SQL選擇一個唯一的字符

Question

在sql中，我想得到所有行，斜杠'/'作為字符只存在一次。

為了exapmle：

[table1]
id | path_name  |
 1 | "/abc/def/"|
 2 | "/abc"     |
 3 | "/a/b/cdfe"|
 4 | "/hello"   |

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

所以在這個例子中，我想只有第二和第四行......

我該如何形成這個陳述？

Answer 1

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

要么

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

Answer 2

要查找只有一個斜杠的表達式：

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

說明：第一個檢查斜杠是否至少出現一次。 第二個檢查它沒有出現兩次。

至少一次，但不到兩次，恰好是一次。

如果只想要以斜杠開頭的那些，則應將第一個模式更改為'/%' 。

Answer 3

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

Answer 4

或者在更換/沒有任何東西后計算長度。
WHERE ( LENGTH(col) - LENGTH(REPLACE(col, '/', '')) )=1

（感謝如何計算SQL列中的字符實例 ）

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'%/%/%'