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.
Related
This is my first post/question since I couldn't find the answer to what I was looking for no matter how I searched. So here it is:
I have a small table of a few hundred rows with four columns (plus id col) as seen on the example below:
| id | phone1 | phone2 | phone3 | phone4 |
| 1 |059374-6| | 065371-7 |023126-8
| 2 | |026372-3| 024353-7 |
| 3 |...
.
.
.
I need to generate the sequence for all the (telephone) numbers of the specific range they belong. In other words, for the first phone number I need to retrieve the numbers:
059374
059375
059376
and consequently for the rest of the phones in the table.
All the number ranges are given in the above mentioned form and do not exceed the total of 10 numbers (max is something like 099500-9). In addition, they are populated through 4 different columns, as shown in the above example, something I have solved so far by using UNION ALL (in order to put them all in one column and finally find all duplicates which is the ultimate goal).
So, is there any elegant way to generate the numbers needed?
I have searched many threads for similar topics but surprisingly none seem to answer my question.
Finally, if there is any better way than combining all cols into one with UNION ALL and then GROUP BY col1 for finding the duplicates would be highly appreciated.
Create a table that contains the numbers from 0 to 9. Extract the starting and ending digits of each number, and join with the number table to find all the numbers in the range.
WITH digits AS (
SELECT 0 AS dig
UNION
SELECT 1 AS dig
...
UNION
SELECT 9 AS dig)
SELECT SUBSTR(phone1, 1, 5) + digits.dig
FROM yourTable
JOIN digits ON dig BETWEEN SUBSTR(phone1, 6, 1) AND SUBSTR(phone1, 8, 1)
WHERE phone1 != ''
This shows how to do it for the phone1 column, you can use UNION to repeat it for the other 3 columns.
So I'm currently using MySQL's JSON field to store some data.
So the 'reports' table looks like this:
id | stock_id | type | doc |
1 | 5 | Income_Statement | https://pastebin.com/bj1hdK0S|
The pastebin is the content of the json field
What I want to do is get a number (ebit) from the first object under yearly (2018-12-31) in the JSON and then use that to do a WHERE query on so that it only returns where ebit > 50000000 for example. The issue is that the dates under yearly are not standard (i.e. one might be 2018-12-31, the other might by 2018-12-15). So essentially I want a way to get the data using integer indexes rather than the actual names of the objects, so something like yearly.[0].ebit.
How would I do this in MySQL? Alternatively if it's not possible in MySQL, would it be possible in either PostgeSQL or Mongo? If so, could you give me an example? Most of the data fits well into MySQL only this table has a JSON column which is why I started with MySQL.
so StackOverflow isn't letting my link to pastebin without some code so here's some random code:
if(dog == "poodle") {
print "test"
}
I don't know for MySQL nor MongoDB, but here's a simple version for PostgreSQL JSONB type:
SELECT (doc->'yearly'-> max(years) -> 'ebit')::numeric AS ebit
FROM reports, jsonb_object_keys(doc->'yearly') AS years
GROUP BY reports.doc;
...with simplistic test data:
WITH reports(doc) AS (
SELECT '{"yearly":{"2018-12-31":{"ebit":123},"2017-12-31":{"ebit":1.23}}}'::jsonb
)
SELECT (doc->'yearly'-> max(years) -> 'ebit')::numeric AS ebit
FROM reports, jsonb_object_keys(doc->'yearly') AS years
GROUP BY reports.doc;
...gives:
ebit
------
123
(1 row)
So I've basically selected the latest entry under "yearly" without knowing actual values but assuming that the key date formatting will allow a sort order (in this case it seems to comply with ISO-8601).
Using data type JSON instead of JSONB would preserve object key order but is not as efficient in PostgreSQL further down the road and wouldn't help here either.
IF you want to then select only those reports entries having their latest ebit greater than a certain value, just pack it into a sub-select or a CTE. I usualy prefer CTE's because they are better to read, so here we go:
WITH
reports (id, doc) AS (
VALUES
(1, '{"yearly":{"2018-12-31":{"ebit":123},"2017-12-31":{"ebit":1.23}}}'::jsonb),
(2, '{"yearly":{"2018-12-23":{"ebit":50},"2017-12-22":{"ebit":"1200.00"}}}'::jsonb)
),
r_ebit (id, ebit) AS (
SELECT reports.id, (reports.doc->'yearly'-> max(years) -> 'ebit')::numeric AS ebit
FROM reports, jsonb_object_keys(doc->'yearly') AS years
GROUP BY reports.id, reports.doc
)
SELECT id, ebit
FROM r_ebit
WHERE ebit > 100;
However, as you already see, it is not possible to filter the original rows using this strategy. A pre-processing step would make sense here so that the JSON format actually is filter-friendly.
ADDENDUM
To add the possibility of selecting the values for the n-th completed fiscal year, we need resort to window functions and we also need to reduce the resulting set to only return a single row per actual group (in the demonstration case: reports.id):
WITH reports(id, doc) AS (VALUES
(1, '{"yearly":{"2018-12-31":{"ebit":123},"2017-12-31":{"ebit":1.23},"2016-12-31":{"ebit":"23.42"}}}'::jsonb),
(2, '{"yearly":{"2018-12-23":{"ebit":50},"2017-12-22":{"ebit":"1200.00"}}}'::jsonb)
)
SELECT DISTINCT ON (1) reports.id, (reports.doc->'yearly'-> (lead(years, 0) over (partition by reports.doc order by years desc nulls last)) ->>'ebit')::numeric AS ebit
FROM reports, jsonb_object_keys(doc->'yearly') AS years
GROUP BY 1, reports.doc, years.years ORDER BY 1;
...will behave exactly as using the max aggregate function previously. Increasing the offset parameter within the lead(years, <offset>) function all will select the n-th year backwards (because of descending order of the window partition).
The DISTINCT ON (1) clause is the magic that reduces the result to a single row per distinct column value (first column = reports.id). This is why the NULLS LAST is very important inside the window OVER clause.
Here are results for different offsets (I've added a third historic entry for the first id but not for the second to also show how it deals with absent entries):
N = 0:
id | ebit
----+------
1 | 123
2 | 50
N = 1
id | ebit
----+---------
1 | 1.23
2 | 1200.00
N = 2
id | ebit
----+-------
1 | 23.42
2 |
...which means absent entries will just result in a NULL value.
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)