SQL Find names that contain a letter (without using Like) - sql

I need to write a select statement that returns all last names from a column that contains the letter A. I can't use LIKE. I am trying to do so with SUBSTR.

I don't think substr is the way to go. instr, on the other hand, may do the trick:
SELECT last_name
FROM mytable
WHERE INSTR(last_name, 'A') > 0
EDIT:
As David Bachmann Jeppesen mentioned, Oracle is case sensitive, so if you want to find last names containing any case of "A", you could do something like this:
SELECT last_name
FROM mytable
WHERE INSTR(UPPER(last_name), 'A') > 0

Related

Showing Values in which there is a Substring that starts with a letter in range 'A' to 'X' followed by 2 letter of which second one is 'A' in Oracle

I need to find all the FIRST_NAMES in the table that looks like - There is a substring of 3 letter in the name of which first letter is in range 'A' to 'X' and 3rd one is 'A'. Moreover, the substring should be like '[A-X]_A' and the Name should be like '%[A-X]_A%'. But however ORACLE is showing NO DATA FOUND . Whenever I try with '[]' ORACLE can't find any data. Can anyone give me the solution?
For more information: I am using HR SCHEMA in LIVESQL.oracle.com for the purpose.
You want a regular expression:
where regexp_like(col, '[A-X].A')

Find the last word of a sentence SQL

I'm trying to return a table that contains all the employees who do have the first name starting with an 'A' and the surname starting with a 'R', from the tables 'dep' and 'emp'. I tried to use the INDEXOF function but seems it doesn't work . what can I do?
select emp.nome,emp.sal,emp.ndep,dep.nome,
from emp, dep
where (substring(emp.nome,1,indexof(' ',emp.nome)))like 'A%'
and (substring(emp.nome,lastindexof(' ',emp.nome),emp.nome.lenght)) like 'R%'
order by 1
There's no point doing the first substring, as a string 'ABC DEF' is like 'A%' whether or not you cut the bit off after the space.
The surname, your use of lastindexof causes the space to become part of the name because you forgot to add 1 to skip onto the next character after the space, and ' RST' is never like 'R%'. I swapped your lastindexof to instr, because I wasn't sure if lastindexof is as universally applicable as instr. calling instr with a negative start index causes it to search backward from the end of the string
select emp.nome,emp.sal,emp.ndep,dep.nome,
from
emp
INNER JOIN dep ON /*something_goes_here*/
where emp.nome like 'A%' and
SUBSTR(emp.nome, INSTR(emp.nome, ' ', -1) + 1,emp.nome.length)) like 'R%'
order by 1
You need to finish the query off by putting a clause in describing how the tables should be joined
I would expect a condition like this:
where nom like 'A% R%'
assuming that nom has both the first name and last name.

Get total number of user where username have defferrent case

I have SQL table where username have different cases for example "ACCOUNTS\Ninja.Developer" or "ACCOUNTS\ninja.developer"
I want to find the how many records where username where first in first and last name capitalize ? how can use Regex to find the total ?
x table
User
"ACCOUNTS\James.McAvoy"
"ACCOUNTS\michael.fassbender"
"ACCOUNTS\nicholas.hoult"
"ACCOUNTS\Oscar.Isaac"
Do you want something like this?
select count(*)
from t
where name rlike 'ACCOUNTS\[A-Z][a-z0-9]*[.][A-Z][a-z0-9]*'
Of course, different databases implement regular expressions differently, so the actual comparator may not be rlike.
In SQL Server, you can do:
select count(*)
from t
where name like 'ACCOUNTS\[A-Z][^.][.][A-Z]%';
You might need to be sure that you have a case-sensitive collation.
In most cases in MS SQL string collation is case insensitive so we need some trick. Here is an example:
declare #accts table(acct varchar(100))
--sample data
insert #accts values
('ACCOUNTS\James.McAvoy'),
('ACCOUNTS\michael.fassbender'),
('ACCOUNTS\nicholas.hoult'),
('ACCOUNTS\Oscar.Isaac')
;with accts as (
select
--cleanup and split values
left(replace(acct,'ACCOUNTS\',''),charindex('.',replace(acct,'ACCOUNTS\',''),0)-1) frst,
right(replace(acct,'ACCOUNTS\',''),charindex('.',replace(acct,'ACCOUNTS\',''),0)) last
from #accts
)
,groups as (--add comparison columns
select frst, last,
case when CAST(frst as varbinary(max)) = CAST(lower(frst) as varbinary(max)) then 'lower' else 'Upper' end frstCase, --circumvert case insensitive
case when CAST(last as varbinary(max)) = CAST(lower(last) as varbinary(max)) then 'lower' else 'Upper' end lastCase
from accts
)
--and gather fruit
select frstCase, lastCase, count(frst) cnt
from groups
group by frstCase,lastCase
Your question is a little vague but;
You might be looking for the DISTINCT command.
REF
I don't think you need regex.
Maybe do something like:
Get distinct names from Table X as Table A
Use inputs table A as where clause on Table X
count
union
I hope this helps,
Rhys
Given your example set you can use a combination of techniques. First if the user name always begins with "ACCOUNTS\" then you can use substr to select the characters that start after the "\" character.
For the first name:
Then you can use a regex function to see if it matches against [A-Z] or [a-z] assuming your username must start with an alpha character.
For the last name:
Use the instr function on the substr and search for the character '.' and again apply the regex function to match against [A-Z] or [a-z] to see if the last name starts with an upper or a lower character.
To total:
Select all matches where both first and last match against upper and do a count. Repeat for the lower matches and you'll have both totals.

