php - Getting data from multiple tables in sql -

table 1: invoices (inv_id, inv_value, cust_id)
table 2: customers (cust_id, sales_rep)
table 3: members (member_id, member_cateogry, member_type, cust_id)

note 1: each customer pays multiple invoices. (one-to-many relationship).
note 2: each customer pays one-or-more members (so more 1 member related 1 customer).
note 3: each member has category 1 "represents individual" or 2 "represents group".
note 4: each member has type 1 "represents new" or 2 "represents renew".

i want total of invoice_value field customers who's sales_rep = 1 , member_category = 10 , members_type = 123
ex: total amount of invoices customers paid if sales_rep these customers 1 , members paid new , individual members.

i tried:

select sum(invoices.inv_value) total   invoices,        customers,        members  invoices.cust_id = customers.cust_id ,    members.custid = customers.cust_id ,    members.category = {$category} ,    members_type = {$type} ,    customers.sales_rep = {$id}"; 

and

select     sum(invoices.inv_value) total       members inner join customers on         members.custid = customers.cust_id inner join invoices on         customers.cust_id = invoices.cust_id      customers.sales_rep = {$id} ,        members.category = {$category} ,        members.type = {$type}"; 

but both return double invoice value.
ex.: 1 invoice $120 in invoices table return $240 using these sql queries.

why happen , how fix ???
in advance

this query:

select sum(i.inv_value) total members m inner join      customers c      on m.custid = c.cust_id inner join      invoices      on c.cust_id = i.cust_id c.sales_rep = {$id} ,        m.category = {$category} ,       m.type = {$type}"; 

(don't use implicit join syntax using commas. archaic , less powerful.)

the problem 2 members can have same customer id. can check running:

select m.cust_id, count(*) members m group m.cust_id having count(*) > 1; 

it possible customer ids duplicated in customers.

assuming duplicates in members, change query exists:

select sum(i.inv_value) total customers c      on  inner join      invoices      on c.cust_id = i.cust_id c.sales_rep = {$id} ,        exists (select 1               members m               m.custid = c.cust_id ,                     m.category = {$category} ,                     m.type = {$type}              );