Given this linq query against an EF data context:
var customers = data.Customers.Where(c => c.EmailDomain.StartsWith(term))
You’d expect it to produce SQL like this, right?
SELECT {cols} FROM Customers WHERE EmailDomain LIKE #term+’%’
Well, actually, it does something like this:
SELECT {cols} FROM Customer WHERE ((CAST(CHARINDEX(#term, EmailDomain) AS int)) = 1)
Do you know why?
Also, replacing the Where selector to:
c => c.EmailDomain.Substring(0, term.Length) == term
it runs 10 times faster but still produces some pretty yucky SQL.
NOTE: Linq to SQL correctly translates StartsWith into Like {term}%, and nHibernate has a dedicated LikeExpression.
I don't know about MS SQL server but on SQL server compact LIKE 'foo%' is thousands time faster than CHARINDEX, if you have INDEX on seach column. And now I'm sitting and pulling my hair out how to force it use LIKE.
http://social.msdn.microsoft.com/Forums/en-US/adodotnetentityframework/thread/1b835b94-7259-4284-a2a6-3d5ebda76e4b
The reason is that CharIndex is a lot faster and cleaner for SQL to perform than LIKE. The reason is, that you can have some crazy "LIKE" clauses. Example:
SELECT * FROM Customer WHERE EmailDomain LIKE 'abc%de%sss%'
But, the "CHARINDEX" function (which is basically "IndexOf") ONLY handles finding the first instance of a set of characters... no wildcards are allowed.
So, there's your answer :)
EDIT: I just wanted to add that I encourage people to use CHARINDEX in their SQL queries for things that they didn't need "LIKE" for. It is important to note though that in SQL Server 2000... a "Text" field can use the LIKE method, but not CHARINDEX.
Performance seems to be about equal between LIKE and CHARINDEX, so that should not be the reason. See here or here for some discussion. Also the CAST is very weird because CHARINDEX returns an int.
charindex returns the location of the first term within the second term.
sql starts with 1 as the first location (0 = not found)
http://msdn.microsoft.com/en-us/library/ms186323.aspx
i don't know why it uses that syntax but that's how it works
I agree that it is no faster, I was retrieving tens of thousands of rows from our database with the letter i the name. I did find however that you need to use > rather than = ... so use
{cols} FROM Customer WHERE ((CAST(CHARINDEX(#term, EmailDomain) AS int)) > 0)
rather than
{cols} FROM Customer WHERE ((CAST(CHARINDEX(#term, EmailDomain) AS int)) = 1)
Here are my two tests ....
select * from members where surname like '%i%' --12 seconds
select * from sc4_persons where ((CAST(CHARINDEX('i', surname) AS int)) > 0) --12 seconds
select * from sc4_persons where ((CAST(CHARINDEX('i', surname) AS int)) = 1) --too few results
Related
I need a good expression in order to select correctly parts of a field.
For example, the field can be of the type: "google_organic" or "google_campaign_HereGoesMyCode" . The part I am interested in is "organic" or "campaign" without any other addition.
So far I select with this:
substring(Referer, charIndex('_',Referer)+1, len(Referer))
But in the case of "campaign" I select the whole thing... I don't know how to manage the existence or non-existence of the second underscore...
thank you
One way is to basically create a lastIndex type search using the below SQL and use the result as the length:
len(Referer) – (charindex('_', reverse(Referer))-1)
You can then rewrite your query as follows, although you need the result of the first charIndex so this is fairly intense:
substring(Referer, charIndex('_',Referer)+1, (len(Referer) – (charindex('_', reverse(Referer))-1) - (charIndex('_',Referer)+1))-1 )
I realize that this will now only work if you have 2 underscores. But you can filter which query to run based off a CASE/WHEN statement.
I have a data set that looks something like this:
A6177PE
A85506
A51SAIO
A7918F
A810004
A11483ON
A5579B
A89903
A104F
A9982
A8574
A8700F
And I need to find all the ENDings where they are non-numeric. In this example, that means PE, AIO, F, ON, B and F.
In pseudocode, I'm imagining I need something like
SELECT DISTINCT X FROM
(SELECT SUBSTR(COL,[SOME_CLEVER_LOGIC]) AS X FROM TABLE);
Any ideas? Can I solve this without learning regexp?
EDIT: To clarify, my data set is a lot larger than this example. Also, I'm only interested in the part of the string AFTER the numeric part. If the string is "A6177PE" I want "PE".
Disclaimer: I don't know Oracle SQL. But, I think something like this should work:
SELECT DISTINCT X FROM
(SELECT SUBSTR(COL,REGEXP_INSTR(COL, "[[:ALPHA:]]+$")) AS X FROM TABLE);
REGEXP_INSTR(COL, "[[:ALPHA:]]+$") should return the position of the first of the characters at the end of the field.
For readability, I'd recommend using the REGEXP_SUBSTR function (If there are no performance issues of course, as this is definitely slower than the accepted solution).
