Im trying to construct a query that goes over 3 tables and im COMPLETELY stumped ... my knowledge limit is basic 1 table query and i need some help before i stick my head in a blender.
I have the following query
SELECT * FROM internalrole WHERE introle = $imarole
Im fine with that part .. its the next thats getting me all stressed.
That query returns the following columns ( id, user_id, introle, proven, used )
What i then need to do is take the user_id from the results returned and use it to get the following
SELECT * FROM users WHERE id = user_id(from previous query) AND archive = 0 and status = 8
I need to put that into 1 query, but wait, theres more .... from the results there, i need to check if that user's 'id' is in the availability table, if it is, check the date ( column name is date ) and if it matches todays date, dont return that one user.
I need to put all that in one query :S ... i have NO IDEA how to do it, thinking about it makes my head shake ... If someone could help me out, i would be eternaly grateful.
Cheers,
Use INNER JOIN, which links tables to each other based on a common attribute (typically a primary - foreign key relationship)
say an attribute, 'id', links table1 and table2
SELECT t1.att1, t2.att2
FROM table1 t1
INNER JOIN table2 t2
ON t1.id = t2.id --essentially, this links ids that are equal with each other together to make one large table row
To add more tables, just add more join clauses.
SELECT u.*
FROM internalrole ir
INNER JOIN users u
ON ir.user_id = u.id
AND u.archive = 0
AND u.status = 8
LEFT JOIN availability a
ON ir.user_id = a.user_id
AND a.date = CURDATE()
WHERE ir.introle = $imarole
AND a.user_id IS NULL /* User does NOT exist in availability table w/ today's date */
EDIT: This second query is based on the comments below, asking to show only users who do exist in the availability table.
SELECT u.*
FROM internalrole ir
INNER JOIN users u
ON ir.user_id = u.id
AND u.archive = 0
AND u.status = 8
INNER JOIN availability a
ON ir.user_id = a.user_id
WHERE ir.introle = $imarole
Hmm, maybe something like this
SELECT * FROM users WHERE id IN (SELECT user_id FROM internalrole WHERE introle = $imarole) AND archive = 0 and status = 8;
A handy thing for me to remember is that tables are essentially arrays in SQL.
HTH!
Nested queries are your friend.
SELECT * FROM users WHERE id in (SELECT user_id FROM internalrole WHERE introle = $imarole) AND archive = 0 and status = 8
Alternatively joins:
SELECT * FROM users INNER JOIN internalrole ON users.id = internalrole.user_id WHERE internalrole.user_id = $imarole AND users.archive = 0 and users.status = 8
Related
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'
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
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
#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'
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
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!
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
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.
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
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.
table posts
table users
how would i count posts for specific user logged in. for example when user with id 3 is logged in it should show me 4 posts
I already did it for total posts count:
<?php
$post_query1 = "SELECT count(*) AS total FROM posts ";
$post_result1 = mysqli_query($db, $post_query1);
$post1 = mysqli_fetch_array($post_result1);
?>
Try below example :
select count(*) as total from user as u inner join post as p on p.id_user = u.id_user AND u.id_user = 3
If you want to get only the posts count for the particular user, say user with id = 3, your query should be this:
$query = "SELECT count(*) AS total FROM posts WHERE id_users = 3";
But if you want to get both the posts count as well as the user information and other post information, you will have to run a join query on both the users and posts table. Your query would now become:
$query = "SELECT u.*, p.*, count(p.id_posts) FROM users AS u JOIN posts AS p ON u.id_users = p.id_users WHERE p.id_users = 3";
Some Useful Notes
p.* - * is a wildcard character that means get all the columns in the posts table
u.* - * is a wildcard that means get all the columns in the users table
posts as p - AS is for aliasing. So, we are giving posts table a temporary name.
Here are the different types of the JOINs in SQL:
(INNER) JOIN: Returns records that have matching values in both tables
LEFT (OUTER) JOIN: Return all records from the left table, and the matched records from the right table
RIGHT (OUTER) JOIN: Return all records from the right table, and the matched records from the left table
FULL (OUTER) JOIN: Return all records when there is a match in either left or right table
Note: It is necessary that you have to join two/more tables only with the help of foreign key. Without the foreign key is is meaningless to join two or more tables
Reference 1: https://www.w3schools.com/sql/sql_join.asp
Reference 2: https://www.tutorialspoint.com/mysql/mysql-using-joins.htm
As per the Question what you have asked to join the tables
Query:
SELECT * FROM TABLE 1 JOIN TABLE 2 ON TABLE1.id = TABLE2.id WHERE TABLE2.ID=3
Kindly replace TABLE1 & TABLE2 with the Tables that are to be joined and the id with the foreign key what you have specified in the Table.
Hope so this might be helpful for you to write your own code in future. Happy Coding :)
You have only to use a simple join.
SELECT count(*)
FROM USER u,
post p
WHERE p.id_user = u.id_user
AND u.id_user = 3
So I currently got this query:
SELECT user_expertise.user_id, user_expertise.expertise_id
FROM user_expertise
INNER JOIN user_locations ON user_expertise.user_id = user_locations.user_id
WHERE user_expertise.expertise_id!=$exid AND user_locations.location_id = $_SESSION["user"]["location"]["location_id"]
ORDER BY user_expertise.user_id
$exid is the current id of the expertise and $_SESSION["user"]["location"]["location_id"] is the current location id retrieved from the session. For the sake of this example let's say $exid = 3981 and $_SESSION["user"]["location"]["location_id"] = 24.
