Weighted Keyword Search - sql

Hello: I want to do a "weighted search" on product that are tagged with keywords.
(So: not fulltext search, but n-to-m-relation). So here it is:
Table 'product':
sku - the primary key
name
Table 'keywords':
kid - keyword idea
keyword_de - German language String (e.g. 'Hund','Katze','Maus')
keyword_en - English language String (e.g. 'Dog','Cat','Mouse')
Table 'product_keyword' (the cross-table)
sku \__ combined primary key
kid /
What I want is to get a score for all products that at least "contain" one relevant keyword. If I search for ('Dog','Elephant','Maus') I want that
Dog credits a score of 1.003,
Elephant of 1.002
Maus of 1.001
So least important search term starts at 1.001, everything else 0.001++. That way, a lower score limit of 3.0 would equal "AND" query (all three keywords must be found), a lower score limit of 1.0 would equal an "OR". Anything in between something more or less matching. In particular by sorting according to this score, most relevant search results would be first (regardless of lower limit)...
I guess I will have to do something with
IF( keyword1 == 'dog', 1.001, 0) + IF...
maybe inside a SUM() and probably with a GROUP BY at the end of a JOIN over the cross table, eh? But I am fairly clueless how to tackle this.
What would be feasible, is to get the keyword id's from the keywords beforehand. That's a cheap query. So the keywords table can be left ignored and it's all about the other of the cross and product table...
I have PHP at hand to automatically prepare a fairly lengthy PHP statement, but I would like to avoid further multiple SQL statements. In particular since I will limit the query outcome (most often to "LIMIT 0, 20") for paging mode results, so looping a very large number of in between results through a script would be no good...
DANKESCHÖN, if you can help me on this :-)

I think a lot of this is in the Lucene engine (http://lucene.apache.org/java/docs/index.html), which is available for PHP in the Zend Framework: http://framework.zend.com/manual/en/zend.search.lucene.html.
EDIT:
If you want to do the weighted thing you are talking about, I guess you could use something like this:
select p.sku, sum(case k.keyword_en when 'Dog' then 1001 when 'Cat' then 1002 when 'Mouse' then 1003 else 0 end) as totalscore
from products p
left join product_keyword pk on p.sku = pk.sku
inner join keywords k on k.kid = pk.kid
where k.keyword_en in ('Dog', 'Cat', 'Mouse')
group by p.sku
(Edit 2: forgot the group by clause.)

Related

PostgreSQL - Query keyword patterns in columns in table

we all know in SQL we can query a column (lets say, column "breeds") for a certain word like "dog" via a query like this:
select breeds
from myStackOverflowDBTable
where breeds = 'dog'
However, say I had many more columns with much more data, say millions of records, and I did not want to find a word, but rather the most common keyword pattern or wildcard expression, a query like this:
SELECT *
FROM myStackOverflowDBTable
WHERE address LIKE '%alb%'"
Is there an efficient way to find these 'patterns' inside the columns using SQL? I need to find the most common substring so-to-speak, per the query above, say the wildcard string "alb" appeared the most in a "location" column that had words like Albany, Albuquerque, Alabama, obviously querying the words directly would yield 0 results but querying on that wildcard keyword pattern would yield many, but I want to find the most repeating or most frequent wildcard/keyword pattern/regex expression/substring (however you want to define it) for a given column - is there an easy way to do this without querying a million test queries and doing it manually???
Well, if you want to find three character patterns, you could extract all 3-character patterns, aggregate and count:
select substr(t.address, gs.i, 3) as ngram_3, count(*)
from t cross join lateral
generate_series(1, length(address) - 3, 1) gs(i)
group by ngram_3
order by count(*) desc
limit 100;

