Essentially I need to only LEFT JOIN on the 2 tables if there is one customerid matching in table2. If there is more than 1 record matching table2 it should not count as a match.
Currently I am doing the following:
SELECT DISTINCT
table1.customerid,
table1.name AS customer,
table2.locationid,
table2.locationname
FROM
table1
LEFT JOIN table2 ON table1.customerid = table2.customerid
WHERE
ORDER BY
name ASC
The issue is it matches on all records.
To clarify -- if customerid is in table2 more than once, it should not join on the match, only if customerid is listed once for a record in table2.
How can this be done?
Join with a subquery that only returns customer IDs that have count = 1.
SELECT DISTINCT
table1.customerid,
table1.name AS customer,
table2.locationid,
table2.locationname
FROM table1
LEFT JOIN (
SELECT customerid, MAX(locationid) AS locationid, MAX(locationname) AS locationname
FROM table2
GROUP BY customerid
HAVING COUNT(*) = 1
) AS table2 ON table1.customerid = table2.customerid
ORDER BY name ASC
Note that using LEFT JOIN means that if locationid is not in table2 at all, the query will still return a row for that location, with NULL in the table2 fields. And with this change, it will also return those null rows when there's more than 1 matching row. If you want those rows omitted from the results entirely, use INNER JOIN rather than LEFT JOIN.
Related
I have two tables (table 1 and table 2). Table 1 consists the options and table 2 consists the result for each options.
**table1** table2
id id
optionvalue optionvalueid
time (unixtime)
So when the data is inserted it will be stored in table2. There are 5 optionvalue in table 1 and when data is inserted, then in the table2 it will insert the optionvalueid of table 1 and timestamp in unixtimestamp. Eaach month, I want to count the number of values for each optionvalue. Evene if there is no value for an optionvalue, I still want to see count as zero.
I did the following query but only return the value with rows with data only.
SELECT
po.id,po.optionvalue, COUNT(pr.optionvalueid) as votes,
FROM_UNIXTIME(`time`, '%m-%Y') as ndate
FROM table 1
LEFT JOIN table 2 pr ON po.id=pr.optionvalueid
GROUP BY ndate, po.optionvalue ORDER BY ndate ASC
Is there any other ways to make the query so that it will return all the options even if there is no value.
You can CROSS join table1 to the distinct months of table2 and then LEFT join to table2 to aggregate:
SELECT t.ndate, t1.id, t1.optionvalue, COUNT(t2.optionvalueid) votes
FROM table1 t1
CROSS JOIN (SELECT DISTINCT FROM_UNIXTIME(`time`, '%m-%Y') ndate FROM table2) t
LEFT JOIN table2 t2 ON t1.id = t2.optionvalueid AND t.ndate = FROM_UNIXTIME(t2.`time`, '%m-%Y')
GROUP BY t.ndate, t1.id, t1.optionvalue
ORDER BY t.ndate ASC
I've two tables, table1 contains 22780 rows. Now I left join table1 with table2 (which doesn't contain any duplicates) and I get 23588 rows.
SELECT * FROM Table1
left join Tabelle6 ON CAST(Table1.Customer AS Int) = table2.Customer
Why do I get more rows now? I only need every row from table1 once.
Edit: found my issue, table 2 does contain duplicates. But is there any way to join every row only once and ignore any further matches?
As the comment suggests, the easiest way to handle this would probably be to do SELECT DISTINCT to remove duplicates from your result set:
SELECT DISTINCT
t1.col1,
t1.col2,
t1.Customer,
...
FROM Table1 t1
LEFT JOIN Table2 t2
ON CAST(t1.Customer AS Int) = t2.Customer
But there is another option here. We could also join to a subquery which removes duplicate customers. This would ensure that no record from the first table gets duplicated from matching to more than one record in the second table.
