OrientDB Equal Query not returning results - orientdb

I ran a query like this:
select urlName from User where urlName like 'Nkj-Fm20%';
Which returns one record.
But when I run a query like this:
select * from User where urlName = 'Nkj-Fm20'
No record is found.
Can you help me?

the like operator allow usage of % jolly character, while the = doesn't. So with your second query it returns only the users with the exact urlName Nkj-Fm20.
Ivan

Related

Use postgresql query results to form another query

I am trying to select from one table using the select result from another table. I can run this in two queries but would like to optimize it into just one.
First query.. Select ids where matching other id
select id from lookuptable where paid = '547'
This results in something like this
6316352
6316353
6318409
6318410
6320468
6320469
6320470
6322526
6322527
6324586
6324587
6326648
I would like to then use this result to make another selection. I can do it manually like below. Note, there could be many rows with these values so I've been using a IN statement
select * from "othertable" where id in (6316352,6316353,6318409,6318410,6320468,6320469,6320470,6322526,6322527,6324586,6324587,6326648);
select
ot.*
from
"othertable" as ot
join
lookuptable as lt
on
ot.id = lt.id
where
lt.paid = '547'
The IN operator supports not just value lists but also subqueries, so you can literally write
select * from "othertable" where id in (select id from lookuptable where paid = '547');

OrientDB: select edge where out=(select ??) does not work

I have a problem. I think that this is supposed to work, otherwise someone else would have run into this problem.
The following command works perfectly:
// suppose my record id is #10:0
select from MyEdgeType where out=#10:0
This works.
select from MyNodeType where name="this"
> returns obj with #rid = #10:0
The following does not work:
select from MyEdgeType where out=(select from MyNodeType where name="this")
select from MyEdgeType where out=(select #rid from (select from MyNodeType where name="this")
select from MyEdgeType let $rec = (select fcom MyNodeType...) where out=$rec.rid
... etc.
Nothing works. Nothing. How do I select from edges such that I do not have to know the record id which is incident to the edges I would like to grab ahead of time?
You're comparing a single field on a resultset (it's like comparing a string to an array), try something like this:
select from MyEdgeType where out IN (select from MyNodeType where name="this")
I got this to work.
Since my nodes are unique (this is a constraint), I used the unique property to ID them during the filtration, rather than the record id from a subquery:
select from MyEdgeType where out.unique_identifier=...
worked.

get data from Moodle Database from code

I have a question on how I can extract data from Moodle based on a parameter thats "greater than" or "less than" a given value.
For instance, I'd like to do something like:
**$record = $DB->get_record_sql('SELECT * FROM {question_attempts} WHERE questionid > ?', array(1));**
How can I achieve this, cause each time that I try this, I get a single record in return, instead of all the rows that meet this certain criteria.
Also, how can I get a query like this to work perfectly?
**$sql = ('SELECT * FROM {question_attempts} qa join {question_attempt_steps} qas on qas.questionattemptid = qa.id');**
In the end, I want to get all the quiz question marks for each user on the system, in each quiz.
Use $DB->get_records_sql() instead of $DB->get_record_sql, if you want more than one record to be returned.
Thanks Davo for the response back then (2016, wow!). I did manage to learn this over time.
Well, here is an example of a proper query for getting results from Moodle DB, using the > or < operators:
$quizid = 100; // just an example param here
$cutoffmark = 40 // anyone above 40% gets a Moodle badge!!
$sql = "SELECT q.name, qg.userid, qg.grade FROM {quiz} q JOIN {quiz_grades} qg ON qg.quiz = q.id WHERE q.id = ? AND qg.grade > ?";
$records = $DB->get_records_sql($sql, [$quizid, $cutoffmark]);
The query will return a record of quiz results with all student IDs and grades, who have a grade of over 40.

Laravel 3 Eloquent How to select column as

I'm trying to figure out how to give a column an alias using Eloquent.
So, in other words, how do I execute the following mysql query using Eloquent?
SELECT occupation AS test FROM users WHERE occupation = 'PIMP';
Thx in adv!
Eloquent returns a regular Fluent query. So you can try something like this (assuming your model name is 'User'):
$user = User::where_occupation('pimp')->get(array('occupation as test'));
This is how I have been able to do this in Laravel 5 using select() and passing the col name and the alias to that method (here I'm also using groupby() to restrict it to "DISTINCT" return values then using toarray() to return any array instead of a Collection Object:
$results = asset::select('model_code__c AS option')->whereRAW("model_code__c <> '' AND status = 'A'")->groupby('model_code__c')->get()->toarray();

Need help on this query

I want to have a output count as 2 for the user_mail logged as test1#gmail.com for a query like this,
SELECT Count(user_refemai) from Table_users where userref_mail = user_mail
but, I'm getting the output as 0! What am I doing wrong?
My table_users looks like:
user_id user_mail user_refemail
1 test1#gmail.com NULL
2 test2#gmail.com test1#gmail.com
3 test3#gmail.com test1#gmail.com
you are getting 0 as in your query the condition is false all the time
for this purpose you have to use the cursor or inner queries and then get the count
or pass the parameter to the query for which you want to get the counts.
Shafqat is correct. To build on that, you'd pass in the query parameter like this:
SELECT count(*)
FROM
table_users
WHERE
userref_email = ?
If you need a reporting query instead of for a particular email address, you could use a self join.