I have two tables.
Invoices
ID | Amount
-----------
1 | 123.54
2 | 553.46
3 | 431.34
4 | 321.31
5 | 983.12
Credit Memos
ID | invoice_ID | Amount
------------------------
1 | 3 | 25.50
2 | 95 | 65.69
3 | 51 | 42.50
I want to get a result set like this out of those two tables
ID | Amount | Cr_memo
---------------------
1 | 123.54 |
2 | 553.46 |
3 | 431.34 | 25.50
4 | 321.31 |
5 | 983.12 |
I've been messing with joins and whatnot all morning with no real luck.
Here is the last query I tried, which pulled everything from the Credit Memo table...
SELECT A.ID, A.Amount FROM Invoices AS A
LEFT JOIN Credit_Memos AS B ON A.ID = B.invoice_ID
Any help or pointers are appreciated.
Your query would work fine. Just add Credit_memo.Amount with an alias:
SELECT Inv.ID,Inv.Amount,IFNULL(C.Amount,'') AS Cr_memo
FROM Invoices Inv LEFT JOIN
Credit_Memos C ON Inv.ID=C.invoice_ID
Result:
ID AMOUNT CR_MEMO
1 124
2 553
3 431 25.50
4 321
5 983
See result in SQL FIDDLE.
You almost got the answer Left Outer Join is what you need but you missed to select Cr_memo from Credit_Memos table. Since you don't want to show Null values when there is no Invoices_ID in Credit Memos table use IFNULL to make NULL's as Empty string
SELECT A.ID, A.Amount, IFNULL(B.Cr_memo,'') AS Cr_memo
FROM Invoices AS A
LEFT JOIN Credit_Memos AS B
ON A.ID = B.invoice_ID
The LEFT JOIN keyword returns all rows from the left table (table1), with the matching rows in the right table (table2). The result is NULL in the right side when there is no match.
SELECT A.ID, A.Amount, IFNULL(B.amount,0) AS Cr_memo FROM Invoices AS A
LEFT JOIN Credit_Memos AS B ON A.ID = B.invoice_ID
here is some useful link about left join link and another
Related
BOOKINGS TABLE
id | price | anotherVal
-----------------------------
1 | 10000 | *
2 | 20000 | *
3 | 1000 | *
4 | 8000 | *
BOOKING PAYMENTS TABLE
id | bookingId | amount | currencyId | mxnAmount
--------------------------------------------------
1 | 1 | 100.00 | 1 | 100.00
2 | 1 | 300.00 | 3 | 6400.00
3 | 2 | 500.21 | 1 | 500.21
4 | 4 | 123.95 | 6 |
4 | 4 | 800.00 | 1 | 800.00
I need to get all BOOKINGS_TABLE columns and then for each booking add up the mxnAmount column, but also the result should tell if all rows in BOOKING_PAYMENTS_TABLE had an mxnAmount so i can know if the mxnAmount is final or there's some rows left to be updated, i have a query that works for the first part:
SELECT b.*, SUM(p.mxnAmount) FROM bookings b LEFT JOIN bookingPayments p ON b.id = p.bookingId GROUP BY b.id
I figured i could make us of COUNT() to count all rows in BOOKING_PAYMENTS_TABLE but then how can i get the number for the rows that have an mxnAMOUNT?
SELECT b.*, SUM(p.mxnAmount), COUNT(p.id) FROM bookings b LEFT JOIN bookingPayments p ON b.id = p.bookingId GROUP BY b.id
I tried this:
SELECT b.*, SUM(p.mxnAmount), COUNT(p.id), COUNT(pp.id) FROM bookings b LEFT JOIN bookingPayments p ON b.id = p.bookingId LEFT JOIN bookingPayments pp ON b.id = pp.bookingId WHERE pp.mxnAmount IS NOT NULL GROUP BY b.id
But then the query returns only bookings that have all their payments rows with an mxnAmount, any leads?
