How to make MS Query work with comma separated parameters in Excel cell?
My query is:
SELECT *
FROM ABC
WHERE Id in (?)
When I put id number for example "1" the query works, but I want to put into a parameters cell a few id's 1, 2, 3, 4 etc, but then I'm trying to to this the query doesn't work... How can I put parameter with comma separated values?
there is 2 diff way to do it:
select * from abc where id in ('1','2','3') etc but not in excel - maybe use notepad++
second way :)
select * from abc where (id like '1' or id like '2' or id like '3') etc
:)
You can use IN in you sql query.
SELECT column_name(s)
FROM table_name
WHERE column_name IN (1,2,3,4);
also try to use BETWEEN with comma as parameters.
SELECT column_name(s)
FROM table_name
WHERE column_name BETWEEN value1 AND value2;
Related
I am using oracle database and writing the likeness filter for the persistence layer and want to perform likeness on a column that can possibly have '%' in it's data.
To filter data I am writing the query for likeness using LIKE clause as
select * from table where columnName like '%%%';
which is returning all the values but I only want the rows that contains '%' in the columnName.
Not sure what escape character to use or what to do to filter on the '%' symbol. Any suggestions??
Also, I have to do the same thing using Criteria api in java and have no clues about putting escape character there.
You can use an escape character.
where columnName like '%$%%' escape '$'
REGEXP_LIKE might help in a rather simple manner.
SQL> with test (col) as
2 (select 'abc%def' from dual union all
3 select '%12345&' from dual union all
4 select '%abc12%' from dual union all
5 select '1234567' from dual
6 )
7 select *
8 from test
9 where regexp_like(col, '%');
COL
-------
abc%def
%12345&
%abc12%
SQL>
If I have understood it correctly the answer from the #Littlefoot is correct(and I will up-vote it now). I will just add more details because you are looking for name of the columns of your table "I only want the rows that contains '%' in the columnName".
Here is the table I have created:
CREATE TABLE "table_WITH%" ("numer%o" number
, name varchar2(50)
, "pric%e" number
, coverage number
, activity_date date);
Then this query gives me the correct answer:
SELECT column_name
FROM ALL_TAB_COLUMNS
WHERE TABLE_NAME = 'table_WITH%'
AND regexp_like(COLUMN_NAME, '%');
Gordon's answer using the escape clause of the like command is correct in the general case.
In your specific case (searching for a single character, anywhere in a string), a simpler method is to use instr, e.g.
where instr(columnname, '%') > 0
Following is the query to select column data from table, where column data starts with a OR b OR c. But the answer i am looking for is to Select data which starts with List of Strings.
SELECT * FROM Table WHERE Name LIKE '[abc]%'
But i want something like
SELECT * FROM Table WHERE Name LIKE '[ab,ac,ad,ae]%'
Can anybody suggest what is the best way of selecting column data which starts with list of String, I don't want to use OR operator, List of strings specifically.
The most general solution you would have to use is this:
SELECT *
FROM Table
WHERE Name LIKE 'ab%' OR Name LIKE 'ac%' OR Name LIKE 'ad%' OR Name LIKE 'ae%';
However, certain databases offer some regex support which you might be able to use. For example, in SQL Server you could write:
SELECT *
FROM Table
WHERE NAME LIKE 'a[bcde]%';
MySQL has a REGEXP operator which supports regex LIKE operations, and you could write:
SELECT *
FROM Table
WHERE NAME REGEXP '^a[bcde]';
Oracle and Postgres also have regex like support.
To add to Tim's answer, another approach could be to join your table with a sub-query of those values:
SELECT *
FROM mytable t
JOIN (SELECT 'ab' AS value
UNION ALL
SELECT 'ac'
UNION ALL
SELECT 'ad'
UNION ALL
SELECT 'ae') v ON t.vame LIKE v.value || '%'
I have a table with a field name.
For a given name like ABC I want to get the records with that name and the records which have an L appended at the end.
So for ABC I want all records with name either ABC or ABCL.
I tried getting the records using the following code but it doesn't work.
