Can someone give me a query that will return as a result rows ID 1 & 3?
ID Name Hidden
1 Mika 1,4,2
2 Loca 0
3 Nosta 4
4 Like 2
Something like this
SELECT * FROM table WHERE Hidden HAVING(4)
SELECT * FROM table WHERE FIND_IN_SET('4',Hidden);
docs for FIND_IN_SET
SELECT * FROM table WHERE CONCAT(',',Hidden,',') LIKE '%,4,%'
or you can avoid using LIKE like this
SELECT * FROM table WHERE INSTR(CONCAT(',',Hidden,','), ',4,') > 0
this will not get things like 40, 14, etc, but you need to make sure there are no spaces in the Hidden field (eg, 1, 4, 5 or update the concat and LIKE function accordingly.
SELECT * FROM table WHERE Hidden LIKE '%4%'
the % are wildcards.
Full Text Search might be a reasonable solution for this as long as you use the correct word breaks.
Either go with Full Text Search, as suggested, or
Spin the Hidden values off into a separate table, with the ID of current row.
Eg, Mika would have three entries in this table
ID = 1, Hidden =1
ID = 1, Hidden =4
ID = 1, Hidden =2
Then you could return results against this spin off table.
You may also want to consider normalizing the table and storing these "hidden" values in a separate table with an index on the apropriate column. Depending on the number of rows you have that would be much faster:
ID Hidden
1 1
1 4
1 2
3 4
4 2
and:
SELECT DISTINCT table.* FROM table, hidden_table WHERE table.ID = hidden_table.ID AND hidden_table.hidden = 4
Related
When I use in keyword in sql, there may be some id is missing , but I want treat them like they exist and other columns are null or 0.
For example, suppose I have a table with two columns and some rows:
[id,value1]
1 1
2 4
3 3
5 5
I may write sql like this:
select * from table where id in (1,4,5) order by value1 limit 0,2 ;
When this sql is executed, the return result is [(1,1),(5,5)].
But what I want is [(4,0),(1,1)], because I want to treat the missing id 4 like it exists in the table.
So the question is : Is there some elegant way to achieve it using sql instead of select all rows and sort them in memory.
Use a left join:
select *
from (select 1 as id union all
select 4 union all
select 5
) i left join
table t
using (id)
order by t.value1
limit 0, 2 ;
Note that you are ordering by a value in the existing table, so this depends on the fact that NULL is ordered before other values.
I have a table that contains this kind a structure of a column, how can I make select only from character 4 to 6 to ignore other character that are outside this boundary , I tried LIKE'%544%', RegExp. ect. ??
1 000544001
2 000054400
3 000544010
4 000344010
5 000544011
One way is to use substr():
where substr(col, 4, 3) = '544'
another is to use like:
where col like '___544%'
you can also use mid(col,start,length) statement
like this
select column1 from table1 where mid(column1,4,3);
I want to achieve one thing though I'm not sure if it's possible.
So let's say I have a table with few columns, but only two of them are of interest to me right now. Example of table:
Column 1 | Column 2
blabla | blablahhhhh
wor154 | blablahhhhh
word123 | word12435564
something | some4565
What I want to achieve, is to select all fields where first 5 or more symbols of value of Column 2 don't match with first 5 or more symbols of value of Column 1. So I don't want to select rows where 5 or more symbols of value of Column 1 match 5 or more symbols of value of Column 2. In example, query should return only 2nd and 4th rows
So, is it possible and if it's, how it can be achieved. Thank you.
I'd go with a SUBSTRING():
SELECT col1 FROM table WHERE SUBSTRING(col1, 1, 5) <> SUBSTRING(col2, 1, 5);
You can use something similar to this:
select *
from table1
where substring(column1, 1, 5) != substring(column2, 1, 5)
See SQL Fiddle with Demo
I need to find best matches from a mysql table given a set of attributes.
For example, given ATTRIBUTE1, ATTRIBUTE2 and ATTRIBUTE3, I want to get the results as follows:
rows with all attributes matched
rows with 2 attributes matched
rows with 1 attribute matched
so far I only know how to accomplish the first statement:
SELECT * FROM Users
WHERE ATTRIBUTE1="aValue", ATTRIBUTE2="aValue", ATTRIBUTE3="aValue"
LIMIT 20
N.B. I need 2 lists. A list with fully matching rows and a list with partial matches
you can consider to build an composite index in ATTRIBUTE{1..3}
this will benefits for List A
SELECT *
FROM Users
WHERE ATTRIBUTE1="aValue" AND ATTRIBUTE2="aValue" AND ATTRIBUTE3="aValue"
LIMIT 20
and might help some row in List B
SELECT *,
IF (ATTRIBUTE1="aValue", 1, 0) as a1,
IF (ATTRIBUTE2="aValue", 1, 0) as a2,
IF (ATTRIBUTE3="aValue", 1, 0) as a3
FROM Users
WHERE ATTRIBUTE1="aValue" OR ATTRIBUTE2="aValue" OR ATTRIBUTE3="aValue"
ORDER BY (a1+a2+a3) DESC
LIMIT 20
Is there a way to do something like this?
SELECT * FROM tablename WHERE x CONTAINS "1"
Basically, I want to select data from the database where x contains a specific number. The thing is, the x column in any row could contain "1, 2, 3" and I want to select all those that contain 1, specifically 1, not 11 or anything that contains 1, but specifically a 1.
Here's an example:
id title x
-------------------
1 row1 1,22,3
2 row2 1,5
3 row3 5,91
4 row4 70
And I want my query to return rows 1 and 2. I don't want row 3, as the 1 is inside the number 91. I don't want row 4 because there's no 1 there either.
You can use the FIND_IN_SET function like so:
SELECT * FROM tablename WHERE FIND_IN_SET('1', x)
This will also get optimised to use bit arithmetic if you are calling it on a SET type.
You can try this:
SELECT * FROM tablename WHERE x REGEXP "(^|,)1(,|$)"
Ideally you'd normalize your 'x' column out to a separate table.
But... you could also hack it like this:
SELECT * FROM tablename WHERE x LIKE '%,1' OR x LIKE '1,%' OR x LIKE '%,1,%'
This basically just handles the three different cases where the "1" is the first, last or a middle element in your list. (note if you've got a space after your commas you'd change the last part to '%, 1,%'
EDIT: Actually Dmitriy's REGEXP is nicer, and a'r's FIND_IN_SET looks ideal.