID which implies you want all tag contacts REGARDLESS of finding a match in the names table. Is that really the case, or will the tags table ALWAYS have a "Names" contact ID record. Additionally, your column "n.Last". Is this the person's last name, or last time something done (which I would assume is actually the timestamp)...
So, that being said, I would just do a simple direct join
SELECT DISTINCT
t.contact_id,
n.last
FROM
tags t
JOIN names n
ON t.contact_id = n.contact_id
WHERE
t.tag = '$tag'
ORDER BY
n.last ASC