If you have seen my previous requests you will see I am a TOTAL NOOB with php and mySQL. I wish I wasn't even doing this but I have been dumped in the deep end by people that have let me down. It is 1 day away from "completion" and I am left to pick up and mend the pieces of this system I have which doesn't fully function!
I have tried to explain this as logically and clearly as possible. If you need clarification please let me know and I will do my best!
Thanks in advance for any help as I am now desperate to get this thing done!
I have 3 issues!
ISSUE1
Matching Keys
(x)
TABLE1
| C_ID | C_Eth_O (x) |
+--------------------+-------------------------+
| 234 | 8 |
| 341 | 11 |
| 440 | 2 |
TABLE2
| Eth_ID (x) | C1_Eth_O |
+-----------------------+------------------------+
| 2 | Label2 |
| 8 | Label8 |
| 9 | Label9 |
| 11 | Label11 |
I need to list all the "C1_Eth_O" values in a multi select list. The user will make multiple selections from that list. When submitted I need to get the "Eth_ID" value(s) and return all the value(s) from TABLE1 where any of the selected options "C_Eth_O" = "Eth_ID". Each "C_ID" can only have one "C_Eth_O".
ISSUE2
Matching Keys
(x)
(o)
TABLE1
| C_ID (x) |
+---------------------+
| 234 |
| 341 |
| 440 |
TABLE3
| Ail_ID (o) | Ali_Label |
+-------------------------+-------------------------+
| 1 | Label1 |
| 2 | Label2 |
| 3 | Label3 |
| 4 | Label4 |
| 5 | Label5 |
| 6 | Label6 |
TABLE4
| CA_ID | C1_ID (x) | Ail1_ID (o) |
+---------------------+------------------------+------------------------+
| 1 | 234 | 1 |
| 2 | 341 | 4 |
| 3 | 341 | 6 |
| 4 | 440 | 2 |
I need to list all the "Ali_Label" values from TABLE3 in a multi select list. The user will make multiple selections from that list. When submitted I need to get the "Ail_ID" value(s) from TABLE3 and return all the value(s) from TABLE1 where any of the selected match in TABLE4 "Ali1_ID" = "Ail_ID" & "C1_ID" = "C_ID". Each "C_ID" can have multiple "Ail_ID" values
ISSUE3
Matching Keys
(x)
(o)
(-)
TABLE1
| C_ID (x) |
+---------------------+
| 234 |
| 341 |
| 440 |
TABLE3
| Ail_ID (o) | Ali_Label |
+------------------------+-------------------------+
| 1 | Label1 |
| 2 | Label2 |
| 3 | Label3 |
| 4 | Label4 |
| 5 | Label5 |
| 6 | Label6 |
TABLE5
| R_ID | C1_ID (x) | Cf1_ID (-) |
+-----------------------+-----------------------+-----------------------+
| 1 | 234 | 768 |
| 2 | 234 | 854 |
| 3 | 234 | 768 |
| 4 | 440 | 854 |
TABLE6
| CA_ID | Cf_ID (-) | Ail1_ID (o) |
+---------------------+------------------------+------------------------+
| 1 | 768 | 1 |
| 2 | 854 | 4 |
| 3 | 768 | 6 |
| 4 | 880 | 2 |
I need to list all the "Ali_Label" values from TABLE3 in a multi select list. The user will make multiple selections from that list. When submitted I need to get the "Ail_ID" value(s) from TABLE3 and return all the value(s) from TABLE1 where any of the selected "Ali_ID" = "Ail1_ID" & "C1_ID" = "C_ID" & "Cf1_ID" = "Cf_ID". Each "C_ID" can have multiple "Cf1_ID" values AND each "Cf_ID" can have multiple "Ail1_ID" values.
I have the system working returning simple queries such as
SELECT * FROM table1 WHERE C_ID = 234
but nothing with multiple tables and multiple results per C_ID!
Thanks again!
ISSUE 1 -
SELECT * FROM table1 INNER JOIN table 2 ON table1.C_Eth_O = table2.Eth_ID
ISSUE 2 -
SELECT * FROM table1 INNER JOIN table4 ON table1.C_ID = table4.C1_ID INNER JOIN table3 ON table3.Ail_ID = table4.Ail1_ID
ISSUE 3 -
SELECT * FROM table1 INNER JOIN table5 ON table1.C_ID = table5.C1_ID INNER JOIN table6 ON table5.Cf1_ID = table6.Cf_ID INNER JOIN table3 ON table3.Ail_ID = table6.Ail1_ID
Related
I got to table need to combine into 1
Table 1 :
| ID | FEEDBACK_VALUE |
| 1 | EMAILS |
| 2 | WALK IN |
| 3 | SMS BLAST |
| 4 | SOCIAL MEDIA |
| 5 | NEWSPAPER |
| 6 | FAMILY & FRIEND |
| 7 | OTHERS |
Table 2 :
| ID | FEEDBACK_ID |
| 1 | 1 |
| 2 | 2 |
| 3 | 2 |
| 4 | 7 |
| 5 | 7 |
| 6 | 7 |
| 7 | 4 |
| 8 | 4 |
| 9 | 3 |
Table 3 :
| ID | FEEDBACK_VALUE | FEEDBACK_RECEIVE |
| 1 | EMAILS | 1 |
| 2 | WALK IN | 2 |
| 3 | SMS BLAST | 1 |
| 4 | SOCIAL MEDIA | 2 |
| 5 | NEWSPAPER | 0 |
| 6 | FAMILY & FRIEND | 0 |
| 7 | OTHERS | 3 |
From table 1 and 2, How can i get result like table 3 using mysql? Thanks
You could use a left jojn, and subquery with count group by
select t1.ID, t1.FEEDBACK_VALUE, ifnull( my_count,0) feedback_receive
from table1 t1
left join (
select FEEDBACK_ID, count(*) as my_count
from table 2
group by FEEDBACK_ID
) t on t1.ID = t.FEEDBACK_ID
Just use a subquery as shown below:
SELECT A.*, (SELECT COUNT(*) FROM TABLE2 B WHERE A.ID=B.FEEDBACK_ID) AS FEEDBACK_RECEIVE
FROM TABLE1 A;
See DEMO on SQL Fiddle
Or, if less code is your thing...
