I am trying to perform a SELECT query using a GROUP BY clause, however I also need to access data from multiple rows and somehow concatenate it into a single column.
Here's what I have so far:
SELECT
COUNT(v.id) AS quantity,
vt.name AS name,
vt.cost AS cost,
vt.postage_cost AS postage_cost
FROM vouchers v
INNER JOIN voucher_types vt
ON v.type_id = vt.id
WHERE
v.order_id = 1 AND
v.sold = 1
GROUP BY vt.id
Which gives me the first four columns I need in the following format.
quantity | name | cost | postage_cost
2 X 5 1
2 Y 6 1
However, I would also like a fifth column to be displayed, showing all of the codes associated with each line of the order like this:
code
ABCD, EFGH
IJKL, MNOP
Where the comma separated values are pulled from the voucher table.
Is this possible?
Any advice would be appreciated.
Thanks
This is what GROUP_CONCAT does.
Assuming the column is called code you would just add ,GROUP_CONCAT(v.code) As Codes to your select list.
Related
I have the following situation. I have a table with all info of article. I will like to compare the same column with it self. because I have multiple type of article. Single product and Master product. the only way that I have to differences it, is by SKU. for example.
ID | SKU
1 | 11111
2 | 11112
3 | 11113
4 | 11113-5
5 | 11113-8
6 | 11114
7 | 11115
8 | 11115-1-W
9 | 11115-2
10 | 11116
I only want to list or / and count only the sku that are full unique. follow th example the sku that are unique and no have variant are (ID = 1, 2, 6 and 10) I will want to create a query where if 11113 are again on the column not cout it. so in total I will be 4 unique sku and not "6 (on total)". Please let me know. if this are possible.
Assuming the length of master SKUs are 5 characters, try this:
select a.*
from mytable a
left join mytable b on b.sku like concat(a.sku, '%')
where length(a.sku) = 5
and b.sku is null
This query joins master SKUs to child ones, but filters out successful joins - leaving only solitary master SKUs.
You can do this by grouping and counting the unique rows.
First, we will need to take your table and add a new column, MasterSKU. This will be the first five characters of the SKU column. Once we have the MasterSKU, we can then GROUP BY it. This will bundle together all of the rows having the same MasterSKU. Once we are grouping we get access to aggregate functions like COUNT(). We will use that function to count the number of rows for each MasterSKU. Then, we will filter out any rows that have a COUNT() over 1. That will leave you with only the unique rows remaining.
Take that unique list and LEFT JOIN it back into your original table to grab the IDs.
SELECT ID, A.MasterSKU
FROM (
SELECT
MasterSKU = SUBSTRING(SKU,1,5),
MasterSKUCount = COUNT(*)
FROM MyTable
GROUP BY SUBSTRING(SKU,1,5)
HAVING COUNT(*) = 1
) AS A
LEFT JOIN (
SELECT
ID,
MasterSKU = SUBSTRING(SKU,1,5)
FROM MyTable
) AS B
ON A.MasterSKU = B.MasterSKU
Now one thing I noticed from you example. The original SKU column really looks like three columns in one. We have multiple values being joined with hypens.
11115-1-W
There may be a reason for it, but most likely this violates first normal form and will make the database hard to query. It's part of the reason why such a complicated query is needed. If the SKU column really represents multiple things then we may want to consider breaking it out into MasterSKU, Version, and Color or whatever each hyphen represents.
I have a table like this
id | user_id | code | type | time
-----------------------------------
2 2 fdsa r 1358300000
3 2 barf r 1358311000
4 2 yack r 1358311220
5 3 surf r 1358311000
6 3 yooo r 1358300000
7 4 poot r 1358311220
I want to get the concatenated 'code' column for user 2 and user 3 for each matching time.
I want to receive a result set like this:
code | time
-------------------------------
fdsayooo 1358300000
barfsurf 1358311000
Please note that there is no yackpoot code because the query was not looking for user 4.
You can use GROUP_CONCAT function. Try this:
SELECT GROUP_CONCAT(code SEPARATOR '') code, time
FROM tbl
WHERE user_id in (2, 3)
GROUP BY time
HAVING COUNT(time) = 2;
SQL FIDDLE DEMO
What you are looking for is GROUP_CONCAT, but you are missing a lot of details in your question to provide a good example. This should get you started:
SELECT GROUP_CONCAT(code), time
FROM myTable
WHERE user_id in (2, 3)
GROUP BY time;
Missing details are:
Is there an order required? Not sure how ordering would be done useing grouping, would need to test if critical
Need other fields? If so you will likely end up needing to do a sub-select or secondary query.
Do you only want results with multiple times?
Do you really want no separator between values in the results column (specify the delimiter with SEPARATOR '' in the GROUP_CONCAT
Notes:
You can add more fields to the GROUP BY if you want to do it by something else (like user_id and time).
I think it will be easiest to start with the table I have and the result I am aiming for.
