MSAccess Query a string matching a pattern - sql

I have a table with a string field containing location information. I want to be able to query this table and retrieve all of the tags matching the format xxxxxxAA where xxxxxx is a 6-digit number and AA is two alphabetic characters.
Is there a method of querying this using SQL or is this something that I need to do in VBA?
Sample data:
BGS5 PM RGP5
022051PM
022201PM
030539PM
WAS3N
179546MM
And I want to return the following without knowing the values:
022051PM
022201PM
030539PM
179546MM
thanks in advance
Jason

You can use a query with a Like comparison in the WHERE clause.
SELECT y.text_field
FROM YourTable AS y
WHERE y.text_field Like '######[A-Z][A-Z]'
The # matches a digit.
[A-Z] matches one character from a character class consisting of only letters. That character class is actually upper case letters. However, the comparison is case-insensitive, so will match lower case letters, too.

Related

Getting the Column containing the non-english language in ORACLE

I have above entries in my database, my requirement is to extract the fields containing the non-english language characters ( including if the data containing the combination of english and non-english characters like HotelName field for the ID 45).
I tried by regexp_like function by looking for the alphanumeric and non-alphanumeric, but i have some data with combination of both the condition fails there.
Thanks in Advance
Raghavan
Does this do what you want?
where regexp_like(hotelname, '[^a-zA-Z0-9 ]')
That is, where the hotel name contains any character that is not a "letter" or digit. You may need to take additional characters into account as well, such as commas, periods, and hyphens.

Consider a query to find details of research fields where the first two parts of the ID are D and 2 and the last part is one character (digit)

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.

Pattern Matching with SQL Like with first part letters and second part numbers of varying length

Is there a way to use Pattern Matching with SQL LIKE, to match first part of letters and a second part of variable number of numbers?
For example, I want to select only ABC1002, ABC23, ABC569, CDE48569.
Here is one method:
where col like '[A-Z][A-Z][A-Z][0-9]%' and
col not like '[A-Z][A-Z][A-Z]%[^0-9]%'
The logic says:
The column starts with three letters and a digit.
Nothing other than a digit follows the three letters.

Display certain sequence only in VARCHAR

I have a column error_desc with values like:
Failure occurred in (Class::Method) xxxxCalcModule::endCustomer. Fan id 111232 is not Effective or not present in BL9_XXXXX for date 20160XXX.
What SQL query can I use to display only the number 111232 from that column? The number is placed at 66th position in VARCHAR column and ends 71st.
SELECT substr(ERROR_DESC,66,6) as ABC FROM bl1_cycle_errors where error_desc like '%FAN%'
This solution uses regular expressions.
The challenge I faced was on pulling out alphanumerics. We have to retain only numbers and filter out string,alphanumerics or punctuations in this case, to detect the standalone number.
Pure strings and words not containing numbers can be easily filtered out using
[^[:digit:]]
Possible combinations of alphanumerics are :
1.Begins with a character, contains numbers, may end with characters or punctuations :
[a-zA-Z]+[0-9]+[[:punct:]]*[a-zA-Z]*[[:punct:]]*
2.Begins with numbers and then contains alphabets,may contain punctuations :
[0-9]+[[:punct:]]*[a-zA-Z]+[[:punct:]]*
Begins with numbers then contains punctuations,may contain alphabets :
-- [0-9]+[a-zA-Z][[:punct:]]+[a-zA-Z] --Not able to highlight as code, refer solution's last regex combination
Combining these regular expressions using | operator we get:
select trim(REGEXP_REPLACE(error_desc,'[^[:digit:]]|[a-zA-Z]+[0-9]+[[:punct:]]*[a-zA-Z]*[[:punct:]]*|[0-9]+[[:punct:]]*[a-zA-Z]+[[:punct:]]*|[0-9]+[a-zA-Z]*[[:punct:]]+[a-zA-Z]*',' '))
from error_table;
Will work in most cases.

SQL String contains ONLY

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.