SELECT *
FROM Table1 t1
LEFT JOIN
(
SELECT DISTINCT Customer
FROM Table2
) t2
ON CAST(t1.Customer AS Int) = t2.Customer
Trying to list all what's in table 1 and records under it in table 2
Table one each row has an id , and each row in table 2 has idontable1
select table1.*, count(table2.idintable1)as total
from table1
left join table2 on table1.id=table2.idintable1
WHERE table1.deleted='0' AND table2.deleted=0
group by
table2.idintable1
My current problem is rows from table1 with 0 records in table2 are not displayed
I want them to be displayed
The query that you want is:
select t1.*, count(t2.idintable1) as total
from table1 t1 left join
table2 t2
on t1.id = t1.idintable1 and t2.deleted = 0
where t1.deleted = 0
group by t1.id;
Here are the changes:
The condition on t2.deleted was moved to the on clause. Otherwise, this turns the outer join into an inner join.
The condition on t1.deleted remains in the where clause, because presumably you really do want this as a filter condition.
The group by clause is based on t1.id, because t2.idintable1 will be NULL when there are no matches. Just using t1.id is fine, assuming that id is unique (or a primary key) in table1.
The table aliases are not strictly necessary, but they make queries easier to write and to read.
You should GROUP BY table1.id.
The LEFT JOIN ensures all the rows from table1 appear in the result set. Those that do not have a pair in table2 will appear with NULL in field table2.idintable1. Because of that your original GROUP BY clause produces a single row for all the rows from table1 that do not appear in table2 (instead of one row for each row of table1).
You have fallen into mysql's non-standard group by support trap.
Change your group by to list all columns of table 1:
group by table1.id, table1.name, etc
or list the column positions of all table1 columns in the select:
group by 1, 2, 3, 4, etc
Or use a subquery to get the count vs the id, and join table1 to that.
I want to show two columns summarize data.
table1 - count all fields that the id same as the id on the show_users table.
table2 - sum all values that the id same as the id on the show_users table.
This is my query:
SELECT show_users.id, COUNT(`table1`.id) as sum_fields , SUM(`table2`.count) as count_all
FROM `show_users`
LEFT JOIN `table1` ON `show_users`.id = `table1`.id
LEFT JOIN `table2` ON `show_users`.id = `table2`.id
GROUP by show_users.id
ORDER BY sum_fields DESC
The table2 results are fine, but the table1 count isn't correct values...
Why is that?
SELECT show_users.id, COUNT(DISTINCT `table1`.id) as sum_fields , SUM(`table2`.count) as count_all
Is it possible to get 1 result where I require data from 3 tables.
First table: I will need to grab all the fields (1 row found by a primary key)
Second table: I will need to grab the field 'username' (connected to first table by 'master_id')
Third table: I will need to grab the latest added row with the associated master_id key (table has 'date', 'master_id', 'previous_name').
select top 1 first.*, second.username, third.*
from first
inner join second on first.id = second.master_id
inner join third on first.id = third.master_id
order by
third.date desc
As always there are dozens of ways to skin a cat, I'm not sure if this is optimized as the subquery methods, but it should work.
You can join the three tables together. Then, you can use a "filter" join to keep only the latest Table3 row:
select *
from Table1 t1
join Table2 t2
on t2.master_id = t1.master_id
join Table3 t3
on t3.master_id = t1.master_id
join (
select master_id
, max(date) as max_date
from Table3
group by
master_id
) as filter
on t3.master_id = filter.master_id
and t3.date = filter.max_date
You'll need a correlated subquery for that third table.
SELECT t1.*, username, date, previous_name
FROM FirstTable t1
INNER JOIN SecondTable t2 ON t1.master_id=t2.master_id
INNER JOIN
(SELECT master_id, date, previous_name
FROM ThirdTable AS t3_1
WHERE date = (
SELECT MAX(date)
FROM ThirdTable AS t3_2
WHERE t3_2.master_id=t3_1.master_id)) q1 ON q1.master_id=t1.master_id;
NOTE: Untested.