I'm having trouble with the following SQL query;
SELECT *
FROM Table
WHERE ID='510' AND CHARINDEX(requires,'Absolute Position Anchor') > 0
I've been using this code for a while but today it all of a sudden starts returning rows where the requires column only contains 'Position Anchor'. Has my query been wrong all along or might something have changed in my MS SQL server settings?
You've got the syntax flipped:
CHARINDEX('string to find','string to look in')
Change to:
SELECT *
FROM Table
WHERE ID='510' AND CHARINDEX('Absolute Position Anchor',requires) > 0
Or use LIKE, as Gordon Linoff suggests:
SELECT *
FROM Table
WHERE ID='510' AND requires LIKE '%Absolute Position Anchor%'
Related
How do I get the column "count(division)" instead of getting the actual number of counts?
select * from num_taught;
gets me this
select count(division) from num_taught;
gets me this, but I actually want the third column "count(division)" from the previous image
I want to know this because I'm doing this right now:
sql> select * from num_taught as a, num_taught as b
...> where a.count(division) = b.count(division);
Error: near "(": syntax error
but as you can see, there's a syntax error and I think it's because the code is not referencing the "count(division)" columns but actually finding the count instead.
My end goal is to output the "Titles" that have the same "Division" and have the same count(division).
So for example, the end table would have the rows "Chief Accountant", "Programmer Trainee", "Scrivener", "Technician", "Wizard". Since these are the rows that have a match in division and count(division)
Thanks!
What does DESC num_taught return? I am curious how the third column is populated - is it some kind of pseudo-column? You may want try wrapping the column name with [], see: How to deal with SQL column names that look like SQL keywords?
i.e. try:
select [count(division)] from num_taught;
You need to escape your column name using quotes (in case it's Sqlite like you mentioned in the comments).
select "count(division)" from num_taught;
or:
select * from num_taught as a, num_taught as b
where a."count(division)" = b."count(division)";
If you don't you are using the count-function provided by your Database-system.
It's very unusual to name a column like this, it might be either a trap by your tutor or an error while initializing the table in your case.
I think you just want a count(distinct):
select count(distinct division)
from num_taught;
when i execute the following select statement i am getting the below error.
I am trying to execute the below statement in Oracle server 9.1.
I believe due to this i am getting this error.
I know we can excute the query for single quotes by '%Blondeau D''Uva%'. But i am looking for query which will pass the special character value as parameter.
Kindluy let me know how to escape single quotes in old oracle server
ERROR :
ORA-00933: SQL Command not properly ended
Error at line : 1 Column : 50
Query:
Select * FROM TABLEA where UI_FNAME like q'[%Michael%]' and UI_LNAME like q'[ %Blondeau D'Uva%]';
On oracle the following should work
Select *
FROM TABLEA
where UI_FNAME like '[%Michael%]'
and UI_LNAME like '[ %Blondeau D''Uva%]';
So no duplicate where
And you have to quote the ' between D und Uva
And probably the q before the ' is wrong ... so I have removed it as well.
Ok tried it out with the q operator:
Select *
FROM employees
where first_name like q'[%Michael%]'
and last_name like q'[ %Blondeau D'Uva%]';
No errors no row ...
Two things about your query:
Two times usage of where where there should be just one.
About the second condition. You have used q'[ %Blondeau D'Uva%]' in like clause. I think that won't give you the result you might be looking for. This has nothing to do with your error, but still, it would not hurt to re-check the query.
Try this, this shouldn't run you into any error :
Select * FROM TABLEA
where UI_FNAME like q'[%Michael%]'
and UI_LNAME like q'[%Blondeau D'Uva%]';
Cheers!
As others have tried to give you an answer, but I think you probably need to keep % outside brackets -
Select *
FROM TABLEA
where UI_FNAME like '%[Michael]%'
and UI_LNAME like '%[ Blondeau D''Uva]%';
I have a table with about 200 million records. One of the columns is defined as varchar(100) and it's included in a full text index. Most of the values are numeric. Only few are not numeric.
The problem is that it's not working well. For example if a row contains the value '123456789' and i look for '567', it's not returning this row. It will only return rows where the value is exactly '567'.
What am I doing wrong?
sql server 2012.
Thanks.
Full text search doesn't support leading wildcards
In my setup, these return the same
SELECT *
FROM [dbo].[somelogtable]
where CONTAINS (logmessage, N'28400')
SELECT *
FROM [dbo].[somelogtable]
where CONTAINS (logmessage, N'"2840*"')
This gives zero rows
SELECT *
FROM [dbo].[somelogtable]
where CONTAINS (logmessage, N'"*840*"')
You'll have to use LIKE or some fancy trigram approach
The problem is probably that you are using a wrong tool since Full-text queries perform linguistic searches and it seems like you want to use simple "like" condition.
If you want to get a solution to your needs then you can post DDL+DML+'desired result'
You can do this:
....your_query.... LIKE '567%' ;
This will return all the rows that have a number 567 in the beginning, end or in between somewhere.
99% You're missing % after and before the string you search in the LIKE clause.
es:
SELECT * FROM t WHERE att LIKE '66'
is the same as as using WHERE att = '66'
if you write:
SELECT * FROM t WHERE att LIKE '%66%'
will return you all the lines containing 2 'sixes' one after other
I have ,for example, this table in a Microsoft Access database:
id numeric
context text
numberfield numeric
I want to select every record that ends with 9 in the column"numberfield". This gives a problem because it is a numeric field and as a result I can not use the following SQL:
select * from table where numberfield like "%9"
A solution is that I change the numberfield to a text. But this gives a problem because there are several users and the change might give a problem in the future. Is there an option to select on the ending when it is a number field?
That sound a little fishy.. are you sure you can use that query? Don't know about Access but almost any other DBMS allows it.
If it really doesn't work, you can do this:
select * from table where STR(numberfield) like "*9"
EDIT: Maybe it didn't work because you used % which is used with * in Access :
select * from table where numberfield like "*9"
Numbers are numbers, so use Mod for this:
select * from table where numberfield mod 10 = 9
Instead of casting to string and comparing, just extract the rightmost digit with a MOD operation.
Edit your query as follows:
SELECT *
FROM table
WHERE ((([numberfield] Mod 10)=9));
I'm having some trouble understanding WHY a select statement isn't working in a query I'm making.
I've got the SELECT and FROM lines functioning. With just those, ALL results from my selected table are displayed - 517 or so
What I want to do is display results based on a pattern using LIKE - What I have so far
SELECT *
FROM Tbl_ServiceRequestMatrix
WHERE Tbl_ServiceRequestMatrix.[Application/Form] LIKE 'P%';
This returns 0 results - despite the fact that the column selected DOES have entries that start with 'P'
I also tried utilising brackets, see if that was the issue - still displays 0 results:
SELECT *
FROM Tbl_ServiceRequestMatrix
WHERE ((Tbl_ServiceRequestMatrix.[Application/Form])='p%');
Can any one help me understand why my WHERE ** LIKE statement is causing 0 results to be displayed?
The wildcard character in MS Access is (by default) * instead of %:
WHERE Tbl_ServiceRequestMatrix.[Application/Form] LIKE "P*"
LIKE Statement has different parameters in different sql languages.
In MS Access you need * Instead of % in LIKE Statement.