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
Related
Do you know how to remove below kind of Characters at once on a query ?
Note : .I'm retrieving this data from the Access app and put only the valid data into the SQL.
select DISTINCT ltrim(rtrim(a.Company)) from [Legacy].[dbo].[Attorney] as a
This column is company name column.I need to keep string characters only.But I need to remove numbers only rows,numbers and characters rows,NULL,Empty and all other +,-.
Based on your extremely vague "rules" I am going to make a guess.
Maybe something like this will be somewhere close.
select DISTINCT ltrim(rtrim(a.Company))
from [Legacy].[dbo].[Attorney] as a
where LEN(ltrim(rtrim(a.Company))) > 1
and IsNumeric(a.Company) = 0
This will exclude entries that are not at least 2 characters and can't be converted to a number.
This should select the rows you want to delete:
where company not like '%[a-zA-Z]%' and -- has at least one vowel
company like '%[^ a-zA-Z0-9.&]%' -- has a not-allowed character
The list of allowed characters in the second expression may not be complete.
If this works, then you can easily adapt it for a delete statement.
In an oracle table:
1- a value in a VARCHAR column contains characters that are not letters.
Consider a scenarion where a name in 'last_name' column contains regular characters (A - Z, a - z) as well as characters that are not english letters (e.g. '.', '-', ' ','_', '>' or similar).
The challenge is to select the rows that has names in 'last_name' as '.John' or 'John.' or '-John' or 'Joh-n'
2- Is it possible to have non-date values in a Date defined column? If yes, how can such records be selected in an oracle query?
Thanks!
I believe this will do the trick:
SELECT * FROM mytable WHERE REGEXP_LIKE(last_name, '[^A-Za-z]');
As for your 2nd question, I am unsure. I would be glad if someone else could add on to what I have to answer your 2nd question. I have found this website thought that might be of help. http://infolab.stanford.edu/~ullman/fcdb/oracle/or-time.html
It explains the DATE format.
If I properly understand your goal, you need to select rows with last_name column containing the name 'John', but it may also have additional characters before, after, or even inside the name. In that case, this should be helpful:
select * from tab where regexp_replace(last_name, '[^A-Za-z]+', '') = 'John'
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 have a table with a field that denotes whether the data in that row is valid or not. This field contains a string of undetermined length. I need a query that will only pull out rows where all the characters in this field are N. Some possible examples of this field.
NNNNNNNNNNNNNNNNNNN
NNNNNNNNNNNNNNNNNNNNNNN
NNNNNEEEENNNNNNNNNNNN
NNNNNOOOOOEEEENNNNNNNNNNNN
Any suggestions on a postcard please.
Many thanks
This should do the trick:
SELECT Field
FROM YourTable
WHERE Field NOT LIKE '%[^N]%' AND Field <> ''
What it's doing is a wildcard search, broken down:
The LIKE will find records where the field contains characters other than N in the field. So, we apply a NOT to that as we're only interested in records that do not contain characters other than N. Plus a condition to filter out blank values.
SELECT *
FROM mytable
WHERE field NOT LIKE '%[^N]%'
I don't know which SQL dialect you are using. For example Oracle has several functions you may use. With oracle you could use condition like :
WHERE LTRIM(field, 'N') = ''
The idea is to trim out all N's and see if the result is empty string. If you don't have LTRIM, check if you have some kind of TRANSLATE or REPLACE function to do the same thing.
Another way to do it could be to pick length of your field and then construct comparator value by padding empty string with N. Perhaps something like:
WHERE field = RPAD('', field, 'N)
Oracle pads that empty string with N's and picks number of pad characters from length of the second argument. Perhaps this works too:
WHERE field = RPAD('', LENGTH(field), 'N)
I haven't tested those, but hopefully that give you some ideas how to solve your problem. I guess that many of these solutions have bad performance if you have lot of rows and you don't have other WHERE conditions to select proper index.
How do I do this?
For example, if my column is "cats,dogs,birds" and I want to get any rows where column contains cats?
Using LIKE:
SELECT *
FROM TABLE
WHERE column LIKE '%cats%' --case-insensitive
While LIKE is suitable for this case, a more general purpose solution is to use instr, which doesn't require characters in the search string to be escaped. Note: instr is available starting from Sqlite 3.7.15.
SELECT *
FROM TABLE
WHERE instr(column, 'cats') > 0;
Also, keep in mind that LIKE is case-insensitive, whereas instr is case-sensitive.