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.
Related
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.
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
iam new to sql and i would like to know how can I find the letter or symbol at the end of value in column called Name?
E.g if i would find something i will write select * from 'table' where 'Name' like '%es%' but it will find me all rows contains es
Lets say - lesl, pespe, mess... but how to write select which will select just values with 'es' At the end of word? ... using regex i will use es\Z..... thanks in advance!
You have to remove the last %, so it will only select words ending with es.
select * from table where Name like '%es'
You're currently matching on: ..where 'Name' like '%es%'.
Which equates to anything, then 'es' then anything else.
Removing the last % changes the where to
anything then 'es'.
in short.. you need ..where 'Name' like '%es'
The query ..where 'Name' like '%es' will find columns where name ends with "ES".
But if we need to find column where name ends with either "E" or "S", the query would be
.. where 'Name' LIKE '%[ES]'
You can also use REGEX
select distinct city from station where city REGEXP '[aeiou]$';
Resources: To Know more about REGEXP
if you want to find name start with something like 'test'
use => select name from table where name like 'test%'.
if you want to find name end with something like 'test'
use => select name from table where name like '%test'.
if you want to find name start with s and end with h
use => select name from table where name like 's%'and name like '%h' or simply select name from table where name like 's%h'.
Try pattern [a-z]ed($) with regexp operator/function.
Pattern explanation:
[a-z] - match any letter
es - match es literally
($) - end of string (so make sure it's not followed by any letters)
Full query would be:
select * from test
where ingr regexp '[a-z]es($)'
Sorry I am new to working with databases - I am trying to perform a query
that will get all of the characters that are similar to a string in SQL.
For example,
If I am looking for all users that begin with a certain string, something like S* or Sm* that would return "Smith, Smelly, Smiles, etc..."
Am I on the right track with this?
Any help would be appreciated, and Thanks in advance!
Sql can't do this ... placing the percentage symbol in the middle.
SELECT * FROM users WHERE last_name LIKE 'S%th'
you would need to write a where clause and an and clause.
SELECT * FROM users WHERE last_name LIKE 'S%' and last_name LIKE '%th'
The LIKE operator is what you are searching for, so for your example you would need something like:
SELECT *
FROM [Users]
WHERE LastName LIKE 'S%'
The % character is the wild-card in SQL.
to get all the users with a lastname of smith
SELECT *
FROM [Users]
WHERE LastName ='Smith'
to get all users where the lastname contains smith do this, that will also return blasmith, smith2 etc etc
SELECT *
FROM [Users]
WHERE LastName LIKE '%Smith%'
If you want everything that starts with smith do this
SELECT *
FROM [Users]
WHERE LastName LIKE 'Smith%'
Standard (ANSI) SQL has two wildcard characters for use with the LIKE keyword:
_ (underscore). Matches a single occurrence of any single character.
% (percent sign). Matches zero or more occurrences of any single character.
In addition, SQL Server extends the LIKE wildcard matching to include character set specification, rather like a normal regular expresion character set specifier:
[character-set] Matches a single character from the specified set
[^character-set] Matches a single character not in the specified set.
Character sets may be specified in the normal way as a range as well:
[0-9] matches any decimal digit.
[A-Z] matches any upper-case letter
[^A-Z0-9-] matches any character that isn't a letter, digit or hyphen.
The semantics of letter matching of course, a dependent on the collation sequence in use. It may or may not be case-sensitive.
Further, to match a literal left square bracket ('[]'), you must use the character range specifier. You won't get a syntax error, but you won't get a match, either.
where x.field like 'x[[][0-9]]'
will match text that looks like 'x[0]' , 'x[8]', etc. But
where 'abc[x' like 'abc[x'
will always be false.
you might also like the results of SOUNDEX, depending on your preference for last name similarity.
select *
from [users]
where soundex('lastname') = soundex( 'Smith' )
or upper(lastname) like 'SM%'
Your question isn't entirely clear.
If you want all the users with last name Smith, a regular search will work:
SELECT * FROM users WHERE last_name = 'Smith'
If you want all the users beginning with 'S' and ending in 'th', you can use LIKE and a wildcard:
SELECT * FROM users WHERE last_name LIKE 'S%th'
(Note the standard SQL many wildcard of '%' rather than '*').
If you really want a "sounds like" match, many databases support one or more SOUNDEX algorithm searches. Check the documentation for your specific product (which you don't mention in the 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.