Sorting with many to many relationship

I have a 3 tables person, person_speaks_language and language.
person has 80 records
language has 2 records
I have the following records
the first 10 persons speaks one language
the first 70 persons (include the first group) speaks 2 languages
the last 10 persons dont speaks any language
Following with the example I want sort the persons by language, How I can do it correctly.
I'm trying to use the the following SQL but seems quite strange
SELECT "person".*
FROM "person"
LEFT JOIN "person_speaks_language" ON "person"."id" = "person_speaks_language"."person_id"
LEFT JOIN "language" ON "person_speaks_language"."language_id" = "language"."id"
ORDER BY "language"."name"
ASC
dataset
71,Catherine,Porter,male,NULL
72,Isabelle,Sharp,male,NULL
73,Scott,Chandler,male,NULL
74,Jean,Graham,male,NULL
75,Marc,Kennedy,male,NULL
76,Marion,Weaver,male,NULL
77,Melvin,Fitzgerald,male,NULL
78,Catherine,Guerrero,male,NULL
79,Linnie,Strickland,male,NULL
80,Ann,Henderson,male,NULL
11,Daniel,Boyd,female,English
12,Ora,Beck,female,English
13,Hulda,Lloyd,female,English
14,Jessie,McBride,female,English
15,Marguerite,Andrews,female,English
16,Maurice,Hamilton,female,English
17,Cecilia,Rhodes,female,English
18,Owen,Powers,female,English
19,Ivan,Butler,female,English
20,Rose,Bishop,female,English
21,Franklin,Mann,female,English
22,Martha,Hogan,female,English
23,Francis,Oliver,female,English
24,Catherine,Carlson,female,English
25,Rose,Sanchez,female,English
26,Danny,Bryant,female,English
27,Jim,Christensen,female,English
28,Eric,Banks,female,English
29,Tony,Dennis,female,English
30,Roy,Hoffman,female,English
31,Edgar,Hunter,female,English
32,Matilda,Gordon,female,English
33,Randall,Cruz,female,English
34,Allen,Brewer,female,English
35,Iva,Pittman,female,English
36,Garrett,Holland,female,English
37,Johnny,Russell,female,English
38,Nina,Richards,female,English
39,Mary,Ballard,female,English
40,Adrian,Sparks,female,English
41,Evelyn,Santos,female,English
42,Bess,Jackson,female,English
43,Nicholas,Love,female,English
44,Fred,Perkins,female,English
45,Cynthia,Dunn,female,English
46,Alan,Lamb,female,English
47,Ricardo,Sims,female,English
48,Rosie,Rogers,female,English
49,Susan,Sutton,female,English
50,Mary,Boone,female,English
51,Francis,Marshall,male,English
52,Carl,Olson,male,English
53,Mario,Becker,male,English
54,May,Hunt,male,English
55,Sophie,Neal,male,English
56,Frederick,Houston,male,English
57,Edwin,Allison,male,English
58,Florence,Wheeler,male,English
59,Julia,Rogers,male,English
60,Janie,Morgan,male,English
61,Louis,Hubbard,male,English
62,Lida,Wolfe,male,English
63,Alfred,Summers,male,English
64,Lina,Shaw,male,English
65,Landon,Carroll,male,English
66,Lilly,Harper,male,English
67,Lela,Gordon,male,English
68,Nina,Perry,male,English
69,Dean,Perez,male,English
70,Bertie,Hill,male,English
1,Nelle,Gill,female,Spanish
2,Lula,Wright,female,Spanish
3,Anthony,Jensen,female,Spanish
4,Rodney,Alvarez,female,Spanish
5,Scott,Holmes,female,Spanish
6,Daisy,Aguilar,female,Spanish
7,Elijah,Olson,female,Spanish
8,Alma,Henderson,female,Spanish
9,Willie,Barrett,female,Spanish
10,Ada,Huff,female,Spanish
11,Daniel,Boyd,female,Spanish
12,Ora,Beck,female,Spanish
13,Hulda,Lloyd,female,Spanish
14,Jessie,McBride,female,Spanish
15,Marguerite,Andrews,female,Spanish
16,Maurice,Hamilton,female,Spanish
17,Cecilia,Rhodes,female,Spanish
18,Owen,Powers,female,Spanish
19,Ivan,Butler,female,Spanish
20,Rose,Bishop,female,Spanish
21,Franklin,Mann,female,Spanish
22,Martha,Hogan,female,Spanish
23,Francis,Oliver,female,Spanish
24,Catherine,Carlson,female,Spanish
25,Rose,Sanchez,female,Spanish
26,Danny,Bryant,female,Spanish
27,Jim,Christensen,female,Spanish
28,Eric,Banks,female,Spanish
29,Tony,Dennis,female,Spanish
30,Roy,Hoffman,female,Spanish
31,Edgar,Hunter,female,Spanish
32,Matilda,Gordon,female,Spanish
33,Randall,Cruz,female,Spanish
34,Allen,Brewer,female,Spanish
35,Iva,Pittman,female,Spanish
36,Garrett,Holland,female,Spanish
37,Johnny,Russell,female,Spanish
38,Nina,Richards,female,Spanish
39,Mary,Ballard,female,Spanish
40,Adrian,Sparks,female,Spanish
41,Evelyn,Santos,female,Spanish
42,Bess,Jackson,female,Spanish
43,Nicholas,Love,female,Spanish
44,Fred,Perkins,female,Spanish
45,Cynthia,Dunn,female,Spanish
46,Alan,Lamb,female,Spanish
47,Ricardo,Sims,female,Spanish
48,Rosie,Rogers,female,Spanish
49,Susan,Sutton,female,Spanish
50,Mary,Boone,female,Spanish
51,Francis,Marshall,male,Spanish
52,Carl,Olson,male,Spanish
53,Mario,Becker,male,Spanish
54,May,Hunt,male,Spanish
55,Sophie,Neal,male,Spanish
56,Frederick,Houston,male,Spanish
57,Edwin,Allison,male,Spanish
58,Florence,Wheeler,male,Spanish
59,Julia,Rogers,male,Spanish
60,Janie,Morgan,male,Spanish
61,Louis,Hubbard,male,Spanish
62,Lida,Wolfe,male,Spanish
63,Alfred,Summers,male,Spanish
64,Lina,Shaw,male,Spanish
65,Landon,Carroll,male,Spanish
66,Lilly,Harper,male,Spanish
67,Lela,Gordon,male,Spanish
68,Nina,Perry,male,Spanish
69,Dean,Perez,male,Spanish
70,Bertie,Hill,male,Spanish
Update
the expect results are: each person must be appears only one time using the language order