SELECT x.*
, COUNT(y.id) total
FROM table_1 x
LEFT
JOIN table_2 y
ON y.feedback_id = x.id
GROUP
BY x.id;
I have a query
$cartable = DB::table('requests')
->select('requests.*','crequest.*')
->leftJoin('crequest','crequest.id','=','requests.request_id')
->whereRaw("FIND_IN_SET('$userid', crequest.user_allows)")
->orderBy('forms_cartable.id','desc')->get();
my tables:
requests:
+------+-------------+-------------------+-------------------+
| id | request_name | request_priority | request_status |
+------+-------------+-------------------+-------------------+
| 1 | test1 | low | process |
| 2 | test2 | low | process |
| 3 | test3 | low | process |
+------+--------------+------------------+-------------------+
crequest
+------+-------------+----------------+--------------+
| id | request_id | users_allow | c_status |
+------+-------------+----------------+--------------+
| 3 | 1 | 12,13,15 | done |
| 4 | 1 | 12 | done |
| 5 | 1 | 13 | end |
| 6 | 2 | 42,12,35 | done |
| 7 | 2 | 47,65 | done |
| 8 | 3 | 42 | open |
+------+-------------+----------------+--------------+
I want if there was an existing current user ID(Logged
) in crequest.user_allows return request`s only without crequest detail
look like this:
(if current user id 12 )
request table result (after execute query)
+------+-------------+-------------------+-------------------+
| id | request_name | request_priority | request_status |
+------+-------------+-------------------+-------------------+
| 1 | test1 | low | process |
| 2 | test2 | low | process |
+------+--------------+------------------+-------------------+
but query result all record match in table crequest i need only requests rows if have match in crequest table
I don't know php but you can achieve it in MySql in so many ways. These are 3 ways among them..
Method 1:
SELECT DISINCT R.*
FROM requests R
INNER JOIN crequest CR ON R.id = CR.request_id
WHERE FIND_IN_SET ('12', CR.users_allow);
Method 2:
SELECT *
FROM requests R
WHERE EXISTS (SELECT 1 FROM crequest CR WHERE R.id = CR.request_id
AND FIND_IN_SET ('12', CR.users_allow));
Method 3:
SELECT *
FROM requests R
WHERE R.id in (SELECT request_id FROM crequest WHERE FIND_IN_SET ('12', CR.users_allow))
I have a table with concatenated values within both rows, I am therefore uniquely retrieve ranking for each row in the tables.
UPDATE
The other tables has been added to question
NamesTable
NID | Name |
1 | Mu |
2 | Ni |
3 | ices |
GroupTable
GID | GName |
1 | GroupA |
2 | GroupB |
3 | GroupC |
MainTable
| NID | Ages | Group |
| 1 | 84 | 1 |
| 2 | 64 | 1 |
| 3 | 78 | 1 |
| 1 | 63 | 2 |
| 2 | 25 | 2 |
| 3 | 87 | 2 |
| 1 | 43 | 3 |
| 2 | 62 | 3 |
| 3 | 37 | 3 |
Now the first Name is equated to the first age in the table, I am able to equate them using php and foreach statements, Now the problem is with the ranking of the ages per each group. I am ranking the names uniquely on each row or group.
