I'm new to these types of MySQL Queries.
Basically, I have two tables in a MySQL Database. One is called groupmembers and the other is called users
groupmembers contains rows with two columns, a groupid and a userid
users contains rows with also two columns, a userid and a username
I am trying to write a query that searches the groupmembers table with rows with a groupid of 15, but where the userid in groupmembers links with the one in users and searches that one too.
For example:
SELECT * FROM groupmembers WHERE groupid = 15 (something to find the user row by using the userids) AND users.username = '%testquery%'
If you need any more information on this, just let me know because I am bad at explaining things!
Try:
SELECT * FROM groupmembers g INNER JOIN users u ON
g.userid = u.userid WHERE g.groupid = 15 and u.username LIKE '%testquery%'
ORDER BY u.userid
Related
I have two tables:
employees:
id, CMS_user_id, practice_group_id, ...
and
users:
id, level, ...
I want to select all employees where practice_group_id is 2 but only if the respective user has a level of 1 according to the users table. I researched and I have a feeling it has something to do with the UNION keyword eventually, but I can't quite figure it out.
In "human language", the query would be like this:
"select all from employees where practice_group_id is 2 and then check the CMS_user_id from the employee and check in the table users whether the respective user with the id that equals CMS_user_id has a level of 1"
A JOIN will match the corresponding rows between two tables. Then, filtering can be done using WHERE.
For example:
select e.*
from employees e
join users u on u.id = e.CMS_user_id
where e.practice_group = 2 and u.level = 1
I want to have some help creating my query to get information from three different tables sharing information in common.
My first table is:
auctions
id title description user_id(who posted it)
My second table is:
bids
id user_id bid auction_id owner_id
My third table is:
users
id username X XX XXX XXXX
...and my SQL is as follows however it's not returning any results:
SELECT auction_bids.user_id AS applicant, auction_bids.*, auctions.title FROM auction_bids, auctions
WHERE auctions.user_id=".$_SESSION['userid']."
INNER JOIN users ON auction_bids.user_id = users.id
WHERE auction_bids.owner_id = ".$_SESSION['userid']."
What I need is to capture the auction's title, username who bidded on the auction and the bid. the auction has to have a bid and posted by the user who owns the $_SESSION['userid'].
Any help is appreciated.
You have two different 'where' statements, which may just need combining;
SELECT auction_bids.user_id AS applicant, auction_bids.*, auctions.title FROM auction_bids, auctions
INNER JOIN users ON auction_bids.user_id = users.id
WHERE auction_bids.owner_id = ".$_SESSION['userid']." AND auctions.user_id=".$_SESSION['userid']."
However, I'm not sure this is really what you want, as it will return only records where the specific user both 'owns' the item AND has bidded on it (both based on the userid session), rather than displaying all records from different people who have bidded on an item 'owned' by the user.
Something like: ?
SELECT auction_bids.user_id AS applicant, auction_bids.*, auctions.title FROM auction_bids, auctions
INNER JOIN users ON auction_bids.user_id = users.id,
WHERE auction.owner_id = ".$_SESSION['userid']."
Hope this points you in the right direction!
you have 2 where clauses, that is incorrect. I have revised your query based on your requirements.
SELECT auction_bids.user_id AS applicant, auction_bids.*, auctions.title
FROM auction_bids, auctions
INNER JOIN users ON auction_bids.owner_id = users.id
WHERE auction_bids.owner_id = ".$_SESSION['userid']."
AND auctions.user_id=auctions_bids.owner_id
I am somewhat new to coding and have been trying to write what I thought would be a straightforward sql query. Please help :)
Table 1: Users
id = user idt
username = user name
Table 2: Orders
orderid = order id
order_to = user id of person buying
order_from = user id of person selling
oder_details = text
Basically I want to:
"Select Username(from), Username(to), order_details FROM mytables WHERE Order id = 1;"
And get the result as 1 row, I'm not sure how to proceed. I thought I could do this with concatenation or something... Can anyone help?
You need to use JOIN to link the tables together.
SELECT fu.username AS fromUser, tu.username AS toUser, o.order_details
FROM Orders o
INNER JOIN Users fu
ON fu.id = o.order_to
INNER JOIN Users tu
ON tu.id = o.order_from
WHERE o.orderid = '1';
Because you have two different users that you need the username from, you need to JOIN the Users table twice to get both user's usernames. Each table needs to have it's own alias fu and tu to allow MySQL to differentiate between them. The same goes for the column names in your SELECT statement so that when you fetch the results with PHP, php can differentiate between the two usernames.
You are looking for a JOIN. This can be done with a keyword or through WHERE clauses. For example,
SELECT * FROM Orders JOIN Users ON Orders.order_to=Users.id
The documentation can be found here: http://dev.mysql.com/doc/refman/5.0/en/join.html.
I'll leave it as an exercise to figure out how to JOIN the order_from.
I'm not sure how I can do this but I have two tables:
--users--
id_users (index)
name_users
--friend--
id_friend (index)
id_user (connects to id_users)
linked_friend (also links to id_users as this is the id of the requested friend)
for a friendship to be requested one row exists with the user who requested in friend.id_user col and the user they are requesting being in the friend.linked_friend col. To confirm the friendship link another row is created with the info reversed so that there are two rows for each friendship with each user being in the friend.id_user AND the friend.linked_friend col. (let me know if I could do this in a better way)
so what I need to do is for each user list all users on the system that are in a confirmed friendship (two rows) and aren't themselves.
SELECT id_user, name_user FROM user WHERE id_user <> $userId
obviously removes themselves from the list but I'm at a loss as to how to select users that have both rows.
I hope that is clear enough.
Thanks
You could join the friend table twice to demand that rows in both direction exist:
select u1.name
, u2.name
from users u1
join friend f1
on u1.id_users = f1.id_user
join friend f2
on f2.id_user = f1.linked_friend
and f2.linked_friend = f1.id_user
join users u2
on u2.is_users = f2.id_user
and u1.id_users < u2.id_users -- Display friends once
I agree with Jack Pettinger that this design is fairly clunky. There should be only one row in the friends table for each relation. It should have a unique constraint to enforce that, like:
alter table Friends add constraint CHK_Friend check (friend1_id < friend2_id);
create index UX_Friends on Friends (friend1_id, friend2_id);
you have to use Join if you want to fetch data from two or more tables.
Try this
select * from users left join friend on users.id_users = friend.id_user
// you also give where condition for particular user_id
I have the userids of people who have joined a group in one table but not their names thatlie in another table. So I think I need to do a join. I'm starting with groupid that describes the grouop.
Table 1, groupmem has groupid and userid for the members.
Table 2, users has userid and username
The users table has every user. The groupmem only has some who have joined groups.
SQL statement should be something like following but can't get it to work.
select users.name
FROM users,groupmem
LEFT JOIN users
on groupmem.userid=users.userid
WHERE groupmem.groupid = 22
22 being some value..
Thinking maybe where clause is in wrong place or I am using wrong type of join but or not using on correctly but, in any case, can't get it to work. Thx for any suggestions:
Try:
SELECT u.username
FROM `users` u
LEFT JOIN `groupmem` g
ON u.userid = g.userid
WHERE g.groupid = 22
This should do the trick:
SELECT users.name
FROM groupmem
LEFT JOIN users ON groupmem.userid = users.userid
WHERE groupmem.groupid = 22
Your query seems to be quite right - assuming I guessed your table-structure right.