For explain the case further, I'll take a new and small dataset, using only the person id and the language name
1,English
2,English
3,English
4,English
19,English
1,Spanish
2,Spanish
3,Spanish
4,Spanish
5,Spanish
14,Spanish
15,Spanish
16,Spanish
19,Spanish
21,Spanish
25,Spanish
I'm using the same order but if I use a limit for example LIMIT 8 the results will be
1,English
2,English
3,English
4,English
19,English
1,Spanish
2,Spanish
3,Spanish
And the expected result is
1,English
2,English
3,English
4,English
19,English
5,Spanish
14,Spanish
15,Spanish
What I'm trying to do
What I'm trying to do is sorting, paginating and filtering a list of X that may have a many-to-many relationship with Y, in this case X is a person and Y is the language. I need do it in a general way. I found a trouble if I want ordering the list by some Y properties.
The list will show in this way:
firstname, lastname, gender , languages
Daniel , Boyd , female , English Spanish
Ora , Beck , female , English
Anthony , Jensen , female , Spanish
....
I only need return a array with the IDs in the correct order
this is the main reason I need that the results only appears the person one time is because the ORM (that I'm using) try to hydrate each result and if I paginate the results using offset and limit. the results maybe aren't the expected. I'm doing assumptions many to many relationships
I can't use the string_agg or group_concat because I dont know the real data, I dont know if are integers or strings
If you want each person to appear only once, then you need to aggregate by that person. If you then want the list of languages, you need to combine them in some way, concatenation comes to mind.
The use of double quotes suggests Postgres or Oracle to me. Here is Postgres syntax for this:
SELECT p.id, string_agg(l.name) as languages
FROM person p LEFT JOIN
person_speaks_language psl
ON p.id = psl.person_id LEFT JOIN
language l
ON psl.language_id = l.id
GROUP BY p.id
ORDER BY COUNT(l.name) DESC, languages;
Similar functionality to string_agg() exists in most databases.
There is nothing wrong with Bertie Hill appearing in two rows, with one language each, that is the Tabular View of Data per the Relational Model. There are no dependencies on data values or number of data values. It is completely correct and un-confused.
But here, the requirement is confused, because you really want three separate lists:
speaks one language
speaks two languages [or the number of languages currently in the language file]
speaks no language [on file] ) ...
But you want those three lists in one list.
Concatenating data values is never, ever a good idea. It is a breach of rudimentary standards, specifically 1NF. It may be common, but it is a gross error. It may be taught by the so-called "theoreticians", but it remains a gross error. Even in a result set, yes.
It creates confusion, such as I have detailed at the top.
With concatenated strings, as the number of languages changes, the width of that concatenated field will grow, and eventually exceed space, wherever it appears (eg. the width of the field on the screen).
Just two of the many reasons why it is incorrect, not expandable, sub-standard.
By the way, in your "dataset" (it isn't the result set produced by your code), the sexes appear to be nicely mixed up.
Therefore the answer, and the only correct one, even if it isn't popular, is that your code is correct (it can be cleaned it up, sure), and you have to educate the user re the dangers of sub-standard code or reports.
You can sort by person.name (rather than by language.name) and then write smarter SQL such that (eg) the person.name is not repeated on the second and subsequent row for persons who speak more than one language, etc. That is just pretty printing.
The non-answer, for those who insist on sub-standard code that will break one day when, is Gordon's response.
Response to Comments
In the Relational Model:
There is no order to the rows, that is deemed a physical or implementation aspect, which we have no control over, and which changes anyway, and which we are warned not to rely upon. If order is sought in the output result set, then we must us ORDER BY, that is its purpose in life.
The data has meaning, and that meaning is carried in Relational Keys. Meaning cannot be carried in surrogates (ie. ID columns).
Limiting myself to the files (they are not tables) that you have given, there is no such thing in the data as:
the first 10 persons who speaks one language
Obtaining persons who speak one language is simple, I believe you already understand that:
SELECT person.first_name,
person.last_name
FROM person P,
(SELECT person_id
FROM person_speaks_language
GROUP BY person_id
HAVING COUNT(*) = 1 -- change this for 2 languages, etc
) AS PL
WHERE P.person_id = PL.person_id
But "first" ? "first" by what criteria ? Record creation date ?
ORDER BY date_created -- if it exists in the data
Record ID does not give first anything: as records are added and deleted, any "order" that may exist initially is completely lost.