Results which is expected
| Names | Ages | Group | Ranking |
| Mu,Ni,ices | 84,64,78 | 1 | 1,3,2 |
| Mu,Ni,ices | 63,25,87 | 2 | 2,3,1 |
| Mu,Ni,ices | 43,62,37 | 3 | 2,1,3 |
In my quest to solving this, I am using GROUP_CONCAT, and I have been able to come to this level in the below query
SELECT
GROUP_CONCAT(Names) NAMES,
GROUP_CONCAT(Ages) AGES,
GROUP_CONCAT(Group) GROUPS,
GROUP_CONCAT( FIND_IN_SET(Ages, (
SELECT
GROUP_CONCAT( Age ORDER BY Age DESC)
FROM (
SELECT
GROUP_CONCAT(Ages ORDER BY Ages DESC ) Age
FROM
`MainTable` s
GROUP by `Group`
) s
)
)) rank
FROM
`MainTable` c
GROUP by `Group`
This actually gives me the below results.
| Names | Ages | Group | Ranking |
| 1,2,3 | 84,64,78 | 1 | 7,9,8 |
| 1,2,3 | 63,25,87 | 2 | 5,6,4 |
| 1,2,3 | 43,62,37 | 3 | 2,1,3 |
The only problem is that the ranking Values increase from 1 to 9 instead of ranking each row uniquely. Its there any idea that can help me cross over and fix this? I would be grateful for your help. Thanks
I have three tables and they are the following
User Table
+---------+-----------+--------+
| user_id | user_name | branch |
+---------+-----------+--------+
| 1 | John | 1 |
| 2 | Jim | 2 |
| 3 | Jern | 3 |
| 4 | Jack | 1 |
| 5 | Jery | 2 |
| 6 | Tom | 3 |
| 7 | Sona | 1 |
| 8 | Tina | 3 |
+---------+-----------+--------+
Branch Table
+-----------+----------------+
| branch_id | branch_name |
+-----------+----------------+
| 1 | IT |
| 2 | SALES |
| 3 | Administration |
+-----------+----------------+
Enquiry Table
+------------+---------------+---------+
| enquiry_id | enquiry_name | user_id |
+------------+---------------+---------+
| 1 | enqury_test1 | 1 |
| 2 | enqury_test2 | 2 |
| 3 | enqury_test3 | 1 |
| 4 | enqury_test4 | 3 |
| 5 | enqury_test5 | 2 |
| 6 | enqury_test6 | 5 |
| 7 | enqury_test7 | 1 |
| 8 | enqury_test8 | 2 |
| 9 | enqury_test9 | 4 |
| 10 | enqury_test10 | 6 |
| 11 | enqury_test11 | 2 |
| 12 | enqury_test12 | 7 |
+------------+---------------+---------+
From the above tables its clear that, each branch contains a number of users.
These users post multiple enquiries.
I need to get the total number of enquiries in each branch as
branch id => number of enquiries
I have tried various queries. But i couldn't get the result. Any one can help?
I am using MySQL and i need a single query to do this.
Thanks in advance
You need count and group by
select
b.branch_id,
count(e.user_id) as `total_enq`
from Branch b
left join User u on u.branch = b.branch_id
left join Enquiry e on e.user_id = u.user_id
group by b.branch_id
The query you have to perform to get you desired result is like this :-
$query = "SELECT u.branch, COUNT(u.user_id) AS `total_enquires`
FROM enquiry e INNER JOIN user u ON e.user_id = u.user_id
GROUP BY u.branch"
This will help you,and i think you don't need to join branch table as user table already contain branch_id.
This is the query
SELECT `branch`,`branch_name`,count(`user`.`user_id`),count(`enquiry_id`) FROM `user` inner join `branch` on `user`.`branch`=`branch`.`branch_id` inner join `enquiry` on `user`.`user_id`=`enquiry`.`user_id` group by `branch`
try it here
http://sqlfiddle.com/#!9/cf3eb/1
SELECT
bt.branch_id
,COUNT(enquiry_id) AS total_enquiry
FROM
enquiry_table et
INNER JOIN user_table ut on ut.user_id = et.user_id
INNER JOIN branch_table bt ON bt.branch_id = ut.branch
WHERE
1=1
GROUP BY
bt.branch_id
you can try this
I want to delete rows based on condition. Let me explain in detail,
table1
-----------------
| t1id | t1detail |
|--------------- |
| 1 | iii |
| 2 | jjj |
| 3 | iik |
| 4 | jjk |
-----------------
table2
------------------------
| t2id | t1id | t1detail |
|----------------------- |
| 1 | 1 | iii |
| 2 | 1 | jja |
| 3 | 3 | iib |
| 4 | 1 | jjc |
| 5 | 2 | iid |
| 6 | 3 | jje |
| 7 | 4 | iif |
| 8 | 2 | jjg |
| 9 | 3 | iih |
| 10 | 4 | jj3 |
------------------------
Now I need to delete table1 id(t1id) at the same time in table2 all id assigned from tablet1 should be deleted.
for example, Suppose I need to delete t1id = 1 mean, it should delete from table1 and table2
in table1
1 | iii
in table2
1 | 1 | iii
2 | 1 | jja
4 | 1 | jjc
should be deleted. Kindly advice me to do in mysql
I tried with
$query = "DELETE FROM table1 INNER JOIN table2 ON table1.t1id = table2.t1id WHERE table1.t1id = {$id};
$result = mysql_query($query, $connection);
but not successfull
I'm going to assume that you don't have foreign keys setup (maybe using MyISAM storage?). If that's the case you will need to specify both tables to delete from in your query.
DELETE table1, table2
FROM table1
JOIN table2
ON table1.t1id = table2.t1id
WHERE table1.t1id = {$id}
If you had foreign keys setup you could use cascading delete, which would solve your issue as well.