I have 3 tables in my database, in which team1 and team2 ids in Matches Table are equal to team_id in TeamNames Table.
also group in Matches Table is equal to group_id in GroupNames Table
Matches Table
-----------------------------------
| team1 | team2 | group | count |
| 3 | 5 | 1 | 1 |
| 1 | 2 | 3 | 0 |
-----------------------------------
GroupNames Table
-----------------------
| group_id | name |
| 1 | Finals |
| 3 | Semi-Final |
-----------------------
TeamName Table
-----------------------
| team_id | name |
| 5 | Flowers |
| 2 | Rainbow |
-----------------------
What I need to get is:
SELECT team1 , team1_name , team2 , team2_name , group , group_name WHERE count=1
I tried joining tables, but as each of the team1 and team2 should be related to unique id in TeamName Table I failed, getting group name was easy, but I failed getting all the above in single query
Questions:
Is this possible in single query?
Can this be done using CodeIgniter's "Active Record Class"?
Answer for q1:yes
Answer for q2:yes
Hope this may help you
$this->db->from('Matches m');
$this->db->select('m.team1,m.team2,m.group,m.count,tn1.name team1_name,tn2.name team2_name,gn.name group_name');
$this->db->join('TeamName tn1','tn1.team_id = m.team1');
$this->db->join('TeamName tn2','tn2.team_id = m.team2');
$this->db->join('GroupNames gn','gn.group_id = m.group');
$this->db->where('m.count',1);
$results=$this->db->get()->result();
yes. with INNER JOIN.
Here is an example to combine your three tables, if you want you can combine more.
SELECT * FROM matches ma INNER JOIN groups gr ON ma.groupid = gr.groupid INNER JOIN teams te ON ma.teamid = te.teamid WHERE ma.count = 1
Good luck!
Martin
Related
I have these tables...
GROUP_MEMBERS
+---------------------------------+
| id | group_id | member_id |
+---------------------------------+
| 1 | 1 | 1 |
| 2 | 1 | 4 |
+---------------------------------+
MEMBERS
+-------------------------------------+
| id | first | last | role_id |
+-------------------------------------+
| 1 | Jack | Jones | 1 |
| 2 | Jane | Doe | 2 |
| 3 | Bob | Bee | 2 |
| 4 | Jen | Nee | 1 |
+-------------------------------------+
GROUPS
+-----------------+
| id | name |
+-----------------+
| 1 | group1 |
| 2 | group2 |
+-----------------+
As it is, I am using the following query...
SELECT
(members.id) AS memid,
members.first,
members.last,
members.role_id
FROM
members
LEFT JOIN group_members ON
members.id = group_members.member_id
WHERE
group_members.member_id IS NULL
GROUP BY
members.id;
This outputs the members (Jane and Bob) who are not in the 'GROUP_MEMBERS' table as it should, but what I am trying get working is if I am on and another group ($_GET['group_id']), how can I show all members that do not have rows that match group_id and member_id on the 'GROUP_MEMBERS' table...
i.e if group_id = '2' show all members
I have tried adding in WHERE clause... AND group_members.group_id IS NULL.. but it shows nothing then.
Does anyone have a query which would get the output I'm looking for?
Thanks
[EDITED]
Just to clarify...
If my url had 'group_id=1'
I should see:
Bob
Jane
If my url has 'group_id=2'
I should see:
Jack
Jane
Bob
Jen
So it only shows 'members' that do not exist( with the 'group_id' in the url) in the 'GROUP_MEMBERS' table
If I have understood the question correctly, you are looking for something like I have made on this fiddle:
DB Fiddle
The query I use is:
$sql = 'SELECT * FROM groups
RIGHT JOIN group_members ON groups.id = group_id
RIGHT JOIN members ON member_id = members.id
WHERE group_id <> ? OR group_id is NULL;'
$group_id = $_GET['group_id'];
$query = $mysqli->prepare($sql);
$query->bind_param('i', $group_id);
In short, this query will select from the groups table, ensuring that we will select every group in your database.
Then we will join the other two tables completely (using the RIGHT JOIN).
Finally, we are going to select every member that isn't the specified the one provided by the URL, or any member that is not in a group.
You can use a sub-query
how can I show all members that do not have rows that match 'group_id'
$group_id= $_GET['group_id'];
$q = "SELECT * FROM MEMBERS WHERE MEMBERS.id NOT IN(
SELECT member_id FROM GROUP_MEMBERS WHERE group_id='$group_id'
);";
Explanation
SELECT member_id FROM GROUP_MEMBERS WHERE group_id='$grID'
this will get all the members in this group by a given ID
then you select all members that are not among them.