Name | Date
A | 03/01/2012
A | 03/01/2012
B | 02/01/2012
A | 02/01/2012
B | 02/01/2012
A | 02/01/2012
B | 01/01/2012
B | 01/01/2012
A | 01/01/2012
I want the result of my query to be:
Name | 01/01/2012 | 02/01/2012 | 03/01/2012
A | 1 | 2 | 2
B | 2 | 2 | 0
So basically I want to count the number of rows that have the same date, but for each individual name. So a simple group by of dates won't do because it would merge the names together. And then I want to output a table that shows the counts for each individual date using php.
I've seen answers suggest something like this:
SELECT
NAME,
SUM(CASE WHEN GRADE = 1 THEN 1 ELSE 0 END) AS GRADE1,
SUM(CASE WHEN GRADE = 2 THEN 1 ELSE 0 END) AS GRADE2,
SUM(CASE WHEN GRADE = 3 THEN 1 ELSE 0 END) AS GRADE3
FROM Rodzaj
GROUP BY NAME
so I imagine there would be a way for me to tweak that but I was wondering if there is another way, or is that the most efficient?
I was perhaps thinking if the while loop were to output just one specific name and date each time along with the count, so the first result would be A,01/01/2012,1 then the next A,02/01/2012,2 - A,03/01/2012,3 - B,01/01/2012,2 etc. then perhaps that would be doable through a different technique but not sure if something like that is possible and if it would be efficient.
So I'm basically looking to see if anyone has any ideas that are a bit outside the box for this and how they would compare.
I hope I explained everything well enough and thanks in advance for any help.
You have to include two columns in your GROUP BY:
SELECT name, COUNT(*) AS count
FROM your_table
GROUP BY name, date
This will get the counts of each name -> date combination in row-format. Since you also wanted to include a 0 count if the name didn't have any rows on a certain date, you can use:
SELECT a.name,
b.date,
COUNT(c.name) AS date_count
FROM (SELECT DISTINCT name FROM your_table) a
CROSS JOIN (SELECT DISTINCT date FROM your_table) b
LEFT JOIN your_table c ON a.name = c.name AND
b.date = c.date
GROUP BY a.name,
b.date
SQLFiddle Demo
You're asking for a "pivot". Basically, it is what it is. The real problem with a pivot is that the column names must adapt to the data, which is impossible to do with SQL alone.
Here's how you do it:
SELECT
Name,
SUM(`Date` = '01/01/2012') AS `01/01/2012`,
SUM(`Date` = '02/01/2012') AS `02/01/2012`,
SUM(`Date` = '03/01/2012') AS `03/01/2012`
FROM mytable
GROUP BY Name
Note the cool way you can SUM() a condition in mysql, becasue in mysql true is 1 and false is 0, so summing a condition is equivalent to counting the number of times it's true.
It is not more efficient to use an inner group by first.
Just in case anyone is interested in what was the best method:
Zane's second suggestion was the slowest, I loaded in a third of the data I did for the other two and it took quite a while. Perhaps on smaller tables it would be more efficient, and although I am not working with a huge table roughly 28,000 rows was enough to create significant lag, with the between clause dropping the result to about 4000 rows.
Bohemian's answer gave me the least amount to code, I threw in a loop to create all the case statements and it worked with relative ease. The benefit of this method was the simplicity, besides creating the loop for the cases, the results come in without the need for any php tricks, just simple foreach to get all the columns. Recommended for those not confident with php.
However, I found Zane's first suggestion the quickest performing and despite the need for extra php coding it seems I will be sticking with this method. The disadvantage of this method is that it only gives the dates that actually have data, so creating a table with all the dates becomes a bit more complicated. What I did was create a variable that keeps track of what date it is supposed to be compared to the table column which is reset on each table row, when the result of the query is equal to that date it echoes the value otherwise it does a while loop echoing table cells with 0 until the dates do match. It also had to do a check to see if the 'Name' value is still the same and if not it would switch to the next row after filling in any missing cells with 0 to the end of that row. If anyone is interested in seeing the code you can message me.
Results of the two methods over 3 months of data (a column for each day so roughly 90 case statements) ~ 12,000 rows out of 28,000:Bohemian's Pivot - ~0.158s (highest seen ~0.36s)Zane's Double Group by - ~0.086s (highest seen ~0.15s)
I have a watchlist system that I've coded, in the overview of the users' watchlist, they would see a list of records, however the list shows duplicates when in the database it only shows the exact, correct number.
I've tried GROUP BY watch.watch_id, GROUP BY rec.record_id, none of any types of group I've tried seems to remove duplicates. I'm not sure what I'm doing wrong.
SELECT watch.watch_date,
rec.street_number,
rec.street_name,
rec.city,
rec.state,
rec.country,
usr.username
FROM
(
watchlist watch
LEFT OUTER JOIN records rec ON rec.record_id = watch.record_id
LEFT OUTER JOIN members usr ON rec.user_id = usr.user_id
)
WHERE watch.user_id = 1
GROUP BY watch.watch_id
LIMIT 0, 25
The watchlist table looks like this:
+----------+---------+-----------+------------+
| watch_id | user_id | record_id | watch_date |
+----------+---------+-----------+------------+
| 13 | 1 | 22 | 1314038274 |
| 14 | 1 | 25 | 1314038995 |
+----------+---------+-----------+------------+
GROUP BY does not "remove duplicates". GROUP BY allows for aggregation. If all you want is to combine duplicated rows, use SELECT DISTINCT.