I want to retrieve only the users that do not have the expertise_id of $exid (3981) attached to them. The current problem is that users that have this id attached to them get displayed when they also got another one attached to them. Let's say that user with user_id 22 has 3981 and 6523. In this case I don't want him to be part of the results but he is. At first he isn't selected because he has 3981 attached to him but then he is selected because he also has 6523 attached to him.
Update: so I got a little mad from the code used to perform the queries and reworked one of them in a new one so I can pass in a complete query string. Now I can try all of your ideas without getting a headache.
Try this:
SELECT p.user_id, p.expertise_id FROM (
SELECT s.user_id, s.expertise_id,t.expertise_id,
CASE WHEN s.expertise_id = t.expertise_id THEN 1 ELSE 0 as ind
FROM user_expertise s
INNER JOIN user_locations ss ON s.user_id = ss.user_id
CROSS JOIN user_expertise t
WHERE t.user_id = $exid
AND s.user_id <> t.user_id
AND ss.location_id = $_SESSION["user"]["location"]["location_id"]) p
GROUP BY p.user_id, p.expertise_id
HAVING MAX(p.ind) = 0
You can apply conditions to tables directly, no need to put conditions on joined result. It is also way more readable.
SELECT user_expertise.user_id, user_expertise.expertise_id
FROM user_expertise
WHERE 1=1
AND user_expertise.user_id NOT IN
(SELECT user_id FROM user_expertise WHERE expertise_id=$exid)
AND user_expertise.user_id IN
(SELECT user_id from user_locations where location_id = $_SESSION["user"]["location"]["location_id"])
ORDER BY user_expertise.user_id
I am having a little issue here.
So I have two tables and I need to fetch data in what i think is a complex way.
So below is a summary of the two tables
clients
client_id
name
booked [default is 0]
accommodation
accommodation_id
client_id
date
price
What I would like to have is select all client id's from tbl
clients where booked is 1
then using the client_ids select all rows in accommodation whose
client_id is an of those returned in step 1
What i had in mind proved difficult for me
$select_accomodation = "SELECT * FROM `accommodation` WHERE `booked` = 1";
if($select_accomodation_run = #mysql_query($select_accomodation))
{
//awesome code that does no 2
}
What is the best possible way to accomplish tasks 1 and 2. Hopefully in one mysql statement
If you just want to select all accommodations for booked clients you could do
SELECT a.*
FROM accommodation a
INNER JOIN clients c ON a.client_id = c.client_ID
WHERE c.booked = 1
Try this:
select t1.client_id, t2.accommodation_id, t2.client_id, t2.data, t2.price from clients t1 JOIN accommodation t2 on t1.client_id = t2.client_id WHERE t1.booked = 1
My thought, is first write a subquery that gets the Ids you want for part 1, which is:
SELECT client_id FROM clients WHERE booked = 1
Then, you can use that subquery inside another query for the accomodations table using the IN clause
SELECT a.* FROM accomodation a WHERE a.client_id IN (SELECT c.client_id FROM clients c WHERE c.booked = 1);
I am making a cookie based favorite system and need to join data from two tables based on the unique user id stored in the cookie so I can tell what items that use has marked as favorites.
I know I need to do a JOIN but have not used them much and dont really have my head around them yet.
Existing query that selects the items from the db:
$query = mysql_query("SELECT *, UNIX_TIMESTAMP(`date`) AS `date` FROM songs WHERE date >= DATE_SUB( NOW( ) , INTERVAL 2 WEEK ) ORDER BY date DESC");
My favorites table is setup as: ID FAVORITE USERID where ID is the primary key, FAVORITE is the song ID from table songs and USERID is a hash stored in a cookie.
I need to join in all the rows from the favorites table where the USERID field matches the cookie hash variable.
I also need to gather the total number of rows in favorites that match the song id so I can display a count of the number of people who set the item as favorite so I can display how many people like it. But maybe need to do that as a separate query?
This should do it, I would imagine:
$user_id = intval($_COOKIE['user_id']);
$query = mysql_query(sprintf("
SELECT *
FROM songs s
INNER JOIN favorites f
ON f.favorite = s.id
WHERE f.userid = %s
", $user_id));
You should probably read up on the different types of joins.
And then to get the total amount of rows returned, you can just call mysql_num_rows on the result:
$favorite_song_count = mysql_num_rows($query);
EDIT: To select all songs but note which are favorited, you would do this:
$query = mysql_query(sprintf("
SELECT s.*, f.id as favorite_id
FROM songs s
LEFT JOIN favorites f
ON f.favorite = s.id AND f.userid = %s
", $user_id));
By switching it from an INNER JOIN to a LEFT JOIN we are selecting all songs even if they don't have a corresponding record in the favorites table. Any songs that are favorites of the user_id provided will have a non-NULL value for favorite_id.
You can have logical (and, or, ...) operators in join conditions:
select t1.*
from t1
join t2 on t1.id = t2.fid and t2.foo = 'blah'
If you are also querying the total number of times each song has been "favorited" then you need a group by construct also, like this way:
select *, count(f.id)
from songs as s
left join favorites as f on s.id = f.favorite and f.userid = <hash>
group by s.id