SELECT * FROM MEMBERS WHERE MEMBERS.id NOT IN()
this one will give members data except the ids inside the brackets
the sub query will get the ids of members in a given group
no need for joining the three tables since you are using id of the group existing in GROUP_MEMBERS and linking the GROUP and MEMBERS
one side note
if you have a group name and what all users not in this group you then will need to use the GROUPS table
SELECT * FROM MEMBERS WHERE MEMBERS.id NOT IN(
SELECT member_id FROM GROUP_MEMBERS WHERE group_id = (
SELECT id from GROUPS WHERE name = '$Group_Name'
)
);
you may use WHERE group_id IN (...) it will work the same
This is a demonstration, I created same database with same data and tested the queries
+----+-------+-------+---------+
| id | first | last | role_id |
+----+-------+-------+---------+
| 1 | Jack | Jones | 1 |
| 2 | Jane | Doe | 2 |
| 3 | Bob | Bee | 2 |
| 4 | Jen | Nee | 2 |
+----+-------+-------+---------+
+----+--------+
| id | name |
+----+--------+
| 1 | group1 |
| 2 | group2 |
+----+--------+
+----+----------+-----------+
| id | group_id | member_id |
+----+----------+-----------+
| 1 | 1 | 1 |
| 2 | 1 | 4 |
+----+----------+-----------+
I run the sub-query as above and the results as expected,
MariaDB []> select * from members where id not in
(select member_id from group_members where group_id = 1);
+----+-------+------+---------+
| id | first | last | role_id |
+----+-------+------+---------+
| 2 | Jane | Doe | 2 |
| 3 | Bob | Bee | 2 |
+----+-------+------+---------+
similar for when you have group name
MariaDB []> select * from members where id not in
(select member_id from group_members where group_id =
(select id from groups where name='group1'));
+----+-------+------+---------+
| id | first | last | role_id |
+----+-------+------+---------+
| 2 | Jane | Doe | 2 |
| 3 | Bob | Bee | 2 |
+----+-------+------+---------+
unfortunately i have to do this in mysql / php . I looked for three days, and there is like 10.000 explantions of this but NONE (and I repeat NONE) works for me. I tried it all. I have to ask, sorry.
I have two tables - articles and control.
table "articles"
------------------
art_id | name |
------------------
1 | aaa |
2 | bbb |
3 | ccc |
4 | ddd |
table "control"
--------------------------------------------
con_id | art_id | data |
--------------------------------------------
1 | 1 | something-a |
2 | 2 | something-b |
3 | 1 | something-a |
4 | 2 | something-c |
5 | 3 | something-f |
art_id exists in both tables. Now what i wanted - for query:
"select * from articles order by art_id ASC" displayed in a table
to have also one cell displaying the count for each of art_id's from table CONTROL...
and so i tried join, left join, inner join - i get errors ... I also tried for each get only one result (for example 2 for everything)... this is semi-right but it displays the array of correct results and it's not even with join!!! :
$query = "SELECT art_id, count(*) as counting
FROM control GROUP BY art_id ORDER BY con_id ASC";
$result = mysql_query($query);
while($row=mysql_fetch_array($result)) {
echo $row['counting'];
}
this displays 221 -
-------------------------------------------------
art_id | name | count (this one from control) |
-------------------------------------------------
1 | aaa | 221 |
2 | bbb | 221 |
3 | ccc | 221 |
and it should be:
for art_id(value1)=2,
for art_id(2)=2,
for art_id(3)=1
it should be simple - like a count of values from CONTROL table displayed in query regarding the "articles" table...
The result query on page for table articles should be:
"select * from articles order by art_id ASC"
-------------------------------------------------
art_id | name | count (this one from control) |
-------------------------------------------------
1 | aaa | 2 |
2 | bbb | 2 |
3 | ccc | 1 |
So maybe i should go with JOIN or with join plus for each... Tried tha too, but then i'm not sure what is the proper thing to echo... all-in-all i'm completely lost here. Please help. Thank you.
So imagine this in two steps:
Get the counts per art_id from the control table
Using your articles table, pick up the counts from step 1
That will give you a query that looks like this:
SELECT a.art_id, a.name, b.control_count
FROM articles a
INNER JOIN
(
SELECT art_id, COUNT(*) AS control_count
FROM control
GROUP BY art_id
) b
ON a.art_id = b.art_id;
Which will give you the results you're looking for.
However, instead of using a subquery, you can do it all in one shot:
SELECT a.art_id, a.name, COUNT(b.art_id) AS control_count
FROM articles a
INNER JOIN control b
ON a.art_id = b.art_id
GROUP BY a.art_id, a.name;
SQL Fiddle demo
SELECT *, (SELECT COUNT(control.con_id) FROM control WHERE control.art_id = articles.art_id) AS count_from_con FROM articles ORDER BY art_id DESC;
If I understood your question right, this query should do the trick.
Edit: Created the tables you have described, and it works.
