I have a column in one of my table where I store multiple ids seperated by comma's.
Is there a way in which I can use this column's value in the "IN" clause of a query.
The column(city) has values like 6,7,8,16,21,2
I need to use as
select * from table where e_ID in (Select city from locations where e_Id=?)
I am satisfied with Crozin's answer, but I am open to suggestions, views and options.
Feel free to share your views.
Building on the FIND_IN_SET() example from #Jeremy Smith, you can do it with a join so you don't have to run a subquery.
SELECT * FROM table t
JOIN locations l ON FIND_IN_SET(t.e_ID, l.city) > 0
WHERE l.e_ID = ?
This is known to perform very poorly, since it has to do table-scans, evaluating the FIND_IN_SET() function for every combination of rows in table and locations. It cannot make use of an index, and there's no way to improve it.
I know you said you are trying to make the best of a bad database design, but you must understand just how drastically bad this is.
Explanation: Suppose I were to ask you to look up everyone in a telephone book whose first, middle, or last initial is "J." There's no way the sorted order of the book helps in this case, since you have to scan every single page anyway.
The LIKE solution given by #fthiella has a similar problem with regards to performance. It cannot be indexed.
Also see my answer to Is storing a delimited list in a database column really that bad? for other pitfalls of this way of storing denormalized data.
If you can create a supplementary table to store an index, you can map the locations to each entry in the city list:
CREATE TABLE location2city (
location INT,
city INT,
PRIMARY KEY (location, city)
);
Assuming you have a lookup table for all possible cities (not just those mentioned in the table) you can bear the inefficiency one time to produce the mapping:
INSERT INTO location2city (location, city)
SELECT l.e_ID, c.e_ID FROM cities c JOIN locations l
ON FIND_IN_SET(c.e_ID, l.city) > 0;
Now you can run a much more efficient query to find entries in your table:
SELECT * FROM location2city l
JOIN table t ON t.e_ID = l.city
WHERE l.e_ID = ?;
This can make use of an index. Now you just need to take care that any INSERT/UPDATE/DELETE of rows in locations also inserts the corresponding mapping rows in location2city.
From MySQL's point of view you're not storing multiple ids separated by comma - you're storing a text value, which has the exact same meaing as "Hello World" or "I like cakes!" - i.e. it doesn't have any meaing.
What you have to do is to create a separated table that will link two objects from the database together. Read more about many-to-many or one-to-many (depending on your requirements) relationships in SQL-based databases.
Rather than use IN on your query, use FIND_IN_SET (docs):
SELECT * FROM table
WHERE 0 < FIND_IN_SET(e_ID, (
SELECT city FROM locations WHERE e_ID=?))
The usual caveats about first form normalization apply (the database shouldn't store multiple values in a single column), but if you're stuck with it, then the above statement should help.
This does not use IN clause, but it should do what you need:
Select *
from table
where
CONCAT(',', (Select city from locations where e_Id=?), ',')
LIKE
CONCAT('%,', e_ID, ',%')
but you have to make sure that e_ID does not contain any commas or any jolly character.
e.g.
CONCAT(',', '6,7,8,16,21,2', ',') returns ',6,7,8,16,21,2,'
e_ID=1 --> ',6,7,8,16,21,2,' LIKE '%,1,%' ? FALSE
e_ID=6 --> ',6,7,8,16,21,2,' LIKE '%,6,%' ? TRUE
e_ID=21 --> ',6,7,8,16,21,2,' LIKE '%,21,%' ? TRUE
e_ID=2 --> ',6,7,8,16,21,2,' LIKE '%,2,%' ? TRUE
e_ID=3 --> ',6,7,8,16,21,2,' LIKE '%,3,%' ? FALSE
etc.
