I have two tables users and linkage. I am creating a link between one user to another. Means user A will be linked to user B and vise versa.
Now I want to get the details of linked users for a particular entered user_id. Means If user A is finding his linked Id's then the details of linked id's should be seen.
Linkage has three columns its id, user_id and linked_contact_id.
Users has columns as user_id, user_name,pass etc..
I tried one join but I only get the linked Id's from this not the details of id's.
SELECT * FROM Users INNER JOIN linkage ON linkage.user_id = Users.user_id WHERE linkage.linked_contact_id = 1
output
user_id linked_contact_id
1 4
1 1
1 5
How can I get this?Please help.. Thank you...
You will need to join on the users table a second time:
SELECT u.*, u2.* FROM Users u
INNER JOIN linkage l ON l.user_id = u.user_id
INNER JOIN Users u2 ON l.linked_contact_id = u2.user_id
WHERE l.linked_contact_id = 1
Please note that as you are selecting the same columns twice (in u.*, u2.*), you will probably have to list out each field with an alias to distinguish between them.
Something wrong on your ratio. But it's okay.
Try LEFT JOIN or RIGHT JOIN
SELECT * users
LEFT JOIN linkage
ON linkage.user_id = users.user_id
WHERE linkage.linked_contact_id = 1
Get Users information
SELECT users.*,
linkage.*,
contact.user_id as contact_user_id,
contact.user_name AS contact_user_name
FROM users
INNER JOIN linkage ON linkage.user_id = users.user_id
INNER JOIN users AS contact ON linkage.linked_contact_id = contact.user_id
WHERE linkage.linked_contact_id = 1
Edit
Here Screenshot of Query Output
Related
I have two tables representing users and their friends: users and friendships
users has id and name
friendships has id, friend1_id and friend2_id
By searching a user's name, I want to get their associated friends' names.
You can try this:
SELECT u2.name from users u1
INNER JOIN users u2 ON u1.name = [your_name_to_search]
INNER JOIN friendships fs
ON u2.id = fs.friend2_id AND u1.id = fs.friend1_id
Explanation:
I used 2 joins in this approach: the first users self join will assure you take the id corresponding to [your_name_to_search], and then the second join with friendships will extract the names of friend_2_id matching with the friend_1_id taken from the first join.
You can test this approach on this db-fiddle
Join friendships to 2 copies of users.
The 1st copy will be filtered for the user name that you search for and the 2nd will return all the user's friends:
select uf.name
from users u
inner join friendships fs on u.id in (fs.friend1_id, fs.friend2_id)
inner join users uf on uf.id in (fs.friend1_id, fs.friend2_id) and uf.id <> u.id
where u.name = ?
Replace ? with the name of the user whose friends you want in the results.
Trying to JOIN a few tables to supply the user some information on other users he subscribed to this session.
There is a list of other users with their pictures
this works:
SELECT DISTINCT profile_picupload.imageFull, goals_list.goal, goal03.other,
users.username, users.email FROM profile_picupload
JOIN anal_muses
ON anal_muses.username = profile_picupload.username
JOIN goal03 ON goal03.username = anal_muses.username
JOIN goals_list ON goal03.goal=goals_list.id
JOIN users ON users.email = anal_muses.username
it gives me the desired list.
But - i need the list from anal_muses for this session only - so i changed the code to this:
SELECT DISTINCT profile_picupload.imageFull, goals_list.goal, goal03.other,
users.username, users.email FROM profile_picupload
right JOIN anal_muses
ON anal_muses.username = profile_picupload.username
AND anal_muses.sessionId='$sessionId'
JOIN goal03 ON goal03.username = anal_muses.username
JOIN goals_list ON goal03.goal=goals_list.id
JOIN users ON users.email = anal_muses.username
table profile_picupload - holds only one picture for each user
while table anal_musers holds many rows for each user ( depends on the number of people he subscribed to)
so when i added AND anal_muses.sessionId='$sessionId' I needed to change the first join to RIGHT JOIN
but now i get a list with NULL values for profile_picupload .imageFull
.....
is there a way around this?
Thank you :-)
I am currently trying to join a table (tasks) with another table (users). I need to select different rows from the users table based on values in the tasks table:
tasks(user1, user2, user3);
users(id, name, email);
Previously I have linked data using a LEFT JOIN like so:
SELECT * FROM tasks LEFT JOIN users ON tasks.user1 = users.id;
Thats pretty simple, so I have tried using an AND like so:
SELECT * FROM tasks LEFT JOIN users ON tasks.user1 = users.id AND tasks.user2 = users.id;
but that returns the users fields as NULL, which makes sense as tasks.user1 is different to tasks.user2
I am pretty stumped on how to do this now, all suggestions welcome!
You'll need to repeat the join and alias the users table diffeently for each joined column:
SELECT * FROM tasks
LEFT JOIN users u1 ON tasks.user1 = u1.id
LEFT JOIN users u2 ON tasks.user2 = u2.id
...
use this one:
SELECT *
FROM tasks
LEFT JOIN users
ON tasks.user1 = users.id
OR tasks.user2 = users.id;
Let's jump right into it: I've got two simple tables set up in my MySQL database, a users table and a matches table. The users table holds, well, users. The matches table is meant to establish many-to-many connections between users and contains just two userID's.