You cannot extract meaning out of, or assign meaning to something that, by definition, has no meaning. If the Record ID is relevant, ie. you are going to use it for some purpose, then it is not a Record ID, name the field for what it actually is.
I fail to see, I do not understand, the relevance of the difference between the "dataset" and the updated "small dataset". The "dataset" size is irrelevant, the field headings are irrelevant, what the result set means, is relevant.
The problem is not some "limitation" in the Relational Model, the problem is (a) your fixed view of data values, and (b) your lack of understanding about what the Relational Model is, what it does, understanding of which makes this whole question disappear, and we are left with a simple SQL (as tagged) "how to" question. Eg. If I had a Relational Database, with persons and languages, with no ID columns, there is nothing that I cannot do with it, no report that I cannot produce from it, from the data.
Please try to use an example that conveys the meaning in the data, in what you are trying to do.
the expect results are: each person must be appear only one time
They already appear only once (for each language)
using the language order
Well, there is no order in the language file. We can give it some order, whatever order is meaning-ful, to you, in the result set, based on the data. Eg. language.name. Of course, many persons speak each language, so what order would you like within language.name? How about last_name, first_name. The Record IDs are meaningless to the user, so I won't display them in the result set. NULL is also meaningless, and ambiguous, so I will make the meaning here explicit. This is pretty much what you have, tidied up:
SELECT [language] = CASE name
WHEN NULL THEN "[None]"
ELSE name
END,
last_name,
first_name
FROM person P
LEFT JOIN person_speaks_language PL
ON P.id = PL.person_id
LEFT JOIN language L
ON PL.language_id = L.id
ORDER BY name,
last_name,
first_name
But then you have:
And the expected result is
The example data of which contradicts your textual descriptions:
the expect results are: each person must be appear only one time using the language order
So now, if I ignore the text, and examine the example data re what you want
(which is a horrible thing to do, because I am joining you in the incorrect activity of focussing on the data values, rather than understanding the meaning),
it appears you want the person to appear only once, full stop, regardless of how many languages they speak. Your example data is meaningless, so I cannot be asked to reproduce it. See if this has some meaning.
SELECT last_name,
first_name,
[language] = ( -- correlated subquery
SELECT TOP 1 -- get the "first" language
CASE name -- make meaning of null explicit
WHEN NULL THEN "[None]"
ELSE name
END
FROM person_speaks_language PL
JOIN language L
ON PL.language_id = L.id
WHERE P.id = PL.person_id -- the subject person
ORDER BY name -- id would be meaningless
)
FROM person P -- vector for person, once
ORDER BY last_name,
first_name
Now if you wanted only persons who speak a language (on file):
SELECT last_name,
first_name,
[language] = ( -- correlated subquery
SELECT TOP 1 -- get the "first" language
name
FROM person_speaks_language PL
JOIN language L
ON PL.language_id = L.id
WHERE P.id = PL.person_id -- the subject person
ORDER BY name -- id would be meaningless
)
FROM person P,
(
SELECT DISTINCT person_id -- just one occ, thanks
FROM person_speaks_language PL -- vector for speakers
) AS PL_1
WHERE P.id = PL_1.person_id -- join them to person fields
There, not an outer join anywhere to be seen, in either solution. LEFT or RIGHT will confuse you. Do not attempt to "get everything", so that you can "see" the data values, and then mangle, hack and chop away at the result set, in order to get what you want from that. No, forget about the data values and get only what you want from the record filing system.
Response to Update
I was trying to explain the case with a data set, I think I made things tougher than they actually were
Yes, you did. Reviewing the update then ...
The short answer is, get rid of the ORM. There is nothing in it of value:
you can access the RDB from the queries that populate your objects directly. The way we did for decades before the flatulent beast came along. Especially if you understand and implement Open Architecture Standards.
Further, as evidenced, it creates masses of problems. Here, you are trying to work around the insane restrictions of the ORM.
Pagination is a straight-forward issue, if you have your data Normalised, and Relational Keys.
The long answer is ... please read this Answer. I trust you will understand that the approach you take to designing your app components, your design of windows, will change. All your queries will be simplified, you get only what you require for the specific window or object.
The problem may well disappear entirely (except for possibly the pagination, you might need a method).
Then please think about those architectural issues carefully, and make specific comments of questions.