I figured i could make us of COUNT() to count all rows in BOOKING_PAYMENTS_TABLE but then how can i get the number for the rows that have an mxnAMOUNT?
Just COUNT() that particular column: this gives you the number of non-null values in the column for each group:
SELECT b.*, SUM(p.mxnAmount), COUNT(p.id), COUNT(p.mxnAmount)
FROM bookings b
LEFT JOIN bookingPayments p ON b.id = p.bookingId
GROUP BY b.id
If you want to know if any mxmamount in the group is missing, you can do:
MAX(p.id IS NOT NULL AND p.mxnAmount IS NULL) has_missing_mxnAmount
I've been trying for two days, without luck.
I have the following simplified tables in my database:
customers:
| id | name |
| 1 | andrea |
| 2 | marco |
| 3 | giovanni |
access:
| id | name_id | date |
| 1 | 1 | 5000 |
| 2 | 1 | 4000 |
| 3 | 2 | 1500 |
| 4 | 2 | 3000 |
| 5 | 2 | 1000 |
| 6 | 3 | 6000 |
| 7 | 3 | 2000 |
I want to return all the names with their last access date.
At first I tried simply with
SELECT * FROM customers LEFT JOIN access ON customers.id =
access.name_id
But I got 7 rows instead of 3 as expected. So I understood I need to use GROUP BY statemet as the following:
SELECT * FROM customers LEFT JOIN access ON customers.id =
access.name_id GROUP BY customers.id
As far I know, GROUP BY combines using a random row. In fact I got unordered access dates with several tests.
Instead I need to group every customer id with its corresponding latest access! How this can be done?
You have to get the latest date from the access table with a group by on the the name_id, then join this result with the customer table. Here is the query:
select c.id, c.name, a.last_access_date from customers c left join
(select id, name_id, max(access_date) last_access_date from access group by name_id) a
on c.id=a.name_id;
Here is a DEMO on sqlfiddle.
I think this is what you'd like to achieve:
SELECT c.id, c.name, max(a.date) last_access
FROM customers c
LEFT JOIN access a ON c.id = a.name_id
GROUP BY c.id, c.name
The LEFT join will return all entries in table customers regardless if the join criteria (c.id = a.name_id) is satisfied. This means that you might get some NULL entries.
Example:
Simply add a new row in the customers table (id: 4, name: manuela). The output will have 4 rows and the newest row will be (id: 4, last_access: null)
I would do this using a correlated subquery in the ON clause:
SELECT a.*, c.*
FROM customers c LEFT JOIN
access a
ON c.id = a.name_id AND
a.DATE = (SELECT MAX(a2.date) FROM access a2 WHERE a2.name_id = a.name_id);
If this statement is true:
I need to group every customer id with its corresponding latest access! How this can be done?
Then you can simply do:
select a.name_id, max(a2.date)
from access a
group by a.name_id;
You do not need the customers table because:
All customers are in access, so the left join is not necessary.
You need no columns from customers.
I have many forms that users fill out. Each form contains a list of questions. In this first table is the form id and the id's of the questions.
form_id | question_id
1 | 1
1 | 2
1 | 3
2 | 4
2 | 5
This table has two forms one which has 3 questions and the other 2. I have a second table which has the answers that the users have given for the questions.
user_id | form_id | question_id | answer
476 | 1 | 1 | "answer1"
476 | 1 | 3 | "answer2"
693 | 1 | 1 | "answer3"
693 | 1 | 2 | "answer4"
235 | 2 | 5 | "answer5"
In this example, 2 users have filled out form 1 and 1 user has filled in form 2. But none have filled in all the questions. Is it possible to write a query which combines the two tables and will give me the answers that the user have given including the questions that they didn't answer? I'd like the results to look like this.
user_id | form_id | question_id | answer
476 | 1 | 1 | "answer1"
476 | 1 | 2 | NULL
476 | 1 | 3 | "answer2"
693 | 1 | 1 | "answer3"
693 | 1 | 2 | "answer4"
693 | 1 | 3 | NULL
235 | 2 | 4 | NULL
235 | 2 | 5 | "answer5"
The problem that I have when I use a left join like this
select * from template t
left join answers a on a.template_id = t.template_id
AND a.question_id = t.question_id
AND t.template_id = t.template_id;
is that the row that results is missing user_id.
