mysql 一个复杂的sql

SELECT  r.material_id,r.material_category,r.material_name,r.material_quality,
 r.standard,r.product_factory,r.material_type,r.measure_unit,(r.num+r.num_sub) as num,
(r.num+r.num_sub)-num_sd-num_yd as current_js_num

 FROM
-- 单独计算各个数值
(
SELECT material_base.*
 , 
-- 合同数量; parent_id 
(
SELECT SUM(num) from scm_material_contract
 GROUP BY material_id,form_id,audit_status HAVING  material_id=material_base.material_id
and form_id=‘1111‘ and  audit_status=‘1‘
)
AS num,
(
SELECT SUM(num) from scm_material_contract_add
 GROUP BY material_id,parent_id,audit_status HAVING  material_id=material_base.material_id
and parent_id=‘1111‘ and  audit_status=‘1‘
)
AS num_sub,
(
SELECT SUM(current_js_num) from scm_material_settle_sd
 GROUP BY material_id,parent_id,audit_status HAVING  material_id=material_base.material_id
and parent_id=‘1111‘ and  audit_status=‘1‘
)
AS num_sd,
(
SELECT SUM(current_js_num) from scm_material_settle_yd
 GROUP BY material_id,parent_id,audit_status,chongxiao_status HAVING  material_id=material_base.material_id
and parent_id=‘1111‘ and  audit_status=‘1‘
and chongxiao_status !=‘1‘
)
AS num_yd

FROM(

SELECT material_id,material_category,material_name,material_quality,standard,product_factory,material_type,measure_unit FROM scm_material_base WHERE
material_id in(
-- 找出所有的物料编号
SELECT material_id from scm_material_contract where form_id=‘1111‘ and audit_status=‘1‘
union 
SELECT material_id from scm_material_contract_add ma INNER JOIN scm_purchase_contract_add ac ON ma.form_id=ac.sub_contract_id
where ac.contract_id=‘1111‘ and ac.audit_status=‘1‘
) 
     )as material_base
)as r