What is want to query is a list of names of all matched users for the user with userID 1 but I can't wrap my head around it. The problem is that the userID (in this case 1) could be in either one field and I don't have a clue in which one.
Just to clarify; I mean something like this (please don't mind the weird pseudo-code):
SELECT users.name
FROM matches
INNER JOIN users
ON userId = (userId1 OR userId2 DEPENDS ON WHERE)
WHERE userId1 = '1'
OR userId2 = '1';
Could you please tell me if this is possible with MySQL and if so, what I should look for/if you would be so kind, give a simple example.
Thanks a lot.
The user of or in a join condition often prevents MySQL from using an index. The use of union or union all makes the query rather cumbersome. You can do what you want with left outer join:
SELECT coalesce(u1.name, u2.name) as name
FROM matches m LEFT JOIN
users u1
ON u.userId = m.userId1 AND m.userId2 = '1' LEFT JOIN
users u2
ON u.userId = m.userId2 AND m.userId1 = '1'
WHERE '1' in (m.userId1, m.userId2);
This should take advantage of indexes on users for looking up the values. If you want distinct names, then add the distinct keyword.
Try:
SELECT DISTINCT u.name
FROM matches m
INNER JOIN users u
ON (u.userId = m.userId1 AND m.userId2 = '1')
OR (u.userId = m.userId2 AND m.userId1 = '1')
Added DISTINCT to avoid duplicate rows.
See this fiddle.
Here's one way to do it that avoids excessive JOIN logic (to make sure SQL can use indexes on users.userId, matches.userId1, matches.userId2)
SELECT u.`name`
FROM
matches AS m
JOIN users AS u
ON m.userId1=u.userId
AND m.userId2='1'
UNION
SELECT u.`name`
FROM
matches AS m
JOIN users AS u
ON m.userId2=u.userId
AND m.userId1='1'
Something like this:
Select UserId1, UserId2
From Matches
Where UserId1 = 1
Union
Select UserId2, UserId1
From Matches
Where UserId2 = 1
Notice the order of the UserIds have been changed in the Select clause. This will give you a single list of matches with you searched user '1' in a single column and all their matches in the other.
This approach will require you then link in your users table as follows:
Select searchmatches.UserId1, searchmatches.UserId2, leftuser.Name, rightuser.name
From (
Select UserId1, UserId2
From Matches
Where UserId1 = 1
Union
Select UserId2, UserId1
From Matches
Where UserId2 = 1
) searchmatches
inner join users leftuser userMatches.UserId1 = leftuser.UserId
inner join users rightuser userMatches.UserId2 = rightuser.UserId
Hope that Helps! If you want you can remove one of the inner joins to the users table as you know who the left user is as you searched on them!
Sorry but I am not sure how to ask this question but I am working on a help desk application where I have tickets being created in one table. I also have another table that stores the users. My problem is with the tickets table, I have listed the user that created the ticket, the tech who will solve the ticket and a user that over sees the ticket. All three users reference the users table. So how do I can I query the tickets table and get all three users that reference the same table storing the users?
Table1: Tickets
1) Ticketnumber
2) EnteredBy User 100
3) Issue
4) FixedBy User 102
5) FixedByNotes
6) ResponsilbeUser User 103
Table2: Users
1) UserID
2) UserName
What I can do now is something like this:
Select Ticketnumber, EnteredBy, Issue, UserName FROM Tickets INNER JOIN Users
ON Tickets.EnteredBy = Users.UserID
Thanks Steve
You can extend current query to somewhat as follows:
Select Ticketnumber, Issue, Reporter.UserName, Developer.UserName, Manager.UserName FROM Tickets
INNER JOIN Users AS Reporter ON Tickets.EnteredBy = Reporter.UserID
INNER JOIN Users AS Developer ON Tickets.FixedBy = Developer.UserID
INNER JOIN Users AS Manager ON Tickets.ResponsibleUser = Manager.UserID
You need alias for joint tables if you want to get all names:
Select Ticketnumber, Issue, Informers.UserName, Fixers.UserName, Supervisors.UserName FROM Tickets
INNER JOIN Users Informers ON Tickets.EnteredBy = Users.UserID
INNER JOIN Users Fixers ON Tickets.FixedBy = Users.UserID
INNER JOIN Users Supervisors ON Ticket.ResponsibleUser = Users.UserID
WHERE...
Sorry as i am not able to understand your words, but if I assumed your need correctly.. just for a try this could help you..
if you need either of them i.e. all users who has either entered or fixed or saw an issue you can find by..
Select t.Ticketnumber, t.EnteredBy, t.Issue, u.UserID ,u.UserName FROM Tickets t
INNER JOIN Users u ON t.EnteredBy = u.UserID or t.FixedBy = u.UserID
or t.ResponsibleUser = u.UserID;
And If you need all users who has entered, fixed and saw an issue you can find by..
Select t.Ticketnumber, t.EnteredBy, t.Issue, u.UserID ,u.UserName FROM Tickets t
INNER JOIN Users u ON t.EnteredBy = u.UserID and t.FixedBy = u.UserID
and t.ResponsibleUser = u.UserID;