Yes, the specified result can be returned by a query.
One way to achieve this is a join to an inline view, and an "outer join" operation to the second table.
The "trick" is getting a distinct list of user_id and form_id from the second table, using a query, for example:
SELECT user_id, form_id
FROM second_table
GROUP BY user_id, form_id
And then using that query as an inline view (wrapping it in parens, assigning a table alias, and referencing it like it was a table in an outer query.
All that's required after that is an "outer join" to the second table.
For example:
SELECT r.user_id
, q.form_id
, q.question_id
, a.answer
FROM first_table q
JOIN ( SELECT p.user_id, p.form_id
FROM second_table p
GROUP BY p.user_id, p.form_id
) r
ON r.form_id = q.form_id
LEFT
JOIN second_table a
ON a.user_id = r.user_id
AND a.form_id = r.form_id
AND a.question_id = q.question_id
ORDER
BY r.user_id
, q.form_id
, q.question_id
Note that the keyword "LEFT" specifies an outer join operation, returning all rows from the left side, along with matching rows from the right side. A typical "inner" join would exclude rows that didn't find a matching row from the table on the right side.
use
left join
something like:
select * from table1 left join table2 on table1.form_id= table2.form_id
How to select 2 table with condition and show all data
store_profile
id + store_name
1 | Accessorize.me
2 | Active IT
3 | Edushop
4 | Gift2Kids
5 | Heavyarm
6 | Bamboo
store_fee
id + store_id + date_end
1 | 1 | 27-6-2013
2 | 2 | 29-8-2013
3 | 3 | 02-6-2013
4 | 4 | 20-4-2013
Below is my previous query
$query = "select sp.id, sp.store_name, sf.id, sf.store_id, sf.date_end from store_profile sp, store_fee sf where sf.store_id=sp.id"
and the result is something like this :
1 | Accessorize.me 27-6-2013
2 | Active IT 29-8-2013
3 | Edushop 02-6-2013
4 | Gift2Kids 20-4-2013
but what i want is show all store name including date_end but if no date_end still can show store name with empty date_end
You want to use an outer join. With an outer join, columns on the joining table do not need to match the conditional columns in the joined table to get results:
SELECT * FROM store_profile sp LEFT JOIN store_fee sf ON (sf.store_id = sp.id)
Use a left join:
select sp.id, sp.store_name, sf.id, sf.store_id, sf.date_end
from store_profile sp left join store_fee sf on sf.store_id=sp.id
The syntax you are using is interpreted as INNER JOIN, which means that stores without a corresponding entry in store_profile won't show up. You want to use LEFT JOIN:
SELECT sp.id, sp.store_name, sf.id, sf.store_id, sf.date_end
FROM store_profile sp
LEFT JOIN store_fee sf
ON sf.store_id=sp.id
LEFT JOIN means that all records in the first table will be returned, even if there's not a match in the second table.
I feel like I was always taught to use LEFT JOINs and I often see them mixed with INNERs to accomplish the same type of query throughout several pieces of code that are supposed to do the same thing on different pages. Here goes:
SELECT ac.reac, pt.pt_name, soc.soc_name, pt.pt_soc_code
FROM
AECounts ac
INNER JOIN 1_low_level_term llt on ac.reac = llt.llt_name
LEFT JOIN 1_pref_term pt ON llt.pt_code = pt.pt_code
LEFT JOIN 1_soc_term soc ON pt.pt_soc_code = soc.soc_code
LIMIT 100,10000
Thats one I am working on:
I see a lot like:
SELECT COUNT(DISTINCT p.`case`) as count
FROM FDA_CaseReports cr
INNER JOIN ae_indi i ON i.isr = cr.isr
LEFT JOIN ae_case_profile p ON cr.isr = p.isr
This seems like the LEFT may as well be INNER is there any catch?