SELECT * FROM tbl
WHERE name like "ABC[|L]"
I am using TSQL.
How can pattern match these names?
Use this SQL:
SELECT * FROM tbl WHERE name IN ('ABC', 'ABCL')
If you are using this SQL within a Stored Procedure / Function, something like following would work. Assuming, you are passing in the value for name in a #name variable.
SELECT * FROM tbl WHERE name IN (#name, #name + 'L')
Use the IN operator.
SELECT *
FROM tbl
WHERE name IN ('ABC', 'ABCL')
In case you will be using variables instead of literal strings, try this:
select *
from tbl
where name like (#YourName + #ExtraLetter)
or name = #YourName
Let's say I have multiple 6 character Alphanumeric strings. abc123, abc231, abc456, cba123, bac231, and bac123.
Basically I want a select statement that can search and list all the abc instances.
I just want a select statement that can list all instances with keyword "abc".
Use LIKE and wildcards %
SELECT *
FROM yourtable
WHERE yourfield LIKE '%abc%'
Input
yourfield
abc123
abc231
abc456
cba123
bac231
bac123
Output:
yourfield
abc123
abc231
abc456
SQL Fiddle:
Could you try like this way
SELECT *
FROM yourtable
WHERE field LIKE '%a%' Or field LIKE '%b%' Or field LIKE '%c%'
Select * from tablename
where yourfield like 'yourconstantcharacters%'
You can use LIKE operator in sql
Starting With Case
Select * From Table Where Column LIKE 'abc%';
Ending With Case
Select * From Tablo Where Column Like '%abc';
Contains With Case
Select * From Table Where Column LIKE '%abc%';
So If you want to use escape character in LIKE condition , you must make small changes on your condition
For Example :
SELECT * FROM Table WHERE column LIKE '!%' escape '!';
More information for you is here
What is the most efficient and elegant SQL query looking for a string containing the words "David", "Moses" and "Robi". Assume the table is named T and the column C.
Select * from table where
columnname like'%David%' and
columnname like '%Moses%' and columnname like'%Robi%'
In SQL Server 2005+ with Full-Text indexing switched on, I'd do the following:
SELECT *
FROM T
WHERE CONTAINS(C, '"David" OR "Robi" OR "Moses"');
If you wanted your search to bring back results where the result is prefixed with David, Robi or Moses you could do:
SELECT *
FROM T
WHERE CONTAINS(C, '"David*" OR "Robi*" OR "Moses*"');
Here is what I uses to search for multiple words in multiple columns - SQL server
Hope my answer help someone :) Thanks
declare #searchTrm varchar(MAX)='one two three ddd 20 30 comment';
--select value from STRING_SPLIT(#searchTrm, ' ') where trim(value)<>''
select * from Bols
WHERE EXISTS (SELECT value
FROM STRING_SPLIT(#searchTrm, ' ')
WHERE
trim(value)<>''
and(
BolNumber like '%'+ value+'%'
or UserComment like '%'+ value+'%'
or RequesterId like '%'+ value+'%' )
)
If you care about the sequence of the terms, you may consider using a syntax like
select * from T where C like'%David%Moses%Robi%'
Oracle SQL :
select *
from MY_TABLE
where REGEXP_LIKE (company , 'Microsodt industry | goglge auto car | oracles database')
company - is the database column name.
results - this SQL will show you if company column rows contain one of those companies (OR phrase)
please note that : no wild characters are needed, it's built in.
more info at : http://www.techonthenet.com/oracle/regexp_like.php
if you put all the searched words in a temporaray table say #tmp and column col1, then you could try this:
Select * from T where C like (Select '%'+col1+'%' from #temp);
Maybe EXISTS can help.
and exists (select 1 from #DocumentNames where pcd.Name like DocName+'%' or CD.DocumentName like DocName+'%')
Oracle SQL:
There is the "IN" Operator in Oracle SQL which can be used for that:
select
namet.customerfirstname, addrt.city, addrt.postalcode
from schemax.nametable namet
join schemax.addresstable addrt on addrt.adtid = namet.natadtid
where namet.customerfirstname in ('David', 'Moses', 'Robi');