MYSQL一个优化的过程

背景：优化一个20数据的表查询
###优化前
EXPLAIN
SELECT DISTINCT
ect.contract_no AS contractCode,
ect.contract_name AS contractName,
NULL AS supplierId,
ect.b_name AS supplierName
FROM
eva_test ect
LEFT JOIN td_outline_test toa
ON ect.contract_no = toa.contract_no
WHERE toa.contract_no IS
ORDER BY ect.id;

###优化后
EXPLAIN
SELECT
ect.contract_no AS contractCode,
ect.contract_name AS contractName,
NULL AS supplierId,
ect.b_name AS supplierName
FROM
eva_test ect
LEFT JOIN td_outline_test toa
ON ect.contract_no = toa.contract_no
WHERE toa.contract_no IS NULL
ORDER BY ect.id; ###查询合同编号的数量大于1的合同

###使用EXISTS ，在使用优化后
EXPLAIN
SELECT
ect_out.contract_no AS contractCode,
ect_out.contract_name AS contractName,
NULL AS supplierId,
ect_out.b_name AS supplierName
FROM
eva_test ect_out
WHERE NOT EXISTS
(SELECT
toa.contract_no
FROM
td_outline_test toa
WHERE ect_out.contract_no = toa.contract_no)
ORDER BY ect_out.id;

###使用eva_test.contract_no字段的索引,不推荐使用
SELECT
ect_out.contract_no AS contractCode,
ect_out.contract_name AS contractName,
NULL AS supplierId,
ect_out.b_name AS supplierName
FROM
eva_test ect_out
WHERE ect_out.contract_no NOT IN
(SELECT DISTINCT
ect.contract_no
FROM
eva_test ect
LEFT JOIN td_outline_test toa
ON ect.contract_no = toa.contract_no
WHERE toa.contract_no IS NOT NULL) ;

####辅助查询
SELECT
contract_no,
COUNT(contract_no) AS COUNT
FROM
eva_test
GROUP BY contract_no
HAVING COUNT(contract_no) > 1
ORDER BY countDESC;