Is there any catch? Yes there is -- left joins are a form of outer join, while inner joins are a form of, well, inner join.
Here's examples that show the difference. We'll start with the base data:
mysql> select * from j1;
+----+------------+
| id | thing |
+----+------------+
| 1 | hi |
| 2 | hello |
| 3 | guten tag |
| 4 | ciao |
| 5 | buongiorno |
+----+------------+
mysql> select * from j2;
+----+-----------+
| id | thing |
+----+-----------+
| 1 | bye |
| 3 | tschau |
| 4 | au revoir |
| 6 | so long |
| 7 | tschuessi |
+----+-----------+
And here we'll see the difference between an inner join and a left join:
mysql> select * from j1 inner join j2 on j1.id = j2.id;
+----+-----------+----+-----------+
| id | thing | id | thing |
+----+-----------+----+-----------+
| 1 | hi | 1 | bye |
| 3 | guten tag | 3 | tschau |
| 4 | ciao | 4 | au revoir |
+----+-----------+----+-----------+
Hmm, 3 rows.
mysql> select * from j1 left join j2 on j1.id = j2.id;
+----+------------+------+-----------+
| id | thing | id | thing |
+----+------------+------+-----------+
| 1 | hi | 1 | bye |
| 2 | hello | NULL | NULL |
| 3 | guten tag | 3 | tschau |
| 4 | ciao | 4 | au revoir |
| 5 | buongiorno | NULL | NULL |
+----+------------+------+-----------+
Wow, 5 rows! What happened?
Outer joins such as left join preserve rows that don't match -- so rows with id 2 and 5 are preserved by the left join query. The remaining columns are filled in with NULL.
In other words, left and inner joins are not interchangeable.
Here's a rough answer, that is sort of how I think about joins. Hoping this will be more helpful than a very precise answer due to the aforementioned math issues... ;-)
Inner joins narrow down the set of rows returns. Outer joins (left or right) don't change number of rows returned, but just "pick up" additional columns if possible.
In your first example, the result will be rows from AECounts that match the conditions specified to the 1_low_level_term table. Then for those rows, it tries to join to 1_pref_term and 1_soc_term. But if there's no match, the rows remain and the joined in columns are null.
An INNER JOIN will only return the rows where there are matching values in both tables, whereas a LEFT JOIN will return ALL the rows from the LEFT table even if there is no matching row in the RIGHT table
A quick example
TableA
ID Value
1 TableA.Value1
2 TableA.Value2
3 TableA.Value3
TableB
ID Value
2 TableB.ValueB
3 TableB.ValueC
An INNER JOIN produces:
SELECT a.ID,a.Value,b.ID,b.Value
FROM TableA a INNER JOIN TableB b ON b.ID = a.ID
a.ID a.Value b.ID b.Value
2 TableA.Value2 2 TableB.ValueB
3 TableA.Value3 3 TableB.ValueC
A LEFT JOIN produces:
SELECT a.ID,a.Value,b.ID,b.Value
FROM TableA a LEFT JOIN TableB b ON b.ID = a.ID
a.ID a.Value b.ID b.Value
1 TableA.Value1 NULL NULL
2 TableA.Value2 2 TableB.ValueB
3 TableA.Value3 3 TableB.ValueC
As you can see, the LEFT JOIN includes the row from TableA where ID = 1 even though there's no matching row in TableB where ID = 1, whereas the INNER JOIN excludes the row specifically because there's no matching row in TableB
HTH
Use an inner join when you want only the results that appear in both tables that matches the Join condition.
Use a left join when you want all the results from Table A, but if Table B has data relevant to some of Table A's records, then you also want to use that data in the same query.
Use a full join when you want all the results from both Tables.
For newbies, because it helped me when I was one: an INNER JOIN is always a subset of a LEFT or RIGHT JOIN, and all of these are always subsets of a FULL JOIN. It helped me understand the basic idea.