I had a table like bellow image. I want 2,3 letters from the column "name"
The result is like bc,da,bc,bc,bc,bc
SELECT SUBSTRING(name,2,2) AS ShortComp FROM table;
You can also use the string function LEFT() to return the leftmost specified number of characters, like this:
SELECT LEFT(name, 2) AS ShortComp FROM table;
Related
I have a variable varchar that always takes in 10 digits. How can I use the LIKE operator to find/use only the first 5 digits of the variable?
my query:
variable IN VARCHAR2
SELECT * FROM items WHERE name LIKE SUBSTRING(variable, 1, 5)
... WHERE name LIKE '12345%'
will match any string that starts 12345. the '%' is a wildcard. You can also use the wildcard to match anywhere in the string: ... WHERE name LIKE '%12345%' will match a string with 12345 anywhere within it.
Edit for completeness: WHERE name LIKE '%12345' will match any string that ends with those five characters.
Try this:
SELECT * FROM items WHERE name LIKE (SUBSTRING(variable, 1, 5) + '%')
I guess you can use LEFT() like this:
SELECT * FROM items WHERE LEFT(name,5)=LEFT(variable,5);
Or if you you want to use LIKE with a wildcard, you can do this:
SELECT * FROM items WHERE name LIKE CONCAT(LEFT(variable,5),'%')
A few more example in the Demo fiddle
Edit: The above solution is for MySQL/MariaDB because earlier the tag of this question have MySQL but it's also my fault for not recognizing OP description of the datatype VARCHAR2. I might as well just post a suggestion related to the rdbms.
So, my first suggestion there using LEFT() however Oracle don't have that function, therefore:
SELECT * FROM items WHERE SUBSTR(name,1,5)=SUBSTR(variable,1,5);
or using concatenation operator
SELECT * FROM items WHERE name LIKE SUBSTR(variable,1,5)||'%'
Demo fiddle
I have a column as varchar2 datatype, the data in it is in format:
100323.3819823.222
100.323123.443422
1001010100.233888
LOL12333.DDD33.44
I need to remove the whole part after the first occurrence of '.'
In the end it should look like this:
100323
100
1001010100
LOL12333
I cant seem to find the exact substring expression due to the fact that there is not any fix length of the first part.
One way is to use REGEXP_SUBSTR:
SELECT REGEXP_SUBSTR(column_name,'^[^.]*') FROM table
The other way is to combine SUBSTR with INSTR, which is a bit faster, but will result in NULL if the data doesn't contain a dot, so you'll have to add a switch if needed:
SELECT SUBSTR(column_name, 1, INSTR(column_name,'.') - 1) FROM table
For oracle you can try this:
select substr (i,1,Instr(i,'.',i)-1) from Table name.
I need to return the fields that have more than one . in a specific column.
Now I have this query:
select *
from table
where column ~ '\.{2,}?';
But for some reason it returns nothing. If I use something like 'A{2,}?' it works. Apparently the problem is the dot.
It returns null since the dots are not next two each other. You have to consider the occurrences of the characters in the order of your regex meta characters. You could try this instead:
select *
from table
where column ~ '\.\d{3}\.';
Or instead of just focusing on the dot characters start parsing the string as a whole and consider the numbers as well:
where column ~ '^\d{3}\.\d{3}\.';
Why not just use like?
where column like '%.%.%'
Is it possible to check if a specific substring which is in SQL Server column, is contained in a user provided string?
Example :
SELECT * FROM Table WHERE 'random words to check, which are in a string' CONTAINS Column
From my understanding, CONTAINS can't do such kind of search.
EDIT :
I have a fully indexed text and would like to search (by the fastest method) if a string provided by me contains words that are present in a column.
You can use LIKE:
SELECT * FROM YourTable t
WHERE 'random words ....' LIKE '%' + t.column + '%'
Or
SELECT * FROM YourTable t
WHERE t.column LIKE '%random words ....%'
Depends what did you mean, first one select the records that the column has a part of the provided string. The second one is the opposite.
Just use the LIKE syntax together with % around the string you are looking for:
SELECT
*
FROM
table
WHERE
Column LIKE '%some random string%'
This will return all rows in the table table in which the column Column contains the text "some random string".
1) If you want to get data starting with some letter you can use % this operator like this in your where clause
WHERE
Column LIKE "%some random string"
2) If you want to get data contains any letter you can use
WHERE
Column LIKE "%some random string%"
3)if you want to get data ending with some letter you can use
WHERE
Column LIKE "some random string%"
How can I get the character at position 4 in a field?
e.g.
field contents = "hello"
I want to return the value of position 2 = "l"
In SQL Server you can use SUBSTRING
SELECT SUBSTRING('hello', 3, 1)
Take care: index is 1-based.
In Oracle, use SUBSTR
Syntax is SUBSTR(<string to parse>,<start position>,(<length>)) - i.e.
SELECT SUBSTR('hello',3,1)
Start position and length are one-, not zero-based. Zero is accepted, but will be interpreted as 1.
following Query will search from specific index for char and will return result
select * from tbPatientMaster
where SUBSTRING (fname,CHARINDEX ('.',fname,0)+1,LEN (fname)) like 'a%'