I tryed some wildcards in Access in Where statement, but they don't works. For example:
This query SELECT staff.* FROM staff;returns:
I tryed to do a query with wildcard SELECT staff.* FROM staff WHERE (staff.s_name LIKE "A%");
but it returns an empty table:
What is the reason? My wildcard doesn't work
(s_name is the second column)
(look that "firstname" is the tag of "s_name" only for the view)
Wildcard character in Access is *, not % unlike in SQL Server.
See MSDN for details.
No, no, no, use '*', not '%'. Or, use 'Like'.
http://www.techrepublic.com/article/10-tips-for-using-wildcard-characters-in-microsoft-access-criteria-expressions/
https://www.techonthenet.com/access/queries/like.php
https://www.techonthenet.com/access/queries/like2007.php
For instance:
Like 'm*'
Result: all values that start with m
Like 'm'
Result: all values that contain m
Like '*m'
Result: all values that end with m
Related
I have a variable varchar that always takes in 10 digits. How can I use the LIKE operator to find/use only the first 5 digits of the variable?
my query:
variable IN VARCHAR2
SELECT * FROM items WHERE name LIKE SUBSTRING(variable, 1, 5)
... WHERE name LIKE '12345%'
will match any string that starts 12345. the '%' is a wildcard. You can also use the wildcard to match anywhere in the string: ... WHERE name LIKE '%12345%' will match a string with 12345 anywhere within it.
Edit for completeness: WHERE name LIKE '%12345' will match any string that ends with those five characters.
Try this:
SELECT * FROM items WHERE name LIKE (SUBSTRING(variable, 1, 5) + '%')
I guess you can use LEFT() like this:
SELECT * FROM items WHERE LEFT(name,5)=LEFT(variable,5);
Or if you you want to use LIKE with a wildcard, you can do this:
SELECT * FROM items WHERE name LIKE CONCAT(LEFT(variable,5),'%')
A few more example in the Demo fiddle
Edit: The above solution is for MySQL/MariaDB because earlier the tag of this question have MySQL but it's also my fault for not recognizing OP description of the datatype VARCHAR2. I might as well just post a suggestion related to the rdbms.
So, my first suggestion there using LEFT() however Oracle don't have that function, therefore:
SELECT * FROM items WHERE SUBSTR(name,1,5)=SUBSTR(variable,1,5);
or using concatenation operator
SELECT * FROM items WHERE name LIKE SUBSTR(variable,1,5)||'%'
Demo fiddle
How can I make Access treat *1* as a string instead of using wildcard in the SQL
SELECT * FROM TABLE1 WHERE ID IN ("*1*","*2*");?
Sadly, this is a limitation of Access. You can use [*] to search for any single character, so:
TABLE1 WHERE ID IN ("[*]1[*]","[*]2[*]")
Would match any single leading and trailing character with 1 or 2 in the middle. Not what you want, but closer.
In this expression:
WHERE ID IN ("*1*", "*2*")
MS Access does treat the values as strings and not wildcards. The wildcards are only used for LIKE.
If you want them to be treated as wildcards, you need to use LIKE. That would require OR:
WHERE ID LIKE "*1*" OR ID LIKE "*2*"
Or more simply as:
WHERE ID LIKE "*[1-2]*"
I have this query in access but it doesn't return any result when I use the Like function. It does return result when I set it to equal to.
Select * from myTable where name like "*Al#pp*"
note that the # sign is part of the spelling to lookup and not a wildcard, but I think access is taking it as a wildcard.
Since # is not supposed to be a wildcard then you have to escape it. Try:
Select * from myTable where name like "*Al[#]pp*"
I want to write an SQL SERVER statement that searches for the beginning of a word in a string that starts with something.
For example, if I search for 'em' on the Company record, I should get:
Emily, Inc
The Emmmy
NOT
Forget Them
Lemming, LLC
I can do this in PHP by extracting/slicing the string into an array and searching the beginning of each words.
But how I would write this query in SQL server without resorting to Stored procedures/functions?
JW's answer will work for entries where Em is at the very beginning of the field.
If you also need to retrieve values where the word is not the first in the string, try:
SELECT * FROM tableName WHERE CompanyName LIKE 'Em%' or CompanyName LIKE '% Em%'
(This is assuming all word are separated by a space.)
use LIKE
SELECT * FROM tableName WHERE CompanyName LIKE 'Em%'
Another option is CONTAINS, something like:
SELECT ... WHERE CONTAINS(CompanyName, 'Em*');
For MS Access:
SELECT * FROM Table1 WHERE Company_Name LIKE 'Word*'
For more standard DBMSs:
SELECT * FROM Table1 WHERE Company_Name LIKE 'Word%'
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).