How to split the string and grouping them by splited token?
I want to get that grouping splited token's each count.
I have a varchar column and it store a string which can split by ',' .
below is the row data of the column. (column name is LogData)
[LogData]
1,2,3,4
1,3,1,9
2,1,3
6,2
And then i want to show(select) like below.
[token] [count]
1 : 4
2 : 3
3 : 3
4 : 1
6 : 1
9 : 1
If possible, then may i have a answer about this with some explanation? (I'm not skilled in db)
Using the and adapting the comment from undefined_variable the correct query looks like this:
SELECT value,COUNT(*) FROM
(SELECT SUBSTRING_INDEX(SUBSTRING_INDEX(t0.logdata, ',', n.n), ',', -1) value
FROM t0 CROSS JOIN
(
SELECT a.N + b.N * 10 + 1 n
FROM
(SELECT 0 AS N UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) a
,(SELECT 0 AS N UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) b
ORDER BY n
) n
WHERE n.n <= 1 + (LENGTH(t0.logdata) - LENGTH(REPLACE(t0.logdata, ',', '')))
ORDER BY value) nt0 GROUP BY value
Related
I have a table in mysql which looks like below.
id cust_id date data
1 1 1/1/2018 a b c d e f g
2 1 2/1/2018 h I j k l m n
Here in this example data column is having huge data seperated by space like a b c d, I would like to show case as in row like below
id cust_id date data
1 1 1/1/2018 a
1 1 1/1/2018 b
1 1 1/1/2018 c
1 1 1/1/2018 d
2 2 2/1/2018 h
2 2 2/1/2018 i
2 2 2/1/2018 j
2 2 2/1/2018 k
I have checked few option like using unpivot function, but unable to achieve my output.
Thanks in advance !!
select
tablename.id,
tablename.date
,SUBSTRING_INDEX(SUBSTRING_INDEX(tablename.data, ' ', numbers.n), ' ', -1) name
from
(
SELECT #row := #row + 1 as n FROM
(select 0 union all select 1 union all select 3 union all select 4 union all select 5 union all select 6 union all select 6 union all select 7 union all select 8 union all select 9) t,
(select 0 union all select 1 union all select 3 union all select 4 union all select 5 union all select 6 union all select 6 union all select 7 union all select 8 union all select 9) t1,
(select 0 union all select 1 union all select 3 union all select 4 union all select 5 union all select 6 union all select 6 union all select 7 union all select 8 union all select 9) t2,
(SELECT #row:=0) r
) numbers INNER JOIN Table1 tablename
on CHAR_LENGTH(tablename.data)
-CHAR_LENGTH(REPLACE(tablename.data, ' ', ''))>=numbers.n-1
order by
id, n
Check link for output
http://sqlfiddle.com/#!9/fa0dcb/1
EXPLANATION:
First go through the inner query i.e.
select 0
union all
select 1
union all
select 3
union all
select 4
union all
select 5
union all
select 6
union all
select 6
union all
select 7
union all
select 8
union all
select 9
This will generate a table of 10 rows with 10 numbers.
Now the other query :
SELECT #row := #row + 1 as n FROM
(select 0 union all select 1 union all select 3 union all select 4 union all select 5 union all select 6 union all select 6 union all select 7 union all select 8 union all select 9) t,
(select 0 union all select 1 union all select 3 union all select 4 union all select 5 union all select 6 union all select 6 union all select 7 union all select 8 union all select 9) t1
Since above query is generating row numbers from below table 't' and table 't1' which is separated by ',' means that they are producing Cartesian product of their total rows.
For example: t have 10 rows and t1 also have 10 rows so, there Cartesian product produces 100 rows. So #row variable incremented 100 times and gives 100 rows of 100 numbers from 1 to 100.
The below query:
SUBSTRING_INDEX(SUBSTRING_INDEX(tablename.data, ' ', numbers.n), ' ', -1)
this one will take "a b c d e f g h" one by one.
For example:
take numbers.n = 1
then inner substring_index will find index of first space and will return string before that index i.e. 'a'
and then outer substring_index will find the space from the end of the resulting string and will give the last character from the string i.e. 'a'.
Now if you
take numbers.n = 2
then inner substring_index will find index of first space and will return string before that index i.e. 'a b'
and then outer substring_index will find the space from the end of the resulting string and will give the last character from the string i.e. 'b'
Always try to breakdown the query like this and you will able to understand the query in simpler way.
I want to a MySQL query to select the records separately from following table
ID AgentID Name Return Date
1 1,2,3 A 2016-05-22,2016-02-1,2016-1-15
2 2,4 B 2016-03-22,2016-04-1
Expecting Answer
ID AgentID Name Return Date
1 1 A 2016-05-22
1 2 A 2016-02-1
1 3 A 2016-1-15
2 2 B 2016-03-22
2 4 B 2016-04-1
You can use MySQL SUBSTRING_INDEX(). It will return the sub-string from the given comma separated string before a specified number of occurrences of the delimiter.