Writing a query to include certain values but exclude others when looking for a latest time period

I am trying to write a query that looks for a people that have a certain code with the latest period (year) but not if they have another code with that latest period(year). I'll be explicit just so my example makes sense.
I want people who have the code A1,A2,A3,A4,A5 but not AG,AP,AQ. There are people who have an A1 code for a period (like 2014) and an AG code for a the same period. I'd like to exclude them. Not everyone has a code so the field value could be NULL.
Is there a way to express this in a different way (i.e. less characters) than the way I did?
SELECT
people.firstName
FROM
people
WHERE EXISTS (
SELECT *
FROM codes
WHERE
codes.people_id = people.id
AND period = (SELECT MAX(period) FROM codes codes2 WHERE codes2.people_id = codes.people_id)
AND code LIKE 'A[1-5]'
)
AND NOT EXISTS (
SELECT *
FROM codes
WHERE
codes.people_id = people.id
AND period = (
SELECT MAX(period)
FROM codes codes2
WHERE codes2.people_id = codes.people_id
)
AND code LIKE 'A[GPQ]'
)
Schema is as follows:
People
id (PK)
firstName
Codes
people_id (FK) many to one relation with People table
code (e.g. "A1", "A2", "AG")
period (e.g. "2013", "2014")
There are so many ways you could do that, I'm not an SQL expert but I can't see your query being too bad, if you want to try and reduce the number of sub-queries you could consider using the GROUP BY clause along with a SUM Aggregate function in a HAVING clause.
I started updating your code as follows:
SELECT
people.firstName
FROM
people
LEFT JOIN codes AS a15 ON a15.people_id = people.id AND a15.code LIKE 'A[1-5]'
LEFT JOIN codes AS agpq ON agpq.people_id = people.id AND agpq.code LIKE 'A[GPQ]'
GROUP BY
people.firstName
HAVING
SUM(CASE WHEN a15.code IS NULL THEN 0 ELSE 1 END) > 0
AND SUM(CASE WHEN agpq.code IS NULL THEN 0 ELSE 1 END) = 0
This however doesn't take into account anything to do with period specific requirements described. You could add the period to the GROUP BY clause or add it to a WHERE or one of the JOIN constraints but I'm not quite sure from your description exactly what you're after (I don't believe this is through any fault of your own, I just can't personally align the code provided to the description).
I would also like to point out that the SUM functions above will not give an accurate count of the number of matching codes. This is because if both A[GPQ] and A[1_5] return at least one row, the number returned by each constraint will be multiplied by the number returned for the other, it can however be used to determine if there are "any" returned items as if the criteria is matched it will have a SUM(...) > 0
I'm sure a more experienced SQL Developer / DBA will be able to poke many holes in my proposed query but it might give them or someone else something to work from and hopefully gives you ideas for alternatives to using sub-queries.