If you need to combine rows that are duplicate in some columns, use GROUP BY but you need to to specify what to do with the other columns. You can either omit them (by not listing them in the SELECT clause) or aggregate them (using functions like SUM, MIN, and AVG). For example:
SELECT watch.watch_id, COUNT(rec.street_number), MAX(watch.watch_date)
... GROUP by watch.watch_id
EDIT
The OP asked for some clarification.
Consider the "view" -- all the data put together by the FROMs and JOINs and the WHEREs -- call that V. There are two things you might want to do.
First, you might have completely duplicate rows that you wish to combine:
a b c
- - -
1 2 3
1 2 3
3 4 5
Then simply use DISTINCT
SELECT DISTINCT * FROM V;
a b c
- - -
1 2 3
3 4 5
Or, you might have partially duplicate rows that you wish to combine:
a b c
- - -
1 2 3
1 2 6
3 4 5
Those first two rows are "the same" in some sense, but clearly different in another sense (in particular, they would not be combined by SELECT DISTINCT). You have to decide how to combine them. You could discard column c as unimportant:
SELECT DISTINCT a,b FROM V;
a b
- -
1 2
3 4
Or you could perform some kind of aggregation on them. You could add them up:
SELECT a,b, SUM(c) "tot" FROM V GROUP BY a,b;
a b tot
- - ---
1 2 9
3 4 5
You could add pick the smallest value:
SELECT a,b, MIN(c) "first" FROM V GROUP BY a,b;
a b first
- - -----
1 2 3
3 4 5
Or you could take the mean (AVG), the standard deviation (STD), and any of a bunch of other functions that take a bunch of values for c and combine them into one.
What isn't really an option is just doing nothing. If you just list the ungrouped columns, the DBMS will either throw an error (Oracle does that -- the right choice, imo) or pick one value more or less at random (MySQL). But as Dr. Peart said, "When you choose not to decide, you still have made a choice."
While SELECT DISTINCT may indeed work in your case, it's important to note why what you have is not working.
You're selecting fields that are outside of the GROUP BY. Although MySQL allows this, the exact rows it returns for the non-GROUP BY fields is undefined.
If you wanted to do this with a GROUP BY try something more like the following:
SELECT watch.watch_date,
rec.street_number,
rec.street_name,
rec.city,
rec.state,
rec.country,
usr.username
FROM
(
watchlist watch
LEFT OUTER JOIN est8_records rec ON rec.record_id = watch.record_id
LEFT OUTER JOIN est8_members usr ON rec.user_id = usr.user_id
)
WHERE watch.watch_id IN (
SELECT watch_id FROM watch WHERE user_id = 1
GROUP BY watch.watch_id)
LIMIT 0, 25
I Would never recommend using SELECT DISTINCT, it's really slow on big datasets.
Try using things like EXISTS.
You are grouping by watch.watch_id and you have two results, which have different watch IDs, so naturally they would not be grouped.
Also, from the results displayed they have different records. That looks like a perfectly valid expected results. If you are trying to only select distinct values, then you don't want ot GROUP, but you want to select by distinct values.
SELECT DISTINCT()...
If you say your watchlist table is unique, then one (or both) of the other tables either (a) has duplicates, or (b) is not unique by the key you are using.
To suppress duplicates in your results, either use DISTINCT as #Laykes says, or try
GROUP BY watch.watch_date,
rec.street_number,
rec.street_name,
rec.city,
rec.state,
rec.country,
usr.username
It sort of sounds like you expect all 3 tables to be unique by their keys, though. If that is the case, you are simply masking some other problem with your SQL by trying to retrieve distinct values.
I need to create a join query for hierarchical data across 2 tables. These tables can have unlimited amounts of data and their structures are as follows:
group_id group_name group_order
1 group 1 2
2 group 2 1
field_id field_name parent_group field_order
1 field 1 1 1
2 field 2 2 2
3 field 3 2 1
I am currently able to get the correct format of data using 2 select queries with the second query inside a loop created from the results of the first query on the groups table.
The structure of the data I require from the result is as follows:
-group 2
- field 3
- field 2
- group 1
- field 1
Is it possible to get these results from one mysql query? I have read through the mysql document on hierarchical data by I am confused about how to incorporate the join.
Thanks for looking
You shouldn't need to think about it in terms of hierarchical data, you should just be able to select your fields and join on your group information. Try something like:
SELECT *
FROM Fields AS F
INNER JOIN Groups AS G
ON G.group_id = F.parent_group
ORDER BY group_order, field_order
Then you will get each fields as a row with the applicable group, also in the correct group order. Your loop should be able to handle the display you need.
one method
something that may convince you change your db schema