Don't know if this is what you want to accomplish. With MySQL there is feature to concatenate values from a group GROUP_CONCAT
You can try something like this:
select * from table where e_ID in (Select GROUP_CONCAT(city SEPARATOR ',') from locations where e_Id=?)
this one in for oracle ..here string concatenation is done by wm_concat
select * from table where e_ID in (Select wm_concat(city) from locations where e_Id=?)
yes i agree with raheel shan .. in order put this "in" clause we need to make that column into row below code one do that job.
select * from table where to_char(e_ID)
in (
select substr(city,instr(city,',',1,rownum)+1,instr(city,',',1,rownum+1)-instr(city,',',1,rownum)-1) from
(
select ','||WM_CONCAT(city)||',' city,length(WM_CONCAT(city))-length(replace(WM_CONCAT(city),','))+1 CNT from locations where e_Id=? ) TST
,ALL_OBJECTS OBJ where TST.CNT>=rownum
) ;
you should use
FIND_IN_SET Returns position of value in string of comma-separated values
mysql> SELECT FIND_IN_SET('b','a,b,c,d');
-> 2
You need to "SPLIT" the city column values. It will be like:
SELECT *
FROM table
WHERE e_ID IN (SELECT TO_NUMBER(
SPLIT_STR(city /*string*/
, ',' /*delimiter*/
, 1 /*start_position*/
)
)
FROM locations);
You can read more about the MySQL split_str function here: http://blog.fedecarg.com/2009/02/22/mysql-split-string-function/
Also, I have used the TO_NUMBER function of Oracle here. Please replace it with a proper MySQL function.
IN takes rows so taking comma seperated column for search will not do what you want but if you provide data like this ('1','2','3') this will work but you can not save data like this in your field whatever you insert in the column it will take the whole thing as a string.
You can create a prepared statement dynamically like this
set #sql = concat('select * from city where city_id in (',
(select cities from location where location_id = 3),
')');
prepare in_stmt from #sql;
execute in_stmt;
deallocate prepare in_stmt;
Ref: Use a comma-separated string in an IN () in MySQL
Recently I faced the same problem and this is how I resolved it.
It worked for me, hope this is what you were looking for.
select * from table_name t where (select (CONCAT(',',(Select city from locations l where l.e_Id=?),',')) as city_string) LIKE CONCAT('%,',t.e_ID,',%');
Example: It will look like this
select * from table_name t where ',6,7,8,16,21,2,' LIKE '%,2,%';
I want to retrieve records from table, all the records are decimal values in (width x Height) format
using like clause I want to to get data but query didn't work for it
Table:
id widthheight
1 17.14X12.41
2 23.4X39.8
select * from test where widthHeight like '%17X12%'
As said above, create two columns 'width' and 'heigth'
select * from (
select
id,
SUBSTRING_INDEX(widthHeight, 'X', 1) width,
SUBSTRING_INDEX(widthHeight, 'X', -1) height
from mytable) x
where floor(x.width)=17 and floor(x.height)=12
Your original query was not working because you should have done this:
select *
from mytable
where widthHeight like '%17%' and widthHeight like '%12%'
Better solution is to use a regex query:
SELECT * FROM test WHERE widthHeight REGEXP '17(\.\d+)?X12(\.\d+)?'
This regexp will match with following values:
17X12
17.14X12.41
17.00X12.00
17.9999999X12.999999
You can playground with this regex on the awesome Regex101. Go to this link.
Note: This query won't be efficient. Consider splitting your data into two columns named height and width that uses DECIMAL column type so you can do more efficient queries.
You should ideally just store the width and height as two separate decimal columns. That being said, you might be able to work with what you have now using some string tricks:
SELECT *
FROM test
WHERE
SUBSTRING_INDEX(widthHeight, 'X', 1) REGEXP '17[[:>:]]' AND
SUBSTRING_INDEX(widthHeight, 'X', -1) REGEXP '12[[:>:]]';
I tried something out. Here is a simple example in SQL Fiddle: Example
There is a column someNumbers (comma-seperated numbers) and I tried to get all the rows where this column contains a specific number. Problem is, the result only contains rows where someNumbers starts with the specific number.
The query SELECT * FROM myTable where 2 in ( someNumbers ) only returns the row with id 2 and not the row with id 1.
Any suggestions? Thank you all.
