I've found some tutorials for ProFTP with MySQL connection. But these handling by general scheme, 99% of them are identically. And do not go into the detail of the configuration.
Is SQLGroupInfo required?
I've noticed the groups are always the same. Therefore, only one database record is needed in the groups table.
I want to fix this by omitting the group table and setting it directly in the configuration.
How can I do this?
The first try was, to register a custom query:
# Hardcoded groups
SQLNamedQuery get-groups SELECT 'ftp' AS `name`, 2001 AS `group`, 'ftpd' AS `members`
SQLGroupInfo sql:/get-groups
But here, i've get following error:
fatal: SQLGroupInfo: badly formatted parameter on line 34 of '/etc/proftpd/sql.conf'
The problem is, i must return correct rows like gid, groupname and members. If wanted, you can remove the FROM $table WHERE [...], thats not required.
Here is the solution:
# Hardcoded groups
SQLNamedQuery get-group-by-name SELECT "`username` AS `groupname`, '1010' AS `gid`, `username` AS `members` FROM `fh_ftp_user` WHERE `username`='%{0}'"
SQLNamedQuery get-group-by-id SELECT "`username` AS `groupname`, '1010' AS `gid`, `username` AS `members` FROM `fh_ftp_user` WHERE `username`='%{0}'"
SQLNamedQuery get-group-by-member SELECT "`username` AS `groupname`, '1010' AS `gid`, `username` AS `members` FROM `fh_ftp_user` WHERE `username`='%{0}'"
SQLGroupInfo custom:/get-group-by-name/get-group-by-id/get-group-by-member
Or
# Hardcoded groups
SQLNamedQuery get-hardcoded-group SELECT "'YourHardcodedGroup' AS `groupname`, 1010 AS `gid`, 'MemberName' AS `members`"
SQLGroupInfo custom:/get-hardcoded-group
Related
See below for what is returned in my automated test for this query:
Select visit_date
from patient_visits
where patient_id = '50'
AND site_id = '216'
ORDER by patient_id
DESC LIMIT 1
08:52:48.406 DEBUG Executing : Select visit_date from patient_visits
where patient_id = '50' AND site_id = '216' ORDER by patient_id DESC
LIMIT 1 08:52:48.416 TRACE Return: [(datetime.date(2017, 2, 17),)]
When i run this in workbench i get
2017-02-17
How can i make the query return this instead of the datetime.date bit above. Some formatting needed?
What you got from the database is python's datetime.date object - and that happens due to the db connector drivers casting the DB records to the their corresponding python counterparts. Trust me, it's much better this way than plain strings the user would have to parse and cast himself later.
Imaging the result of this query is stored in a variable ${record}, there are a couple of things to get to it, in the form you want.
First, the response is (pretty much always) a list of tuples; as in your case it will always be a single record, go for the 1st list member, and its first tuple member:
${the_date}= Set Variable ${record[0][0]}
Now {the_date} is the datetime.date object; there are at least two ways to get its string representation.
1) With strftime() (the pythonic way):
${the_date_string}= Evaluate $the_date.strftime('%Y-%m-%d') datetime
here's a link for the strftime's directives
2) Using the fact it's a regular object, access its attributes and construct the result as you'd like:
${the_date_string}= Set Variable ${the_date.year}-${the_date.month}-${the_date.day}
Note that this ^ way, you'd most certainly loose the leading zeros in the month and day.
The program just selects everything if the carrid is ok even if it is not ok with the lt_spfli. And there aren't any entries with that carrid it gets runtime error. If I try with for all entries he just selects absolutely the entire SFLIGHT.
PARAMETERS: pa_airp TYPE S_FROMAIRP,
pa_carid TYPE S_CARR_ID.
DATA: lt_spfli TYPE RANGE OF SPFLI,
lt_sflight TYPE TABLE OF SFLIGHT.
SELECT CONNID FROM SPFLI
INTO TABLE lt_spfli
WHERE AIRPFROM = pa_airp.
SELECT * FROM SFLIGHT
INTO TABLE lt_sflight
WHERE CARRID = pa_carid AND CONNID in lt_spfli.
I just suppose, that you want every flight connection from a given airport...
Notice, that a RANGE structure has two more fields in front of the actual "compare value". So selecting directly into it will result in a very gibberish table.
Possible Solutions:
Selecting with RANGE
If you really want to use this temporary table, you can have a look at my answer here where I describe the way to fill RANGEs without any overhead. After this step, your current snippet will work the way to wanted it too. Just make sure, that it really has been filled or everything will be selected.
Selecting with FOR ALL ENTRIES
Before you use this variant you should make absolutely sure, that your specified data object is filled. Otherwise it will result in the same mess as the first solution. To do that, you could write:
* select connid
IF lt_spfli[] IS NOT INITIAL.
* select on SFLIGHT
ELSE.
* no result
ENDIF.
Selecting with JOIN
The "correct" approach in this case would be a JOIN like:
SELECT t~*
FROM spfli AS i
JOIN sflight AS t
ON t~carrid = #pa_carid
AND t~connid = i~connid
INTO TABLE #DATA(li_conns)
WHERE i~airpfrom = #pa_airp.
Use a FOR ALL ENTRIES instead of CONNID in lt_SPFLI.
As so:
SELECT *
FROM sflight
FOR ALL ENTRIES IN lt_spfli
WHERE carrid = pa_carid
AND connid = lt_spfli-connid
You are misunderstanding what a "Ranges Table" is. You fill it incorrectly.
This part of your code demonstrates the misunderstanding (with a little debug, you would see the erroneous contents immediately):
DATA: lt_spfli TYPE RANGE OF SPFLI.
