从MySQL路径提取文件名？
要从路径MySQL中提取文件名，可以使用SUBSTRING_INDEX()。语法如下-
SELECT SUBSTRING_INDEX(ypurColumnName, '\\', -1) as anyAliasName FROM yourTableName;
为了理解上述语法，让我们创建一个表。创建表的查询如下-
mysql> create table ExtractFileNameDemo -> ( -> Id int NOT NULL AUTO_INCREMENT PRIMARY KEY, -> AllProgrammingFilePath varchar(100) -> );
现在，您可以使用insert命令在表中插入一些记录。查询如下-
mysql> insert into ExtractFileNameDemo(AllProgrammingFilePath) values('C:\\Users\\John\\AddTwoNumberProgram.java');
mysql> insert into ExtractFileNameDemo(AllProgrammingFilePath) values('E:\\CProgram\\MasterMindGame.c');
mysql> insert into ExtractFileNameDemo(AllProgrammingFilePath) values('F:\\WebApplication\\WebApp.php');
mysql> insert into ExtractFileNameDemo(AllProgrammingFilePath) values('C:\\Users\\John\\Desktop\\AllMySQLScript.sql');使用select语句显示表中的所有记录。查询如下-
mysql> select *from ExtractFileNameDemo;
以下是输出-
+----+------------------------------------------+ | Id | AllProgrammingFilePath | +----+------------------------------------------+ | 1 | C:\Users\John\AddTwoNumberProgram.java | | 2 | E:\CProgram\MasterMindGame.c | | 3 | F:\WebApplication\WebApp.php | | 4 | C:\Users\John\Desktop\AllMySQLScript.sql | +----+------------------------------------------+ 4 rows in set (0.00 sec)
这是从MySQL的路径中提取文件名的查询-
mysql> select SUBSTRING_INDEX(AllProgrammingFilePath, '\\', -1) as AllFileName from ExtractFileNameDemo;
以下是输出-
+--------------------------+ | AllFileName | +--------------------------+ | AddTwoNumberProgram.java | | MasterMindGame.c | | WebApp.php | | AllMySQLScript.sql | +--------------------------+ 4 rows in set (0.00 sec)