PHP - Get only single image from many inside while loop [duplicate] - php

Question

I read many threads about getting only the first row of a left join, but, for some reason, this does not work for me.
Here is my structure (simplified of course)
Feeds
id | title | content
----------------------
1 | Feed 1 | ...
Artists
artist_id | artist_name
-----------------------
1 | Artist 1
2 | Artist 2
feeds_artists
rel_id | artist_id | feed_id
----------------------------
1 | 1 | 1
2 | 2 | 1
...
Now i want to get the articles and join only the first Artist and I thought of something like this:
SELECT *
FROM feeds
LEFT JOIN feeds_artists ON wp_feeds.id = (
SELECT feeds_artists.feed_id FROM feeds_artists
WHERE feeds_artists.feed_id = feeds.id
LIMIT 1
)
WHERE feeds.id = '13815'
just to get only the first row of the feeds_artists, but already this does not work.
I can not use TOP because of my database and I can't group the results by feeds_artists.artist_id as i need to sort them by date (I got results by grouping them this way, but the results where not the newest)
Tried something with OUTER APPLY as well - no success as well.
To be honest i can not really imagine whats going on in those rows - probably the biggest reason why i cant get this to work.
SOLUTION:
SELECT *
FROM feeds f
LEFT JOIN artists a ON a.artist_id = (
SELECT artist_id
FROM feeds_artists fa
WHERE fa.feed_id = f.id
LIMIT 1
)
WHERE f.id = '13815'

Answer 1

If you can assume that artist IDs increment over time, then the MIN(artist_id) will be the earliest.
So try something like this (untested...)
SELECT *
FROM feeds f
LEFT JOIN artists a ON a.artist_id = (
SELECT
MIN(fa.artist_id) a_id
FROM feeds_artists fa
WHERE fa.feed_id = f.feed_id
) a

Answer 2

Version without subselect:
SELECT f.title,
f.content,
MIN(a.artist_name) artist_name
FROM feeds f
LEFT JOIN feeds_artists fa ON fa.feed_id = f.id
LEFT JOIN artists a ON fa.artist_id = a.artist_id
GROUP BY f.id

Answer 3

#Matt Dodges answer put me on the right track. Thanks again for all the answers, which helped a lot of guys in the mean time. Got it working like this:
SELECT *
FROM feeds f
LEFT JOIN artists a ON a.artist_id = (
SELECT artist_id
FROM feeds_artists fa
WHERE fa.feed_id = f.id
LIMIT 1
)
WHERE f.id = '13815'

Answer 4

based on several answers here, i found something that worked for me and i wanted to generalize and explain what's going on.
convert:
LEFT JOIN table2 t2 ON (t2.thing = t1.thing)
to:
LEFT JOIN table2 t2 ON (t2.p_key = (SELECT MIN(t2_.p_key)
FROM table2 t2_ WHERE (t2_.thing = t1.thing) LIMIT 1))
the condition that connects t1 and t2 is moved from the ON and into the inner query WHERE. the MIN(primary key) or LIMIT 1 makes sure that only 1 row is returned by the inner query.
after selecting one specific row we need to tell the ON which row it is. that's why the ON is comparing the primary key of the joined tabled.
you can play with the inner query (i.e. order+limit) but it must return one primary key of the desired row that will tell the ON the exact row to join.
Update - for MySQL 5.7+
another option relevant to MySQL 5.7+ is to use ANY_VALUE+GROUP BY. it will select an artist name that is not necessarily the first one.
SELECT feeds.*,ANY_VALUE(feeds_artists.name) artist_name
FROM feeds
LEFT JOIN feeds_artists ON feeds.id = feeds_artists.feed_id
GROUP BY feeds.id
more info about ANY_VALUE: https://dev.mysql.com/doc/refman/5.7/en/group-by-handling.html

Answer 5

I've used something else (I think better...) and want to share it:
I created a VIEW that has a "group" clause
CREATE VIEW vCountries AS SELECT * PROVINCES GROUP BY country_code
SELECT * FROM client INNER JOIN vCountries on client_province = province_id
I want to say yet, that I think that we need to do this solution BECAUSE WE DID SOMETHING WRONG IN THE ANALYSIS... at least in my case... but sometimes it's cheaper to do this that to redesign everything...
I hope it helps!

Answer 6

Here is my answer using the group by clause.
SELECT *
FROM feeds f
LEFT JOIN
(
SELECT artist_id, feed_id
FROM feeds_artists
GROUP BY artist_id, feed_id
) fa ON fa.feed_id = f.id
LEFT JOIN artists a ON a.artist_id = fa.artist_id

Answer 7

I want to give a more generalized answer. One that will handle any case when you want to select only the first item in a LEFT JOIN.
You can use a subquery that GROUP_CONCATS what you want (sorted, too!), then just split the GROUP_CONCAT'd result and take only its first item, like so...
LEFT JOIN Person ON Person.id = (
SELECT SUBSTRING_INDEX(
GROUP_CONCAT(FirstName ORDER BY FirstName DESC SEPARATOR "_" ), '_', 1)
) FROM Person
);
Since we have DESC as our ORDER BY option, this will return a Person id for someone like "Zack". If we wanted someone with the name like "Andy", we would change ORDER BY FirstName DESC to ORDER BY FirstName ASC.
This is nimble, as this places the power of ordering totally within your hands. But, after much testing, it will not scale well in a situation with lots of users and lots of data.
It is, however, useful in running data-intensive reports for admin.

Answer 8

For some database like DB2 and PostgreSQL, you have to use the key word LATERAL for specifying a sub query in the LEFT JOIN : (here, it's for DB2)
SELECT f.*, a.*
FROM feeds f
LEFT JOIN LATERAL
(
SELECT artist_id, feed_id
FROM feeds_artists sfa
WHERE sfa.feed_id = f.id
fetch first 1 rows only
) fa ON fa.feed_id = f.id
LEFT JOIN artists a ON a.artist_id = fa.artist_id

Answer 9

I know this is not a direct solution but as I've faced this and it's always a huge problem for me, and also using left join select etc. sometimes lead to a heavy process cost in database and server, I prefer doing this kind of left joins using array in php like this:
First get the data in range from second table and while you need just one row from second table, just save them with left join in-common column as key in result array.
SQL1:
$sql = SELECT artist_id FROM feeds_artists fa WHERE fa.feed_id {...RANGE...}
$res = $mysqli->query($sql);
if ($res->num_rows > 0) {
while ($row = $res->fetch_assoc()) {
$join_data[...$KEY...] = $row['artist_id'];
}
Then, get the base data and add detail of left join table from previous array while fetch them like this:
SQL2:
$sql = SELECT * FROM feeds f WHERE f.id {...RANGE...};
$res = $mysqli->query($sql);
if ($res->num_rows > 0) {
while ($row = $res->fetch_assoc()) {
$key = $row[in_common_col_value];
$row['EXTRA_DATA'] = $join_data[$key];
$final_data[] = $row;
}
Now, you'll have a $final_data array with desire extra data from $join_data array. this usually works good for date range data and like this.