SQL选择一个唯一的字符

Question

在sql中，我想得到所有行，斜杠'/'作为字符只存在一次。

exapmle：

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

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

所以在这个例子中，我想只有第二和第四行......

我该如何形成这个陈述？

Answer 1

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

或

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

Answer 2

要查找只有一个斜杠的表达式：

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

说明：第一个检查斜杠是否至少出现一次。第二个检查它没有出现两次。

至少一次，但不到两次，恰好是一次。

如果您只想要以斜线开头的那些，则应将第一个模式更改为'/%'。

Answer 3

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

Answer 4

或更换/没有任何东西后计算长度。
WHERE ( LENGTH(col) - LENGTH(REPLACE(col, '/', '')) )=1

（感谢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'%/%/%'