这是使用递归CTE查询的问题答案：
WITH links AS
( SELECT
loan_id,
client_id as c1,
client_id as c2, 0 as distance
FROM
myTable
-- recursion
UNION ALL
SELECT
t.loan_id,
l.c1 as c1,
tt.client_id as c2,
distance = distance + 1
FROM
links l INNER JOIN
myTable t ON l.c2 = t.client_id
AND l.loan_id != t.loan_id INNER JOIN
myTable tt ON t.loan_id = tt.loan_id
AND t.client_id != tt.client_id
)
SELECT * FROM myTable t
WHERE EXISTS
(SELECT * FROM links
WHERE c2 = t.client_id and c1 = 7);
http://sqlfiddle.com/#!3/8394d/16
我已将distance查询保留在其中以使其更易于理解。
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