SELECT CONNID FROM SPFLI INTO TABLE lt_spfli ...
A "Ranges Table" is an internal table with 4 components (SIGN, OPTION, LOW, HIGH), used in Open SQL to do complex selections on one database column (NB: it can also be used in several ABAP statements to test the value of an ABAP variable).
So, with your SQL statement, you only initialize the first component of the Ranges table, while you should transfer CONNID into the third component.
In "modern" Open SQL, you'd better do:
SELECT 'I' as SIGN, 'EQ' as OPTION, CONNID as LOW FROM SPFLI INTO TABLE #lt_spfli ...
For more information about Ranges Tables, you may refer to the answer here: What actually high and low means in a ranges table
I'm learning about Sphinx with the book "Introduction to Search with Sphinx" and the first query with the test data:
mysql> select * from test1 where match('test');
should produce this output:
but this is the output I am getting:
Why is the weight column missing from my result?
SphinxQL used to do a number of things 'magically', one of them was automatically add columns depending on the query. But it was decided that magic behaviour often confused programs (wasnt sure if the column would be there or not)
... so now, columns are only there if you explicitly ask for them (although can still use * for id+attributes)
mysql> select *,weight() as weight from test1 where match('test');
there are a few topics about this already with accepted answers but I couldn't figure out a solution based on those:
Eg:
Ruby on Rails: must appear in the GROUP BY clause or be used in an aggregate function
GroupingError: ERROR: column must appear in the GROUP BY clause or be used in an aggregate function
PGError: ERROR: column "p.name" must appear in the GROUP BY clause or be used in an aggregate function
My query is:
Idea.unscoped.joins('inner join likes on ideas.id = likes.likeable_id').
select('likes.id, COUNT(*) AS like_count, ideas.id, ideas.title, ideas.intro, likeable_id').
group('likeable_id').
order('like_count DESC')
This is fine in development with sqlite but breaks on heroku with PostgreSQL.
The error is:
PG::GroupingError: ERROR: column "likes.id" must appear in the GROUP BY clause or be used in an aggregate function
If I put likes.id in my group by then the results make no sense. Tried to put group before select but doesn't help. I even tried to take the query into two parts. No joy. :(
Any suggestions appreciated. TIA!
I don't know why you want to select likes.id in the first place. I see that you basically want the like_count for each Idea; I don't see the point in selecting likes.id. Also, when you already have the ideas.id, I don't see why you would want to get the value of likes.likeable_id since they'll both be equal. :/
Anyway, the problem is since you're grouping by likeable_id (basically ideas.id), you can't "select" likes.id since they would be "lost" by the grouping.
I suppose SQLite is lax about this. I imagine it wouldn't group things properly.
ANYWAY(2) =>
Let me propose a cleaner solution.
# model
class Idea < ActiveRecord::Base
# to save you the effort of specifying the join-conditions
has_many :likes, foreign_key: :likeable_id
end
# in your code elsewhere
ideas = \
Idea.
joins(:likes).
group("ideas.id").
select("COUNT(likes.id) AS like_count, ideas.id, ideas.title, ideas.intro").
order("like_count DESC")
If you still want to get the IDs of likes for each item, then after the above, here's what you could do:
grouped_like_ids = \
Like.
select(:id, :likeable_id).
each_with_object({}) do |like, hash|
(hash[like.likeable_id] ||= []) << like.id
end
ideas.each do |idea|
# selected previously:
idea.like_count
idea.id
idea.title
idea.intro
# from the hash
like_ids = grouped_like_ids[idea.id] || []
end
Other readers: I'd be very interested in a "clean" one-query non-sub-query solution. Let me know in the comments if you leave a response. Thanks.
I'm having trouble with the 'Ambiguous column name' issue in Transact-SQL, using the Microsoft SQL 2012 Server Management Studio.
I´ve been looking through some of the answers already posted on Stackoverflow, but they don´t seem to work for me, and parts of it I simply don´t understand or loses the general view of.
Executing the following script :
USE CDD
SELECT Artist, Album_title, track_title, track_number, Release_Year, EAN_code
FROM Artists AS a INNER JOIN CD_Albumtitles AS c
ON a.artist_id = c.artist_id
INNER JOIN Track_lists AS t
ON c.title_id = t.title_id
WHERE track_title = 'bohemian rhapsody'
triggers the following error message :
Msg 209, Level 16, State 1, Line 3
Ambiguous column name 'EAN_code'.
Not that this is a CD database with artists names, album titles and track lists. Both the tables 'CD_Albumtitles' and 'Track_lists' have a column, with identical EAN codes. The EAN code is an important internationel code used to uniquely identify CD albums, which is why I would like to keep using it.
You need to put the alias in front of all the columns in your select list and your where clause. You're getting that error because one of the columns you have currently is coming from multiple tables in your join. If you alias the columns, it will essentially pick one or the other of the tables.
SELECT a.Artist,c.Album_title,t.track_title,t.track_number,c.Release_Year,t.EAN_code
FROM Artists AS a INNER JOIN CD_Albumtitles AS c
ON a.artist_id = c.artist_id
INNER JOIN Track_lists AS t
ON c.title_id = t.title_id
WHERE t.track_title = 'bohemian rhapsody'
so choose one of the source tables, prefixing the field with the alias (or table name)
SELECT Artist,Album_title,track_title,track_number,Release_Year,
c.EAN_code -- or t.EAN_code, which should retrieve the same value
By the way, try to prefix all the fields (in the select, the join, the group by, etc.), it's easier for maintenance.