Dcount function with wildcard characters - ms-access

I have a table containing project and region column.
Project region
Exp UK
Exp1 uk/usa
Dcount("[project]","[tablename]","[region]='uk'")
It will return 1 instead of 2.
Can someone help how can I count this with wildcard.

Use LIKE with wildcard characters.
Consider:
Dcount("[project]", "[tablename]", "[region] LIKE 'uk*'")

Related

MySQL query LIKE %...% query returning other results?

So I'm trying to code a PHP script, but we'll just leave it at the SQL part of things since this is where the issue is arising. I've a SELECT * query which should only grab from the rows where the user ID matches, and the badge ID meets their userID followed by an underscore. Although, it's grabbing results that shouldn't be included?
Here's my SQL query:
SELECT *
FROM `user_badges`
WHERE `user_id` = 1
AND `badge_id` LIKE '%1_%'
That should only return badges that start/contain 1_, it is grabbing all the badges that do contain/start with 1_ but it's also grabbing it215. If I search a different user ID, for example my own, it will grab all the badges with 3_ AND it's also grabbing ACH_RoomDecoFurniCount31 which is confusing because it doesn't contain 3_. Maybe there's more to it? Could someone please point me in the right direction.
You need to escape the _ as it's a wildcard character. Your query would should be like this:
SELECT *
FROM `user_badges`
WHERE `user_id` = 1
AND `badge_id` LIKE '%1\_%'
_ is also a wildcard in SQL - A substitute for a single character
_ is also a wildcard character. It means "any single character" (whereas % is "any sequence of characters").
You could escape/quote that _ or use the LOCATE function instead of a pattern match.
WHERE badge_id LIKE '%1\_%'
WHERE locate('1_', badge_id) > 0
_ is a wildcard "_ matches exactly one character." so what you are saying is:
% : starts with anything(or nothing)
1: contains 1
_: has exactly 1 of % (or anything, or nothing)
http://dev.mysql.com/doc/refman/5.7/en/string-comparison-functions.html

How does % wildcard exactly work in sql when searching

I saw a select statement with something like this in the where part 'LIKE %Blabla%Another%', now I tried it myself and I don't clearly understand how it works. The thing I'm confused about is the % in the middle, I understand how '%Blabla Another%' works, but with the % as a replacement for the space, I got confused.
% means "any sequence of characters, including an empty one". So LIKE '%Blabla%Another%' will match, for example 'XYZBlablaABCAnotherPQR', 'BlablaAnother' and ' Blabla Another '
% means any (including none) chars.
If you have a string 'abcdef' it would match e.g 'a%f' since the string start and ends with a and f.
'%b%e%'would also match as it means any string with b and even, in that order, but not necessarily next to each other
With LIKE you can use the following two wildcard characters in the pattern:
% matches any number of characters, even zero characters.
_ matches exactly one character.
© dev.mysql.com
In SQL, wildcard characters are used with the SQL LIKE operator.
there are the wild cards in sql %,_,[charlist],[^charlist].
the % wild cards is a substitute for zero or more characters.
for example
SELECT * FROM Customers
WHERE City LIKE '%es%';
this will selects all customers with a City containing the pattern "es".
so here you can see that the % wild card act as null or space character.
Here is a listing of the wildcards that you can use and below some examples.
%: A substitute for zero or more characters
_: A substitute for a single character
[charlist]: Sets and ranges of characters to match
[!charlist]: Matches only a character NOT specified within the brackets
The following SQL statement selects all customers with a City containing the pattern "es"
SELECT * FROM Customers WHERE City LIKE '%es%'
The following SQL statement selects all customers with a City starting with any character, followed by "erlin":
SELECT * FROM Customers WHERE City LIKE '_erlin'
The following SQL statement selects all customers with a City starting with "b", "s", or "p":
SELECT * FROM Customers WHERE City LIKE '[bsp]%'
The following SQL statement selects all customers with a City NOT starting with "b", "s", or "p":
SELECT * FROM Customers WHERE City LIKE '[!bsp]%'
Source: http://www.w3schools.com/sql/sql_wildcards.asp

MySQL search within the last 5 characters in a column?

My user table has a column "name" which contains information like this:
Joe Lee
Angela White
I want to search for either first name or last name efficiently. First name is easy, I can do
SELECT * FROM user WHERE name LIKE "ABC%"
But for last name, if I do
SELECT * FROM user WHERE name LIKE "%ABC"
That would be extremely slow.
So I am thinking about counting the characters of the input, for example, "ABC" has 3 characters, and if I can search only the last three characters in name column, that would be great. So I want something like
SELECT * FROM user WHERE substring(name, end-3, end) LIKE "ABC%"
Is there anything in MySQL that can do this?
Thanks so much!
PS. I cannot do fulltext because our search engine doesn't support that.
The reason that
WHERE name LIKE '%ith'
is a slow way to look for 'John Smith' by last name is the same reason that
WHERE Right(name, InStr(name, ' ' )) LIKE 'smi%'
or any other expression on the column is slow. It defeats the use of the index for quick lookup and leaves the MySQL server doing a full table scan or full index scan.
If you were using Oracle (that is, if you worked for a formerly wealthy employer) you could use function indexes. As it is you have to add some extra columns or some other helping data to accelerate your search.
Your smartest move is to split your first and last names into separate columns. Several other people have pointed out good reasons for doing that.
If you can't do that you could try creating an extra column which contains the name string reversed, and create an index on that column. That column will have, for example, 'John Smith' stored as 'htimS nhoJ'. Then you can search as follows.
WHERE nameReversed LIKE CONCAT(REVERSE('ith'),'%')
This search will use the index and be decently fast. I've had good success with it.
You're close. In MySQL you should be able to use InStr(str, substr) and Right(str, index) to do the following:
SELECT * FROM user WHERE Right(name, InStr(name, " ")) LIKE "ABC%"
InStr(name, " ") returns the index of the Space character (you may have to play with the " " syntax). This index is then used in the Right() function to search for only the last name (basically; problems arise when you have multiple names, multiple spaces etc). LIKE "ABC%" would then search for a last name starting with ABC.
You cannot use a fixed index as names that are more than 3 or less than 3 characters long would not return properly as you suggest.
However, as Zane said, it's a much better practise to use seperate fields.
If it is a MyIsam table, you may use Free text search to do the same.
You can use the REGEXP operator:
SELECT * FROM user WHERE name REGEXP "ABC$"
http://dev.mysql.com/doc/refman/5.1/en/regexp.html

