Related
Below is the result set from SELECT query,
mysql> select * from mytable where userid =242 ;
+--------+-----------------------------+------------+---------------------+---------------------+
| UserId | ActiveLinks | ModifiedBy | DateCreated | DateModified |
+--------+-----------------------------+------------+---------------------+---------------------+
| 242 | 1|2|4|6|9|15|22|33|43|57|58 | 66 | 2013-11-28 16:17:25 | 2013-11-28 16:17:25 |
+--------+-----------------------------+------------+---------------------+---------------------+
What I want is to SELECT the records by splitting the Active links columns and associating it with UserId in the below format,
eg,
UserId ActiveLinks
242 1
242 2
242 4
242 6
Can anyone help me with this query , as of now nothing coming to my mind. Thanks
Dealing with lists stored in data is a pain. In MySQL, you can use substring_index(). The following should do what you want:
SELECT userid,
substring_index(substring_index(l.ActiveLinks, '||', n.n), '|', -1) as link
FROM (select 1 as n union all select 2 union all select 3 union all select 4) n join
ipadminuserslinks l
on length(l.ActiveLinks) - length(replace(l.ActiveLinks, '||', '')) + 1 <= n.n
WHERE userid = 242;
The first subquery generates a bunch of numbers, which you need. You may have to increase the size of this list.
The on clause limits the numbers to the number of elements in the list.
As you can probably tell, this is rather complicated. It is much easier to use a junction table, which is the relational way to store this type of information.
I would create a routine which will have the delimiter as an argument.
Another in_var would be the correspondent line.
Every time you call it, it will return the set of values for the UserId called.
It will basically use a loop based on the count of '|' (we call this pipeline)
This way you can implement the solution proposed by #Gordon Linoff without the need to know how many active links you have.
If this is just a list of values that do not relate to anything on another table I would do it the same way as Gordon (if needs be you can cross join the sub query that gets the lists of numbers to easily generate far larger ranges of numbers). One minor issue is that if the range of number is bigger than the number of delimited values on a row then the last value will be repeated (easily removed using DISTINCT in this case, more complicated when there are duplicate values in there that you want to keep).
However if the list of delimited values are related to another table (such as being the id field of another table then you could do it this way:-
SELECT a.UserId, b.link_id
FROM mytable a
LEFT OUTER JOIN my_link_table b
ON FIND_IN_SET(b.link_id, replace(a.ActiveLinks, '|', ','))
Ie, use FIND_IN_SET to join your table with the related table. In this case converting any | symbols used as delimiters to commas to allow FIND_IN_SET to work.
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)
Is it possible to sort in MySQL by "order by" using a predefined set of column values (ID) like order by (ID=1,5,4,3) so I would get records 1, 5, 4, 3 in that order out?
UPDATE: Why I need this...
I want my records to change sort randomly every 5 minutes. I have a cron task to update the table to put different, random sort order in it.
There is just one problem! PAGINATION.
I will have visitors who come to my page, and I will give them the first 20 results. They will wait 6 minutes, go to page 2 and have the wrong results as the sort order has already changed.
So I thought that if I put all the IDs into a session on page 2, we get the correct records even if the sorting had already changed.
Is there any other better way to do this?
You can use ORDER BY and FIELD function.
See http://lists.mysql.com/mysql/209784
SELECT * FROM table ORDER BY FIELD(ID,1,5,4,3)
It uses Field() function, Which "Returns the index (position) of str in the str1, str2, str3, ... list. Returns 0 if str is not found" according to the documentation. So actually you sort the result set by the return value of this function which is the index of the field value in the given set.
You should be able to use CASE for this:
ORDER BY CASE id
WHEN 1 THEN 1
WHEN 5 THEN 2
WHEN 4 THEN 3
WHEN 3 THEN 4
ELSE 5
END
On the official documentation for mysql about ORDER BY, someone has posted that you can use FIELD for this matter, like this:
SELECT * FROM table ORDER BY FIELD(id,1,5,4,3)
This is untested code that in theory should work.
SELECT * FROM table ORDER BY id='8' DESC, id='5' DESC, id='4' DESC, id='3' DESC
If I had 10 registries for example, this way the ID 1, 5, 4 and 3 will appears first, the others registries will appears next.
Normal exibition
1
2
3
4
5
6
7
8
9
10
With this way
8
5
4
3
1
2
6
7
9
10
There's another way to solve this. Add a separate table, something like this:
CREATE TABLE `new_order` (
`my_order` BIGINT(20) UNSIGNED NOT NULL,
`my_number` BIGINT(20) NOT NULL,
PRIMARY KEY (`my_order`),
UNIQUE KEY `my_number` (`my_number`)
) ENGINE=INNODB;
This table will now be used to define your own order mechanism.
