select len(col)-len(replace(col,'/',''))+1

------解决方案--------------------
探讨

SQL code

select len('a/b/c')-len(replace('a/b/c','/',''))+1
    ,len('a/b/c/d')-len(replace('a/b/c/d','/',''))+1
    
    
/*
                
----------- -----------
3           4

(1 行受影……


------解决方案--------------------
SQL code

select (LEN('a/b')+1)/2 as [times]
times
2
select (LEN('a/b/c/d')+1)/2 as [times]
times
4

这个思路源自数学的等差数列，可以证明