...also similar to REGEXP_INSTR, but instead of returning the position of the substring, it returns the substring itself
SELECT DISTINCT SUBSTR(MY_COLUMN,REGEXP_SUBSTR("[a-zA-Z]+$")) FROM MY_TABLE;
(:alpha: is supported also, as #Audun wrote )
Also useful: Oracle Regexp Support (beginning page)
For example
SELECT SUBSTR(col,INSTR(TRANSLATE(col,'A0123456789','A..........'),'.',-1)+1)
FROM table;
I'm trying to retrieve all columns that start with any non alpha characters in SQlite but can't seem to get it working. I've currently got this code, but it returns every row:
SELECT * FROM TestTable WHERE TestNames NOT LIKE '[A-z]%'
Is there a way to retrieve all rows where the first character of TestNames are not part of the alphabet?
Are you going first character only?
select * from TestTable WHERE substr(TestNames,1) NOT LIKE '%[^a-zA-Z]%'
The substr function (can also be called as left() in some SQL languages) will help isolate the first char in the string for you.
edit:
Maybe substr(TestNames,1,1) in sqllite, I don't have a ready instance to test the syntax there on.
Added:
select * from TestTable WHERE Upper(substr(TestNames,1,1)) NOT in ('A','B','C','D','E',....)
Doesn't seem optimal, but functionally will work. Unsure what char commands there are to do a range of letters in SQLlite.
I used 'upper' to make it so you don't need to do lower case letters in the not in statement...kinda hope SQLlite knows what that is.
try
SELECT * FROM TestTable WHERE TestNames NOT LIKE '[^a-zA-Z]%'
SELECT * FROM NC_CRIT_ATTACH WHERE substring(FILENAME,1,1) NOT LIKE '[A-z]%';
SHOULD be a little faster as it is
A) First getting all of the data from the first column only, then scanning it.
B) Still a full-table scan unless you index this column.
in a string array i have a variable amount of values.
how will an sql statement look if i want to select all the records that are equal with the array variables.
it will look something this:
SELECT * FROM users WHERE username='"+ variable amount of elements in the String array like: Steven, Mike, John ..... +"'"
You might be looking for the IN( ) operator.
SELECT * FROM users WHERE username IN ('chris', 'bob', 'bill');
This? (obviously the "OR username='xxx' is repeated for as many items as you require)
SELECT * FROM users WHERE username='item1' OR username='item2' OR username='item3'
Every one above me is corrrect! There are multiple ways of doing this! A simple google search for 'mysql statement variables' yield tons of help full results including:
http://www.daniweb.com/forums/thread126895.html
Just treat your array like a standard varible with [a number] on the end!
Two tips come from this:
Google / Search your problem first 9 times out of 10 someone else has had the same problem and found a working solution!
And be prepared to look at the answers on here with in 5 mins. This forums probably the quickest you'll ever see. The only thing that slows the community down is typing ! :P
Hope that Helps,
Andy
Another approach, although less efficient, can be used if parsing the string apart is problematic...
DECLARE #listOfNames VARCHAR(2000)
SET #ListOfNames = 'JOE,KELLIE,PETE'
SELECT * FROM users WHERE charindex(","+userName+",",","+#listOfNames+",") > 0
This approach is slower than either of the other answers, but can save you from using dynamic SQL to build a SQL query based on the list of names passed in...
I have stored values in my database that look like 5XXXXXX, where X can be any digit. In other words, I need to match incoming SQL query strings like 5349878.
Does anyone have an idea how to do it?
I have different cases like XXXX7XX for example, so it has to be generic. I don't care about representing the pattern in a different way inside the SQL Server.
I'm working with c# in .NET.
You can write queries like this in SQL Server:
--each [0-9] matches a single digit, this would match 5xx
SELECT * FROM YourTable WHERE SomeField LIKE '5[0-9][0-9]'
stored value in DB is: 5XXXXXX [where x can be any digit]
You don't mention data types - if numeric, you'll likely have to use CAST/CONVERT to change the data type to [n]varchar.
Use:
WHERE CHARINDEX(column, '5') = 1
AND CHARINDEX(column, '.') = 0 --to stop decimals if needed
AND ISNUMERIC(column) = 1
References:
CHARINDEX
ISNUMERIC
i have also different cases like XXXX7XX for example, so it has to be generic.
Use:
WHERE PATINDEX('%7%', column) = 5
AND CHARINDEX(column, '.') = 0 --to stop decimals if needed
AND ISNUMERIC(column) = 1
References:
PATINDEX
Regex Support
SQL Server 2000+ supports regex, but the catch is you have to create the UDF function in CLR before you have the ability. There are numerous articles providing example code if you google them. Once you have that in place, you can use:
5\d{6} for your first example
\d{4}7\d{2} for your second example
For more info on regular expressions, I highly recommend this website.
Try this
select * from mytable
where p1 not like '%[^0-9]%' and substring(p1,1,1)='5'
Of course, you'll need to adjust the substring value, but the rest should work...
In order to match a digit, you can use [0-9].
So you could use 5[0-9][0-9][0-9][0-9][0-9][0-9] and [0-9][0-9][0-9][0-9]7[0-9][0-9][0-9]. I do this a lot for zip codes.
SQL Wildcards are enough for this purpose. Follow this link: http://www.w3schools.com/SQL/sql_wildcards.asp
you need to use a query like this:
select * from mytable where msisdn like '%7%'
or
select * from mytable where msisdn like '56655%'