Add your values in there:
my_order | my_number
---------+----------
1 | 1
2 | 5
3 | 4
4 | 3
...and then modify your SQL statement while joining this new table.
SELECT *
FROM your_table AS T1
INNER JOIN new_order AS T2 on T1.id = T2.my_number
WHERE ....whatever...
ORDER BY T2.my_order;
This solution is slightly more complex than other solutions, but using this you don't have to change your SELECT-statement whenever your order criteriums change - just change the data in the order table.
If you need to order a single id first in the result, use the id.
select id,name
from products
order by case when id=5 then -1 else id end
If you need to start with a sequence of multiple ids, specify a collection, similar to what you would use with an IN statement.
select id,name
from products
order by case when id in (30,20,10) then -1 else id end,id
If you want to order a single id last in the result, use the order by the case. (Eg: you want "other" option in last and all city list show in alphabetical order.)
select id,city
from city
order by case
when id = 2 then city else -1
end, city ASC
If i had 5 city for example, i want to show the city in alphabetical order with "other" option display last in the dropdown then we can use this query.
see example other are showing in my table at second id(id:2) so i am using "when id = 2" in above query.
record in DB table:
Bangalore - id:1
Other - id:2
Mumbai - id:3
Pune - id:4
Ambala - id:5
my output:
Ambala
Bangalore
Mumbai
Pune
Other
SELECT * FROM TABLE ORDER BY (columnname,1,2) ASC OR DESC
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.
Is it possible to do a SELECT statement with a predetermined order, ie. selecting IDs 7,2,5,9 and 8 and returning them in that order, based on nothing more than the ID field?
Both these statements return them in the same order:
SELECT id FROM table WHERE id in (7,2,5,9,8)
SELECT id FROM table WHERE id in (8,2,5,9,7)
I didn't think this was possible, but found a blog entry here that seems to do the type of thing you're after:
SELECT id FROM table WHERE id in (7,2,5,9,8)
ORDER BY FIND_IN_SET(id,"7,2,5,9,8");
will give different results to
SELECT id FROM table WHERE id in (7,2,5,9,8)
ORDER BY FIND_IN_SET(id,"8,2,5,9,7");
FIND_IN_SET returns the position of id in the second argument given to it, so for the first case above, id of 7 is at position 1 in the set, 2 at 2 and so on - mysql internally works out something like
id | FIND_IN_SET
---|-----------
7 | 1
2 | 2
5 | 3
then orders by the results of FIND_IN_SET.
Your best bet is:
ORDER BY FIELD(ID,7,2,4,5,8)
...but it's still ugly.
Could you include a case expression that maps your IDs 7,2,5,... to the ordinals 1,2,3,... and then order by that expression?
All ordering is done by the ORDER BY keywords, you can only however sort ascending and descending. If you are using a language such as PHP you can then sort them accordingly using some code but I do not believe it is possible with MySQL alone.
This works in Oracle. Can you do something similar in MySql?
SELECT ID_FIELD
FROM SOME_TABLE
WHERE ID_FIELD IN(11,10,14,12,13)
ORDER BY
CASE WHEN ID_FIELD = 11 THEN 0
WHEN ID_FIELD = 10 THEN 1
WHEN ID_FIELD = 14 THEN 2
WHEN ID_FIELD = 12 THEN 3
WHEN ID_FIELD = 13 THEN 4
END
You may need to create a temp table with an autonumber field and insert into it in the desired order. Then sort on the new autonumber field.
Erm, not really. Closest you can get is probably:
SELECT * FROM table WHERE id IN (3, 2, 1, 4) ORDER BY id=4, id=1, id=2, id=3
But you probably don't want that :)
It's hard to give you any more specific advice without more information about what's in the tables.
It's hacky (and probably slow), but you can get the effect with UNION ALL:
SELECT id FROM table WHERE id = 7
UNION ALL SELECT id FROM table WHERE id = 2
UNION ALL SELECT id FROM table WHERE id = 5
UNION ALL SELECT id FROM table WHERE id = 9
UNION ALL SELECT id FROM table WHERE id = 8;
Edit: Other people mentioned the find_in_set function which is documented here.
You get answers fast around here, don't you…
The reason I'm asking this is that it's the only way I can think of to avoid sorting a complex multidimensional array. I'm not saying it would be difficult to sort, but if there were a simpler way to do it with straight sql, then why not.
One Oracle solution is:
SELECT id FROM table WHERE id in (7,2,5,9,8)
ORDER BY DECODE(id,7,1,2,2,5,3,9,4,8,5,6);
This assigns an order number to each ID. Works OK for a small set of values.
Best I can think of is adding a second Column orderColumn:
7 1
2 2
5 3
9 4
8 5
And then just do a ORDER BY orderColumn