Matching exactly 2 characters in string - SQL

How can i query a column with Names of people to get only the names those contain exactly 2 “a” ?
I am familiar with % symbol that's used with LIKE but that finds all names even with 1 a , when i write %a , but i need to find only those have exactly 2 characters.
Please explain - Thanks in advance
Table Name: "People"
Column Names: "Names, Age, Gender"
Assuming you're asking for two a characters search for a string with two a's but not with three.
select *
from people
where names like '%a%a%'
and name not like '%a%a%a%'
Use '_a'. '_' is a single character wildcard where '%' matches 0 or more characters.
If you need more advanced matches, use regular expressions, using REGEXP_LIKE. See Using Regular Expressions With Oracle Database.
And of course you can use other tricks as well. For instance, you can compare the length of the string with the length of the same string but with 'a's removed from it. If the difference is 2 then the string contained two 'a's. But as you can see things get ugly real soon, since length returns 'null' when a string is empty, so you have to make an exception for that, if you want to check for names that are exactly 'aa'.
select * from People
where
length(Names) - 2 = nvl(length(replace(Names, 'a', '')), 0)
Another solution is to replace everything that is not an a with nothing and check if the resulting String is exactly two characters long:
select names
from people
where length(regexp_replace(names, '[^a]', '')) = 2;
This can also be extended to deal with uppercase As:
select names
from people
where length(regexp_replace(names, '[^aA]', '')) = 2;
SQLFiddle example: http://sqlfiddle.com/#!4/09bc6
select * from People where names like '__'; also ll work

Ignoring spaces between words - an SQL question

I want to grab from a db a record with name='John Doe'.
I'd like my query to also grab 'John(4 spaces between)Doe','John(2 spaces betwewen)Doe' etc. (at least one space however).
I'd also like that the case won't matter, so I can also get 'John Doe' by typing
'john doe' etc.
Try this:
SELECT * FROM table WHERE lower(NAME) like 'john%doe%'
use like with wildcards (e.g. %) to get around the spaces and the lower (orlcase) to be case insensitive.
EDIT:
As the commenters pointed out, there are two shortcomings within this solution.
First: you will select "johnny doe" or worse "john Eldoe", or worse, "john Macadoelt" with this query, so you'll need extra filtering on the application side.
Second: using a function can lead to table scans instead of index scans. This may be avoided, if your dbms supports function based indexes. See this Oracle example
If your database has Replace function
Select * From Table
Where Upper(Replace(name, ' ', '')) = 'JOHNDOE'
The rest of these will return rows where the middle part between John and Doe is anything, not just spaces...
if your database has left function and Reverse, Try either
Select * From Table
Where left(Upper(name), 4) = 'JOHN'
And Left(Reverse(Upper(name), 3)) = 'EOD'
else use substring and reverse
Select * From Table
Where substring(Upper(name), 4) = 'JOHN'
And substring(Reverse(Upper(name), 3)) = 'EOD'
or Like operator
Select * From Table
Where Upper(name) Like 'JOHN%DOE'
In MYSQL:
select user_name from users where lower(trim(user_name)) regexp 'john[[:blank:]]+doe' = 1;
Explanation:
In the above case the user_name matches the regular expression john(one or more spaces)doe. The comparison is case insensitive and trim takes care of spaces before and after the string.
In case the search string contains special characters, they need to be escaped. More in mysql documentation: http://dev.mysql.com/doc/refman/5.5/en/regexp.html
In sql you can use wildcards, that is a character that stands in place of other characters. for example:
select * from table where name like 'john%doe'
will select all names that start with john and end with doe no matter how many characters in between.
This article explains more and gives some options.
The wildcard will match on zero characters, so if you want at least one space then you should do
select * from table where lower(name) like 'john %doe'
In Oracle you can do:
SELECT * FROM table
WHERE UPPER(REGEXP_REPLACE(NAME,'( ){2,}', ' ') = 'JOHN DOE';
No false positives like 'john mordoe' etc.