SQL選擇一個獨特的字符

Question

在sql中，我想獲得所有的行，其中斜槓'/'作爲字符只存在一次。

對於exapmle：

[table1] 
id | path_name | 
1 | "/abc/def/"| 
2 | "/abc"  | 
3 | "https://stackoverflow.com/a/b/cdfe"| 
4 | "/hello" | 

select * from table1 where path_name=.... ; 

所以在這個例子，我想有隻第二和第四行......

如何做我必須成立這種說法？

Answer 1

爲了找到表達式正好與一個斜線：

where path_name like '%/%' and not path_name like '%/%/%' 

說明：第一個檢查該斜線出現至少一次。第二個檢查它沒有出現兩次。

至少一次，但少於兩次只是一次。

如果您只想要以斜線開頭的那些，則應該將第一個圖案更改爲'/%'。

Answer 2

where path_name like '%/%' and path_name not like '%/%/%' 

或

where len(path_name) = len(replace(path_name,'/','')) + 1 

Answer 3

select * from table1 where path_name not like '/%/%' and path_name not like '/%/'; 

Answer 4

或者計數替換後的長度/無。
WHERE ( LENGTH(col) - LENGTH(REPLACE(col, '/', '')) )=1

（感謝How to count instances of character in SQL Column）

Answer 5

1.select * from Table1 where len(path_name)-len(replace(path_name,'/',''))= 1 

2.select * from table1 where len(path_name)= len(replace(path_name,'/',''))+1 

3.select * from table1 where path_name like '%/%' and path_name not like'%/%/%'