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'
Related
]
The name column has both first and last name in one column.
Look at the 'rep' column. How do I filter only those rep names where the last name is starting with 'K'?
The way that table is defined won't allow you to do that query reliably--particularly if you have international names in the table. Not all names come with the given name first and the family name second. In Asia, for example, it is quite common to write names in the opposite order.
That said, assuming you have all western names, it is possible to get the information you need--but your indexes won't be able to help you. It will be slower than if your data had been broken out properly.
SELECT rep,
RTRIM(LEFT(LTRIM(RIGHT(rep, LEN(rep) - CHARINDEX(' ', rep))), CHARINDEX(' ', LTRIM(RIGHT(rep, LEN(rep) - CHARINDEX(' ', rep)))) - 1)) as family_name
WHERE family_name LIKE 'K%'
So what's going on in that query is some string manipulation. The dialect up there is SQL Server, so you'll have to refer to your vendor's string manipulation function. This picks the second word, and assumes the family name is the second word.
LEFT(str, num) takes the number of characters calculated from the left of the string
RIGHT(str, num) takes the number of characters calculated from the right of the string
CHARINDEX(char, str) finds the first index of a character
So you are getting the RIGHT side of the string where the count is the length of the string minus the first instance of a space character. Then we are getting the LEFT side of the remaining string the same way. Essentially if you had a name with 3 parts, this will always pick the second one.
You could probably do this with SUBSTRING(str, start, end), but you do need to calculate where that is precisely, using only the string itself.
Hopefully you can see where there are all kinds of edge cases where this could fail:
There are a couple records with a middle name
The family name is recorded first
Some records have a title (Mr., Lord, Dr.)
It would be better if you could separate the name into different columns and then the query would be trivial--and you have the benefit of your indexes as well.
Your other option is to create a stored procedure, and do the calculations a bit more precisely and in a way that is easier to read.
Assuming that the name is <firstname> <lastname> you can use:
where rep like '% K%'
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 am trying to figure out some commands/code in SQL.
I have database with names, addresses IDs etc, but I have to convert firstname values ending in “jnr” to “(Jnr)” and those ending in “snr” to “(Snr)”.
How do I do this?
update table TABLE_NAME set NAMES = '*xyz*Jnr' where NAMES like '%jnr'
Update or select:
PASTE(column, CHAR_LENGTH(column)-3, 1, UPPER(SUBSTRING(column FROM CHAR_LENGTH(column)-3 FOR 1)
WHERE column LIKE '%jnr' OR column LIKE '%snr'
PASTE is used to put in one character at position 3 from end,
CHAR_LENGTH to get length of column value,
UPPER converts character to upper case,
SUBSTRING is used to pick one character here (j or s),
LIKE is used to find values ending with jnr, or snr.
All ANSI SQL (no dbms specified!)
I want to identify the number of words occurring after a comma, in a full name field in Oracle database table.
The name field contains format of "LAST, FIRST MIDDLE"
Some names may have up to 4 to 5 names, such as "DOE, JOHN A B"
For example, if the Name field = 'WILLIAMS JR, HANK' it would output 1 (for 1 word occurring after the comma.
If the Name field = 'DOE, JOHN A B' i want it to output 3.
I would like to use a regexp_count function to determine this count.
I am using the following code to identify how many words exist in the field and would like to modify it to include this functionality:
REGEXP_COUNT(REPLACE(fieldname, ',',', '), '[^]+')
It would likely have to remove the replace function in order to find the comma, but this was the best I could do so far.
Help is much appreciated!
How about the following:
REGEXP_COUNT( fieldname, "\\w", INSTR(fieldname, ",")+1)
I have updated the code as follows, which appears to be working as desired:
REGEXP_COUNT(fieldname, '[^ ]+', (INSTR(fieldname, ',')+1))
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