Mysql RegExp question selecting from a list of codes

I am trying to match a list of motorcycle models to a series of ebay codes for listing motorcycles in ebay.
So we get a motorcycle model name that will be something like:
XL883C Sportster where the manufacturer is Harley Davidson
I have a list of ebay codes that look like this
MB-100-0 Other
MB-100-1 883
MB-100-2 1000
MB-100-3 1130
MB-100-4 1200
MB-100-5 1340
MB-100-6 1450
MB-100-7 Dyna
MB-100-8 Electra
MB-100-9 FLHR
MB-100-10 FLHT
MB-100-11 FLSTC
MB-100-12 FLSTR
MB-100-13 FXCW
MB-100-14 FXSTB
MB-100-15 Softail
MB-100-16 Sportster
MB-100-17 Touring
MB-100-18 VRSCAW
MB-100-19 VRSCD
MB-100-20 VRSCR
So I want to match the model name against the list above using a regExp pattern.
I have tried the following code:
SELECT modelID FROM tblEbayModelCodes WHERE
LOWER(makeName) = 'harley-davidson' AND fnmodel REGEXP '[883|1000|1130|1200|1340|1450|Dyna|Electra|FLHR|FLHT|FLSTC|FLSTR|FXCW|FXSTB|Softail|Sportster|Touring|VRSCAW|VRSCD|VRSCR].*' LIMIT 1
however when I run the query I would expect the code to match on either MB-100-1 for 883 or MB-100-16 for Sportster but when I run it the query returns MB-100-0 for Other.
I am guessing that I have the pattern incorrect, so can anybody suggest what I might need to do to correct this?
Many thanks
Graham
[chars] matches any of the characters 'c','h','a','r','s'
So by giving it such a long list, it will inevitably match just the first item (single character)
Try this instead
LOWER(makeName) = 'harley-davidson' AND fnmodel REGEXP '(883|1000|1130|1200|1340|1450|Dyna|Electra|FLHR|FLHT|FLSTC|FLSTR|FXCW|FXSTB|Softail|Sportster|Touring|VRSCAW|VRSCD|VRSCR).*' LIMIT 1
You might also consider not using REGEX and using FIND_IN_SET instead.
Not really fully tested, but it should be something like this:
REGEXP '^MB-[0-9]+-[0-9]+[[:space:]]+(883|1000|1130|1200|1340|1450|Dyna|Electra|FLHR|FLHT|FLSTC|FLSTR|FXCW|FXSTB|Softail|Sportster|Touring|VRSCAW|VRSCD|VRSCR)$'
In detail:
^MB- Starts with MB-
[0-9]+ One or more digits
- Dash
[0-9]+ One or more digits
[[:space:]]+ One or more white space
(883|1000|...)$ Ends with one of these
Here's the reference for the regexp dialect spoken by MySQL:
http://dev.mysql.com/doc/refman/5.1/en/regexp.html
Answer to comment:
If you want to match the Sportster row them remove all other conditions. And you may not even need regular expressions:
WHERE fnmodel LIKE '% Sportster'

MySQL query - select postcode matches

I need to make a selection based on the first 2 characters of a field, so for example
SELECT * from table WHERE postcode LIKE 'rh%'
But this would select any record that contains those 2 characters at any point in the "postcode" field right? I am in need of a query that just selects the first 2 characters. Any pointerS?
Thanks
Your query is correct. It searches for postcodes starting with "rh".
In contrast, if you wanted to search for postcodes containing the string "rh" anywhere in the field, you would write:
SELECT * from table WHERE postcode LIKE '%rh%'
Edit:
To answer your comment, you can use either or both % and _ for relatively simple searches. As you have noticed already, % matches any number of characters whereas _ matches a single character.
So, in order to match postcodes starting with "RHx " (where x is any character) your query would be:
SELECT * from table WHERE postcode LIKE 'RH_ %'
(mind the space after _). For more complex search patterns, you need to read about regular expressions.
Further reading:
http://dev.mysql.com/doc/refman/5.1/en/pattern-matching.html
http://dev.mysql.com/doc/refman/5.1/en/regexp.html
LIKE '%rh%' will return all rows with 'rh' anywhere
LIKE 'rh%' will return all rows with 'rh' at the beginning
LIKE '%rh' will return all rows with 'rh' at the end.
If you want to get only first two characters 'rh', use MySQL SUBSTR() function
http://dev.mysql.com/doc/refman/5.1/en/string-functions.html#function_substr
Dave, your way seems correct to me (and works on my test data). Using a leading % as well will match anywhere in the string which obviously isn't desirable when dealing with postcodes.