你可以这样做:
select
substring_index(substring_index(str,',',1),',',-1)AS c1
, CASE WHEN LENGTH(str)-LENGTH(REPLACE(str,',','')) >= 1 THEN substring_index(substring_index(str,',',2),',',-1) ELSE NULL END AS c2
, CASE WHEN LENGTH(str)-LENGTH(REPLACE(str,',','')) >= 2 THEN substring_index(substring_index(str,',',3),',',-1) ELSE NULL END AS c3
, CASE WHEN LENGTH(str)-LENGTH(REPLACE(str,',','')) >= 3 THEN substring_index(substring_index(str,',',4),',',-1) ELSE NULL END AS c4
, CASE WHEN LENGTH(str)-LENGTH(REPLACE(str,',','')) >= 4 THEN substring_index(substring_index(str,',',5),',',-1) ELSE NULL END AS c5
, CASE WHEN LENGTH(str)-LENGTH(REPLACE(str,',','')) >= 5 THEN substring_index(substring_index(str,',',6),',',-1) ELSE NULL END AS c6
, CASE WHEN LENGTH(str)-LENGTH(REPLACE(str,',','')) >= 6 THEN substring_index(substring_index(str,',',7),',',-1) ELSE NULL END AS c7
, CASE WHEN LENGTH(str)-LENGTH(REPLACE(str,',','')) >= 7 THEN substring_index(substring_index(str,',',8),',',-1) ELSE NULL END AS c8
, CASE WHEN LENGTH(str)-LENGTH(REPLACE(str,',','')) >= 8 THEN substring_index(substring_index(str,',',9),',',-1) ELSE NULL END AS c9
, CASE WHEN LENGTH(str)-LENGTH(REPLACE(str,',','')) >= 9 THEN substring_index(substring_index(str,',',10),',',-1) ELSE NULL END AS c10
from test
Demo.
表达式有两个共同部分:
-
LENGTH(str)-LENGTH(REPLACE(str,',','')) >= K - 此子表达式确定字符串是否至少有 K 分隔符
-
substring_index(substring_index(str,',',K),',',-1) - 这个子表达式删除了K-th 分隔符之后的元素