I have been using Like function to filter the records based on the user input, it works for all the characters but its not working for hyphen,below the condition where I'm using this Like function to filter the records having hyphen,
LIKE upper('%-%')
Related
I would like to extract the string using where clause in SAP HANA.For an example,these are 3 strings for name column.
123._SYS_BIC.meag.app.qthor.cidwh_eingangsschicht.backend.dblayer.l2.checks/MasterData_Holdings.
153._SYS_BIC.meag.app.qthor.centralAdministration.backend.dblayer.l2.checks/AuditAndSecurities.
meag.app.qthor.centralAdministration.backend.dblayer.l2.checks/GeneralLedger
After filter the name column using where clause, output in the name column would be shown only the last portion of the string. So, output will be like this. That means whatever we have, just remove from the beginning till '/'.
"MasterData_Holdings"
"AuditAndSecurities"
"GeneralLedger"
You can try using the REPLACE_REGEXPR
I'm not familiar myself with Hana but the function is pretty straight forward and it should be:
select REPLACE_REGEXPR('.+/(.+)' IN fieldName WITH '\1' OCCURRENCE ALL) as field
...
where
... -- your filter
Be aware that this regex '.+/(.+)' will eat everything until the last / so for instance if you have ....checks/MasterData_Holdings/Something it will return only Something
The ID of research fields have three parts, each part separated by a period.
Consider a query to find the details of research fields where the first two parts of the ID are D and 2, and the last part is a single character (digit).
IDs like D.2.1 and D.2.3 are in the query result whereas IDs like D.2.12 or D.2.15 are not.
The SQL query given below does not return the correct result. Explain the reason why it does not return the correct result and give the correct SQL query.
select *
from field
where ID like 'B.1._';
I have no idea why it doesnt work.
Anyone can help on this? Many thanks
D.2.1 and D.2.3 are in the query result whereas IDs like D.2.12 or D.2.15 are not.
An underscore matches any single character in a LIKE filter so B.1._ is looking for the start of the string followed by a B character followed by a . character then a 1 character then a . character then any single character then the end of the string.
You could use:
SELECT *
FROM field
WHERE ID like 'B.1._%';
The % will match any number of characters (including zero) until the end of the string and the preceding underscore will enforce that there is at least one character after the period.
What Oracle SQL query could return the second, third and fourth positions of characters contained within rows of a specific column using the REGEXP_SUBSTR method instead of using SUBSTR method like my example provided below?
SELECT SUBSTR(city,2,3) AS "2nd, 3rd, 4th"
FROM student.zipcode;`
One way that works for me (with test data) is:
SELECT REGEXP_SUBSTR(city, '\S{3}', 2) AS partial FROM student.zipcode;
Note that this is set to find three non-whitespace characters beginning at the second position of the string.
You could also use:
SELECT REGEXP_SUBSTR(city, '.{3}', 2) AS partial FROM student.zipcode;
which will instead match any three characters in the 2nd to 4th position.
However, I'm not sure what advantage this has over simply:
SELECT SUBSTR(city,2,3) AS partial FROM student.zipcode;
The REGEXP_INSTR function is not what you want, as it returns an index (position number) for the search item in the searched string. You can read about it here: http://www.techonthenet.com/oracle/functions/regexp_instr.php
I'm trying to write an active record statement where I will be able to get all rows in the database where the first letter is not equal to an alphabet.
This is my active record query
$this->db->where("employer_name NOT LIKE '[^a-z]%'");
return $this->db->select()->from("v_employers")->get();
the problem with my code is that it gets everything in my database
i only want to get rows where the first letter is not an alphabet
ex(123pencil,$helloworld,7eleven)
and so on
Try this regex in the where clause
WHERE `employer_name` REGEXP '^[^a-zA-Z].*$'
Full query:
$this->db->query("SELECT * FROM v_employers WHERE `employer_name` REGEXP '^[^a-zA-Z].*$'");
If I have a query to return all matching entries in a DB that have "news" in the searchable column (i.e. SELECT * FROM table WHERE column LIKE %news%), and one particular row has an entry starting with "In recent World news, Somalia was invaded by ...", can I return a specific "chunk" of an SQL entry? Kind of like a teaser, if you will.
select substring(column,
CHARINDEX ('news',lower(column))-10,
20)
FROM table
WHERE column LIKE %news%
basically substring the column starting 10 characters before where the word 'news' is and continuing for 20.
Edit: You'll need to make sure that 'news' isn't in the first 10 characters and adjust the start position accordingly.
You can use substring function in a SELECT part. Something like:
SELECT SUBSTRING(column, 1,20) FROM table WHERE column LIKE %news%
This will return the first 20 characters from column column
I had the same problem, I ended up loading the whole field into C#, then re-searched the text for the search string, then selected x characters either side.
This will work fine for LIKE, but not full text queries which use FORMS OF INFLECTION because that may match "women" when you search for "woman".
If you are using MSSQL you can perform all kinds VB-like of substring functions as part of your query.