Try this, It seems to work fine:
SELECT ID
,SUBSTRING_INDEX(SUBSTRING_INDEX(t.AgentID, ',', n.n), ',', -1) Agent
,Name
,SUBSTRING_INDEX(SUBSTRING_INDEX(t.Return_Date, ',', n.n), ',', -1) Return_Date
FROM table1 t CROSS JOIN
(
SELECT a.N + b.N * 10 + 1 n
FROM
(SELECT 0 AS N UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) a
,(SELECT 0 AS N UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) b
ORDER BY n
) n
WHERE n.n <= 1 + (LENGTH(t.Return_Date) - LENGTH(REPLACE(t.Return_Date, ',', '')))
ORDER BY ID;
Check this..
SQL Fiddle HERE
For further Study go on MySQL Split String Function
if your values in tblA and you want to insert into tblB than query like this
insert into tblB (date) select date from tblA;
INSERT INTO second_table (
Field_1,
Field_2,
Field_3)
SELECT Field_1,
Field_2,
Field_3
FROM first_table;
My database is like
Name | IC | Item
--------------------------
lee | xxx | pear,bear
--------------------------
ron | xxx | apple,dog
what should I do to retrieve the 4 values contained in the column "Item" and then separate them?
Do you have only two items separated by comma in Item? Or it may vary?
LE: you can use this SQL split comma separated row
LLE: just played around with that and this what I've done:
create table myTable(name varchar(7), ic varchar(7), item varchar(200));
insert into myTable(name,ic,item) values ('lee','xxx','pear,bear');
insert into myTable(name,ic,item) values ('ron','xxx','apple,dog');
insert into myTable(name,ic,item) values ('a','xxx','gamma');
insert into myTable(name,ic,item) values ('b','xxx','a,b,c,d');
SELECT SUBSTRING_INDEX(SUBSTRING_INDEX(t.item, ',', n.n), ',', -1) value
FROM myTable t CROSS JOIN
(
SELECT a.N + b.N * 10 + 1 n
FROM
(SELECT 0 AS N UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) a
,(SELECT 0 AS N UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) b
ORDER BY n
) n
WHERE n.n <= 1 + (LENGTH(t.item) - LENGTH(REPLACE(t.item, ',', '')))
ORDER BY value;
I want to a MySQL query to select the records separately from following table
ID AgentID Name Return Date
1 1,2,3 A 2016-05-22,2016-02-1,2016-1-15
2 2,4 B 2016-03-22,2016-04-1
Expecting Answer
ID AgentID Name Return Date
1 1 A 2016-05-22
1 2 A 2016-02-1
1 3 A 2016-1-15
2 2 B 2016-03-22
2 4 B 2016-04-1
You can use MySQL SUBSTRING_INDEX(). It will return the sub-string from the given comma separated string before a specified number of occurrences of the delimiter.
Try this, It seems to work fine:
SELECT ID
,SUBSTRING_INDEX(SUBSTRING_INDEX(t.AgentID, ',', n.n), ',', -1) Agent
,Name
,SUBSTRING_INDEX(SUBSTRING_INDEX(t.Return_Date, ',', n.n), ',', -1) Return_Date
FROM table1 t CROSS JOIN
(
SELECT a.N + b.N * 10 + 1 n
FROM
(SELECT 0 AS N UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) a
,(SELECT 0 AS N UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) b
ORDER BY n
) n
WHERE n.n <= 1 + (LENGTH(t.Return_Date) - LENGTH(REPLACE(t.Return_Date, ',', '')))
ORDER BY ID;
Check this..
SQL Fiddle HERE
For further Study go on MySQL Split String Function
if your values in tblA and you want to insert into tblB than query like this
insert into tblB (date) select date from tblA;
INSERT INTO second_table (
Field_1,
Field_2,
Field_3)
SELECT Field_1,
Field_2,
Field_3
FROM first_table;
I need to get unique values from a table. I have a single column with comma separated keywords. I need to derive a single list of all the keywords without duplicates. Getting the count of how often each keyword is present, too.
From what I have researched, it is an UNPIVOTING like function with an unknown number of columns?
For example:
keywords
red, blue, yellow
blue, orange, black, white
brown, black, clear, pink
blue, violet, orange
Result
color | count
red 1
blue 3
yellow 1
orange 2
black 2
white 1
brown 1
clear 1
pink 1
violet 1
Thank you in advance!!
**
Thus far I have tried adding an explode_table type procedure, but realized I can't call that dynamically from a View. Then I have been experimenting with performing a reverse GROUP_CONCAT() on the column. I haven't been able to produce code that performs.
My version of echo_Me's answer:
SELECT SUBSTRING_INDEX(SUBSTRING_INDEX(sKeywords, ',', n.n), ',', -1) value , count(*) as counts
FROM tblPatternMetadata t CROSS JOIN
(SELECT a.N + b.N * 10 + 1 n FROM (SELECT 0 AS N UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) a, (SELECT 0 AS N UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) b
ORDER BY n ) n
WHERE n.n <= 1 + (LENGTH(sKeywords) - LENGTH(REPLACE(sKeywords, ',', ''))) group by value
Try that:
SELECT SUBSTRING_INDEX(SUBSTRING_INDEX(t.keywords, ',', n.n), ',', -1) value , count(*) as counts
FROM table1 t CROSS JOIN
(
SELECT a.N + b.N * 10 + 1 n
FROM
(SELECT 0 AS N UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) a
,(SELECT 0 AS N UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) b
ORDER BY n
) n
WHERE n.n <= 1 + (LENGTH(t.keywords) - LENGTH(REPLACE(t.keywords, ',', '')))
group by value
DEMO HERE