SELECT * FROM articles;
+--------+------+
| art_id | name |
+--------+------+
| 1 | aaa |
| 2 | bbb |
| 3 | ccc |
| 4 | ddd |
+--------+------+
4 rows in set (0.00 sec)
SELECT * FROM control;
+--------+--------+------+
| con_id | art_id | data |
+--------+--------+------+
| 1 | 1 | NULL |
| 2 | 2 | NULL |
| 3 | 1 | NULL |
| 4 | 2 | NULL |
| 5 | 3 | NULL |
+--------+--------+------+
5 rows in set (0.00 sec)
SELECT *, (SELECT COUNT(control.con_id) FROM control WHERE control.art_id = articles.art_id) AS count_from_con FROM articles ORDER BY art_id ASC;
+--------+------+----------------+
| art_id | name | count_from_con |
+--------+------+----------------+
| 1 | aaa | 2 |
| 2 | bbb | 2 |
| 3 | ccc | 1 |
| 4 | ddd | 0 |
+--------+------+----------------+
You haven't quite explained what you want to accomplish with the print out but here is an example in PHP: (Use PDO instead of mysql_)
$pdo = new PDO(); // Make your connection here
$stm = $pdo->query('SELECT *, (SELECT COUNT(control.con_id) FROM control WHERE control.art_id = articles.art_id) AS count_from_con FROM articles ORDER BY art_id ASC');
while( $row = $stm->fetch(PDO::FETCH_ASSOC) )
{
echo "Article with id: ".$row['art_id']. " has " .$row['count_from_con'].' connected rows in control.';
}
Alternatively with the mysql_ extension:
$result = mysql_query('SELECT *, (SELECT COUNT(control.con_id) FROM control WHERE control.art_id = articles.art_id) AS count_from_con FROM articles ORDER BY art_id ASC');
while( $row = mysql_fetch_assoc($result) )
{
echo "Article with id: ".$row['art_id']. " has " .$row['count_from_con'].' connected rows in control.';
}
This should be enough examples to help you accomplish what you need.
I have 2 tables
table A
tag_id | Tag_name
1 | tg1
2 | tg2
3 | tg3
4 | tg4
table B
id | name |tag_id
1 | avq | 1,2,4
2 | bdq | 2
3 | abc | 3,2
4 | vdf | 1,4
5 | zxc | 3
I want to inner join both tables and get its count using tag_id in the following format
`tg1=> 2,tg2=> 3,tg3=> 2,tg4=> 2`
How is it possible in a single MySQL query?
The best option is to normalize the 2nd table and create an association table for storing the tag id and the id of the 2nd table. In the meanwhile the following should do the job but for long run you need to normalize the table else more problems will happen in future
select
t1.Tag_name, count(*) as total
from tableA t1
join tableB t2 on find_in_set(t1.tag_id,t2.tag_id) > 0
group by t1.tag_id ;
You need to create relation table. For example:
Tag table:
+----+----------+
| id | name |
+----+----------+
| 1 | Tag name |
+----+----------+
| 2 | Tag 2 |
+----+----------+
B Table:
+----+-----------+
| id | title |
+----+-----------+
| 1 | Any title |
+----+-----------+
Reference table ex. :
+------+--------+
| b_id | tag_id |
+------+--------+
| 1 | 1 |
+------+--------+
| 1 | 2 |
+------+--------+
In your reference table you put many tags for one B element. In this example you see two tags assigned by reference to b_id = 1
select tag_name, count(position)
from (
select a.tag_name, FIND_IN_SET(a.tag_id,b.tag_id) as position
from a,b
) as tmpTB
where position !=0
group by tag_name
Let's say I have three tables like these:
table name:
id | name
1 | Bob
2 | Alice
3 | Bryan
etc
table pants:
id | size
1 | S
3 | M
table skirt:
id | size
2 | M
How do I merge the three tables using MySQL and obtain a table like this:
result table:
id | name | pants | skirt
1 | Bob | S |
2 | Alice| | M
3 | Bryan| M |
When there's no matching id, the cell will just be blank.
Join the tables with a left join. Like this:
SELECT
tablename.id,
tablename.name,
tablepants.size AS pants,
tableskirt.size as skirt
FROM
tablename
LEFT JOIN tablepants
on tablename.id=tablepants.id
LEFT JOIN tableskirt
ON tablename.id=tableskirt.id
I having two tables
table 1: users
| id | username |
| 1 | john |
| 2 | marry |
| 3 | deep |
| 4 | query |
| 5 | value|
and
table 2:users_2
| table_2_id | user_id |
| 1 | 2,4 |
I need required something like this
| table_2_id | username |
| 1 | marry,query |
anyone can help me for this output in mysql
Is this what you are looking ?
select
`users_2`.`table_2_id` , GROUP_CONCAT(`users`.`username`) as `usernames`
from `users_2`
inner join `users` on FIND_IN_SET(`users`.`id`,`users_2`.`user_id`)
Check output here
http://sqlfiddle.com/#!2/c498bc/3
select a.table_2_id,b.username
from users b,users_2 a
where a.table_2_id=b.id
and b.id in(a.user_id)
group by a.table_2_id
First of all, you should not store a multiple value in a single field. For table users_2, the data should be:
table_2_id user_id
1 2
1 4
After you normalized your table, you can use mysql GROUP_CONCAT() to get the result in the format you mentioned
SELECT
users_2.table_2_id,
GROUP_CONCAT(users.username) AS username
FROM
users_2
JOIN
users ON users.id = users_2.user_id
GROUP BY
users_2.table_2_id
;