Oracle 'Contains' / 'Group' function return incorrect value

I have this query:
SELECT last_name, SCORE(1)
FROM Employees
WHERE CONTAINS(last_name, '%sul%', 1) > 0
It produces output below:
The question is:
Why does the SCORE(1) produce 9? As I recall that CONTAINS function returns number of occurrences of search_string (in this case '%sul%').
I expect the output should be:
Sullivan 1
Sully 1
But when I try this syntax:
SELECT last_name, SCORE(1)
FROM Employees
WHERE CONTAINS(last_name, 'sul', 1) >0;
It returns 0 rows selected.
And can someone please explain me what is the third parameter for?
Thanks in advance :)
The reason your second query is returning no rows is, you are looking for word sul in your search. Contains will not do pattern search unless you tell it to, it searches for words which you specified as your second paramter. To look for patterns, you will have to use wildcards, as you did in your first example.
Now, coming to the third parameter in CONTAINS - it is label and is just used to label the score operator. You should use the third parameter when you use SCORE in your SELECT list. It's importance is more clear when there are multiple SCORE operators
Quoting directly from documentaion
label
Specify a number to identify the score produced by the query.
Use this number to identify the CONTAINS clause which returns this
score.
Example
Single CONTAINS
When the SCORE operator is called (for example, in a SELECT clause),
the CONTAINS clause must reference the score label value as in the
following example:
SELECT SCORE(1), title from newsindex
WHERE CONTAINS(text, 'oracle', 1) > 0 ORDER BY SCORE(1) DESC;
Multiple CONTAINS
Assume that a news database stores and indexes the title and body of
news articles separately. The following query returns all the
documents that include the words Oracle in their title and java in
their body. The articles are sorted by the scores for the first
CONTAINS (Oracle) and then by the scores for the second CONTAINS
(java).
SELECT title, body, SCORE(10), SCORE(20) FROM news WHERE CONTAINS
(news.title, 'Oracle', 10) > 0 OR CONTAINS (news.body, 'java', 20) > 0
ORDER BY SCORE(10), SCORE(20);
The Oracle Text Scoring Algorithm does not score by simply counting the number of occurrences. It uses an inverse frequency algorithm based on Salton's formula.
Inverse frequency scoring assumes that frequently occurring terms in a document set are noise terms, and so these terms are scored lower. For a document to score high, the query term must occur frequently in the document but infrequently in the document set as a whole.
Think of a google search. If you search for the term Oracle you will not find (directly) any result that may help to explain your scoring value questioning, so we can consider this term a "noise" to your expectations. But if you search for the term Oracle Text Scoring Algorithm you will find your answer in the first google result.
And about your other questionings, I think that #Incognito already gives them a good answer.

changing sorting criteria after the first result

I am selecting from a database of news articles, and I'd prefer to do it all in one query if possible. In the results, I need a sorting criteria that applies ONLY to the first result.
In my case, the first result must have an image, but the others should be sorted without caring about their image status.
Is this something I can do with some sort of conditionals or user variables in a MySQL query?
Even if you manage to find a query that looks like one query, it is going to be logicaly two queries. Have a look at MySQL UNION if you really must make it one query (but it will still be 2 logical queries). You can union the image in the first with a limit of 1 and the rest in the second.
Something like this ensures an article with an image on the top.
SELECT
id,
title,
newsdate,
article
FROM
news
ORDER BY
CASE WHEN HasImage = 'Y' THEN 0 ELSE 1 END,
newsdate DESC
Unless you define "the first result" closer, of course. This query prefers articles with images, articles without will appear at the end.
Another variant (thanks to le dorfier, who deleted his answer for some reason) would be this:
SELECT
id,
title,
newsdate,
article
FROM
news
ORDER BY
CASE WHEN id = (
SELECT MIN(id) FROM news WHERE HasImage = 'Y'
) THEN 0 ELSE 1 END,
newsdate DESC
This sorts the earliest (assuming MIN(id) means "earliest") article with an image to the top.
I don't think it's possible, as it's effectively 2 queries (the first query the table has to get sorted for, and the second unordered), so you might as well use 2 queries with a LIMIT 1 in the first.