I am using PhpMysql and this simple query yields an empty result ("MySQL returned an empty result set").
Here is the query: SELECT * FROM TABLE 2 WHERE prenom LIKE '%Scott%'
My table has a row called prenomand there is a 'Scott' value. I have tried with other first names (which I know are in the database) and it does not work either.
The strange thing is that when I use the wildcards between One letter only i.e: SELECT * FROM TABLE 2 WHERE prenom LIKE '%S%' it works..
I need help please!
Ok so the pb was that due to the collation type instead of SCOTT, the file was encoded with spaces S C O T T. I had to reformat the file in utf8 to fix this!
Related
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 am trying to connect a Filemaker DB to Firebird SQL DB in both ways import to FM and export back to Firebird DB.
So far it works using the MBS Plug-in but FM 13 Pro canot handle NULL.
That means that for example Timestamp fields that are empty (NULL) produce a "0" value.
Thats means in Time something like 01.01.1889 00:00:00.
So my idea was to simply ignore fields containing NULL.
But here my poor knowlege stops.
First I thought I can do this with WHERE, but this is ignoring whole records sets:
SELECT * FROM TABLE WHERE FIELD IS NOT NULL
Also I tried to filter it later on like this:
If (IsEmpty (MBS("SQL.GetFieldAsDateTime"; $command; "FIELD") ) = 0 ; MBS("SQL.GetFieldAsDateTime"; $command; "FIELD"))
With no result either.
This is a direct answer to halfbit's suggestion, which is correct but not for this SQL dialect. In a query to provide a replacement value when a field is NULL you need to use COALESCE(x,y). Where if X is null, Y will be used, and if Y is null then the field is NULL. Thats why it is common for me to use it like COALESCE(table.field,'') such that a constant is always outputted if table.field happens to be NULL.
select COALESCE(null,'Hello') as stackoverflow from rdb$database
You can use COALESCE() for more than two arguments, I just used two for conciseness.
I dont know the special SQL dialect, but
SELECT field1, field2, value(field, 0), ...FROM TABLE
should help you:
value gives the first argument, ie, your field if it is NOT NULL or the second argument if it is.
Suppose I have a table named 'Customer' with many columns and I want to display all customers who's name ends with 'Thomas'(Lastname = 'Thomas'). The following query shows an empty result(no rows). Also it didn't show any error.
SELECT * FROM Customer WHERE Lastname = 'Thomas';
While executing the following query give me correct result.
SELECT * FROM Customer WHERE Lastname LIKE '%Thomas%';
I would like to know what is the problem with my first query. I am using sqlite3 with Npm. Below is the result of '.show' command(Just in case of the problem is with config).
sqlite> .show
echo: off
explain: off
headers: on
mode: column
nullvalue: ""
output: stdout
separator: "|"
stats: off
width:
Use Like instead of =
Trim to ensure that there arent spaces messing around
so the query will be
SELECT * FROM Customer WHERE trim(Lastname) LIKE 'Thomas';
depending on your types, probably you dont need point 2, since as can be read in mysql manual
All MySQL collations are of type PADSPACE. This means that all CHAR
and VARCHAR values in MySQL are compared without regard to any
trailing spaces
But the point 1 could be the solution. Actually if you want to avoid problems, you should compare strings with LIKE, instead of =.
If You still have problems, probably you will have to use collates.
SELECT *
FROM t1
WHERE k LIKE _latin1 'Müller' COLLATE latin1_german2_ci; #using your real table collation
more information here But specifically with 'Thomas' you shouldn't need it, since it hasn't got any special characters.
I am not able to make this out: Between eliminates use of >= and <=
...but when i tried this query:
SELECT *
FROM names
WHERE name >= 'Ankit'
AND name <= 'P'
...it gives output:
name
------
Ankit
Mac
Bob
When I tried:
SELECT *
FROM names
WHERE name BETWEEN 'Ankit' AND 'P'
...it gives output:
name
------
Ankit
Can you explain this why?
First, Oracle VARCHAR2 type is case sensitive.
Second, check that you do not have spaces in the beginning of name like this:
" Bob"
" Mac"
Use trim function to check if this causes the problem:
SELECT *
FROM names
WHERE trim(name) BETWEEN 'Ankit' AND 'P'
If this does not help, check that language and sort order are correct for your database.
Edit:
Since above advice did not solve your problem, you could try following:
Maybe you have some other non-printable characters in field. Use Oracle DUMP function to check:
SELECT DUMP(name), name FROM names
You should get something like this:
Typ=1 Len=3: 66,111,98 Bob
...
Verify that Len is correct length.
Check NLS parameters so that they are not inadvertently changed to something that does not work for your database:
SELECT * FROM NLS_SESSION_PARAMETERS
SELECT * FROM NLS_DATABASE_PARAMETERS
SELECT * FROM NLS_INSTANCE_PARAMETERS
Check results of these three queries and verify that parameters on sort, language and character set are correct.
I'm quite certain this has nothing to do with your syntax and everything to do with your DB setup. I've recreated your test scenario and, like others, have no problem with either query returning the results you expect. Did you check your NLS_SESSION_PARAMETERS as mentioned earlier?
SQL code is case insensitive.
String values and string comparisons are case sensitive.
See for yourself:
SELECT CASE WHEN 'a' = 'A' THEN 'string comparison is case insensitive'
WHEN 'a' <> 'A' THEN 'string comparison is case sensitive'
END
FROM dual;
You seem to have used a lowercase 'p' in the top query and an uppercase 'P' in the second. Was this intentional?
SELECT *
FROM names
WHERE name BETWEEN 'Ankit' AND 'P'
also should return all three rows - I just verified that it does so using your example. Are you sure that you made the test correctly? Maybe you inserted data in other session and didn't commit additional rows?
Nut 100% sure about it, but as it's only 3 rows, you could try it:
SELECT *
FROM names
WHERE (name BETWEEN "Ankit" AND "P")
OR (name BETWEEN "ankit" AND "p")