You are storing data in the wrong format! You should not be storing multiple values in a single string column. You should not be storing numbers as strings. Instead, you should have a junction table with one row per id and per number.
Sometimes, you just have no choice, because someone else created a really poorly designed database. For these situations, MySQL has the function find_in_set():
SELECT *
FROM myTable
WHERE find_in_set(2, someNumbers ) > 0;
The right solution, however, is to fix the data model.
While Gordon's answer is a good one, here is a way to do this with like
SELECT * FROM myTable where someNumbers like '2,%' or someNumbers like '%,2,%' or someNumbers like '%,2'
The first like checks if your array starts with the number you are looking for (2). The second one checks if 2 is within the array and the last like tests for appearance at the end.
Note that the commas are essential here, because something like '%2%' would also match ...,123,...
EDIT: As suggested by the OP it may happen that only a single value is present in the row. Consequently, the query must check this case by doing ... someNumbers = '2'
I would suggest this query :
SELECT * FROM myTable where someNumbers like '%2%'
It will select every entry where someNumbers contains '2'
Select * from table_name where coloumn_name IN(value,value,value)
you can use it
i have in my table places named field. there are space separated values(there are problem to store csv value in one field). now i want to fire query like below. how i can do ??
select * from tablename where variablename in places
i did try this way but it shows syntax error.
select * from tablename where variablename in replace(places,' ',',')
### places ###
bank finance point_of_interest establishment
Use FIND_IN_SET
For comma separated
SELECT *
FROM tablename
WHERE ( FIND_IN_SET( 'bank', variablename ) )
Refer : SQL Fiddle
For space separated
SELECT *
FROM tablename
WHERE ( FIND_IN_SET( 'bank', replace(variablename,' ',',') ) )
Refer : SQL Fiddle
The best solution would be to normalise your data structure and do not have a single field storing multiple values.
You can make a query work without normalisation, but any solutions would be lot less optimal from a performance point of view.
Use patter matching with like operator:
... where fieldname like '% searched_value %'
Use the replace() function and combine it with find_in_set():
... where find_in_set('searched_value',replace(fieldname,' ',','))>0
Hi I think your problem comes from the usage of IN
IN for MySql is used like this
SELECT *
FROM table_name
WHERE column_name IN (bank,finance,point_of_interest, establishment);
In case of you want to select places you need to specify each place into value like
I have a table like:
id name
--------
1 clark_009
2 clark_012
3 johny_002
4 johny_010
I need to get results in this order:
johny_002
clark_009
johny_010
clark_012
Do not ask me what I already tried, I have no idea how to do this.
This will do it, very simply selecting the right-most 3 characters and ordering by that value ascending.
SELECT *
FROM table_name
ORDER BY RIGHT(name, 3) ASC;
It should be added that as your data grows, this will become an inefficient solution. Eventually, you'll probably want to store the numeric appendix in a separate, indexed integer column, so that sorting will be optimally efficient.
you should try this.
SELECT * FROM Table order by SUBSTRING(name, -3);
good luck!
You may apply substring_index function to parse these values -
select * from table order by substring_index(name, '_', -1)
You can use MySQL SUBSTRING() function to sort by substring
Syntax : SUBSTRING(string,position,length)
Example : Sort by last 3 characters of a String
SELECT * FROM TableName ORDER BY SUBSTRING(FieldName, -3);
#OR
SELECT * FROM TableName ORDER BY SUBSTRING(FieldName, -3,3);
Example : Sort by first 3 characters of a String
SELECT * FROM TableName ORDER BY SUBSTRING(FieldName, 1,3);
Note : Positive Position/Index start from Left to Right and Negative Position/Index start from Right to Left of the String.
Here is the details about SUBSTRING() function.
If you want to order by the last three characters (from left to right) with variable name lengths, I propose this:
SELECT *
FROM TABLE
ORDER BY SUBSTRING (name, LEN(name)-2, 3)
The index starts at lenght of name -2 which is the third last character.
I'm a little late but just encountered the same problem and this helped me.