Table description
Table Name:- Name condition
Name | Pattern
A | %A% or Name like %a%
B | %B% or Name like %b%
C | %C% or Name like %c%
D | %D% or Name like %d%
E | %E% or Name like %e%
F | %F% or Name like %f%
G | %G% or Name like %g%
Table name:- Employees
Emp_ID | EMP_NAME
1 | Akshay
2 | Akhil
3 | Gautam
4 | Esha
5 | bhavish
6 | Chetan
7 | Arun
[Table description] [1]: https://i.stack.imgur.com/wvOgr.png
Above are my two tables now my query is (in the image)
Select * from Employees,Name_condition where EMP_NAME like Pattern
Here the query is correct syntactically but produces wrong output.
It takes the column Pattern as a string and searches for it in EMP_NAME and it will find nothing.
So my question is how we can take the values present in the Pattern column as a condition and not as a string so that the query will become like this
Select * from Employees,Name_condition where EMP_NAME like ‘%A%’ or Name like ‘%a%’
what i need is when i pass colunm name(Pattern) in the where condition it takes %A% or Name like %a% whole as a string but i want that select * from Employees,Name_condition where EMP_NAME like Pattern Here the column name pattern internally must be replace by the value present in the column and the the query produces o/p like this
Select * from Employees,Name_condition where EMP_NAME like ‘%A%’ or Name like ‘%a%’
Desired Result:-I expect all the rows in my result which includes bhavish but as we see we have a like condition in the column itself like %B% or Name like %b%
What i want is when it matches
where EMP_NAME like Pattern
The value of pattern must internally replaced by
%B% or Name like %b%
and the it produces the output which includes bhavish which starts with b
try:
select *
from employees
where emp_name like '%oh%'
or emp_name like '%a%';
Good luck.
Try this from orafaq:
SELECT * FROM employees
WHERE emp_name LIKE '\%a\%' ESCAPE '\';
It's not that simple (and it shouldn't be). If you really have to use such tables you have to write a piece of PL/SQL to handle your conditions.
Two things you have to read about:
dynamic sql
sql injection (because you want to prevent it)
Try to Put && instead of 'And' in condition
Related
Sorry, I don't know how to describe that as a title.
With a query (example: Select SELECT PKEY, TRUNC (CREATEDFORMAT), STATISTICS FROM BUSINESS_DATA WHERE STATISTICS LIKE '% business_%'), I can display all data that contains the value "business_xxxxxx".
For example, the data field can have the following content: c01_ad; concierge_beendet; business_start; or also skill_my; pre_initial_markt; business_request; topIntMaster; concierge_start; c01_start;
Is it now possible in a temp-only output the corresponding value in another column?
So the output looks like this, for example?
PKEY | TRUNC(CREATEDFORMAT) | NEW_STATISTICS
1 | 13.06.2020 | business_start
2 | 14.06.2020 | business_request
That means removing everything that does not start with business_xxx? Is this possible in an SQL query? RegEx would not be the right one, I think.
I think you want:
select
pkey,
trunc(createdformat) createddate,
regexp_substr(statistics, 'business_\S*') new_statistics
from business_data
where statistics like '% business_%'
You can also use the following regexp_substr:
SQL> select regexp_substr(str,'business_[^;]+') as result
2 from
3 --sample data
4 (select 'skill_my; pre_initial_markt; business_request; topIntMaster; concierge_start; c01_start;' as str from dual
5 union all
6 select 'c01_ad; concierge_beendet; business_start;' from dual);
RESULT
--------------------------------------------------------------------------------
business_request
business_start
SQL>
I'm wondering how can I use a row value as a variable for my like statement? For example
ID | PID | DESCRIPTION
1 | 4124 | Hi4124
2 | 2451 | Test
3 | 1467 | Hello
4 | 9642 | Me9642
I have a table above, I want to return IDs 1 and 4 since DESCRIPTION contains PID.
I'm thinking it would be SELECT * from TABLE WHERE DESCRIPTION LIKE '%PID%' but I can't get it.
You can use CONCAT() to assemble the matching pattern, as in:
select *
from t
where description like concat('%', PID, '%')
We could also try using CHARINDEX here:
SELECT ID, PID, DESCRIPTION
FROM yourTable
WHERE CHARINDEX(PID, DESCRIPTION) > 0;
Demo
Note that I assume in the demo that the PID column is actually text, and not a numeric column. If PID be numeric, we might have to first use a cast in order to use CHARINDEX (or any of the methods given in the other answers).
Use the CONCAT SQL function
SELECT *
FROM TABLE
WHERE DESCRIPTION LIKE CONCAT('%', PID, '%')
I have table called INFO
ID Name
1 John
2 David
3 Kim
... ...
... ...
And I want to select Name with first letter D
SELECT Name FROM INFO
WHERE Name LIKE 'D%';
But I don't get any results. Why? What is wrong?
You should use * instead of %. see here for examples.
SELECT Name FROM INFO
WHERE Name LIKE 'D*';
So, at the moment I have two columns in a table, one of which containing a JSON document, like so:
CID:
2
Name:
{"first" : "bob","1" : "william", "2" : "archebuster", "last" : "smith"}
When I do a search on this column using:
SELECT "CID", "Name"->>(json_object_keys("Name")) AS name FROM "Clients" WHERE
"Name"->>'first' LIKE 'bob' GROUP BY "CID";
I get:
CID | name
--------------
2 | bob
2 | william
2 | archebuster
2 | smith
When really I want:
CID | name
2 | bob william archebuster smith
How would i go about doing this? I'm new to JSON in postgresql.
I've tried string_agg and it wouldn't work, presumably because i'm working in a json column, despite the fact '->>' should type set the result to string
UPDATE:
First, you need to understand, if you include a set-returning function into the SELECT clause, you will create an implicit LATERAL CROSS JOIN.
Your query in reality looks like this:
SELECT "CID", "Name"->>"key" AS name
FROM "Clients"
CROSS JOIN LATERAL json_object_keys("Name") AS "foo"("key")
WHERE "Name"->>'first' LIKE 'bob'
GROUP BY "CID", "Name"->>"key"
If you really want to do that, you can apply an aggregate function here (possibly array_agg or string_agg).
SQLFiddle
I have some table 'TableName' like Id,Name:
1 | something
2 | _something
3 | something like that
4 | some_thing
5 | ...
I want to get all rows from this table where name containes 'some'.
I have 2 ways:
SELECT * FROM TableName WHERE Name like '%some%'
Result is table :
1 | something
2 | _something
3 | something like that
4 | some_thing
But if I use CONTAINS function
SELECT * FROM TableName WHERE CONTAINS(Name,'"*some*"')
I get only
1 | something
3 | something like that
What should I do to make CONTAINS function work properly?
The last time I looked (admittedly SQL Server 2000) CONTAINS didn't support wildcard matching at the beginning of words, only at the end. Also, you might need to check your noise files to see if the "_" character is being ignored.
Also see
How do you get leading wildcard full-text searches to work in SQL Server?
http://doc.ddart.net/mssql/sql70/ca-co_15.htm
If you read this article you will see that * means prefix this means that word must start with this, but like means the word contains key phrase.
Best Regards,
Iordan
Try this:
SELECT * FROM TableName WHERE CONTAINS(Name,'some')