I'm trying to make a routine that first checks a users password, if it's correct it shall return some values from a different table or change some values in a row.
Is this even possible without making two queries that you handle in PHP? First call for the password, check if its correct then allow the user to make the name change.
Here an example of getting the Rows in User with email and password.
DELIMITER $$
CREATE DEFINER=`root`@`localhost` PROCEDURE `get_user_info`(
IN in_Email VARCHAR(45),
IN in_Pass VARCHAR(45)
)
BEGIN
SELECT * FROM User WHERE Email = in_Email AND Pass = in_Pass;
END
And this is what Ive got so far:
DELIMITER 开发者_如何学Go$$
CREATE DEFINER=`root`@`localhost` PROCEDURE `change_pass`(
in_Email VARCHAR(45),
in_PassOld VARCHAR(45),
in_PassNew VARCHAR(45)
)
BEGIN
SET @PassOld = (SELECT Pass From User WHERE Email = in_Email);
IF(@PassOld = in_PassOld) THEN
UPDATE User SET Pass = in_PassNew WHERE Email = in_Email;
END IF;
ENDND IF;
END
Thanks for all the help!
You should really hash those passwords, use the following code
DELIMITER $$
CREATE DEFINER=`root`@`localhost` PROCEDURE `change_pass`(
in_Email VARCHAR(45),
in_PassOld VARCHAR(45),
in_PassNew VARCHAR(45)
)
BEGIN
DECLARE KnowsOldPassword INTEGER;
SELECT count(*) INTO KnowsOldPassword
FROM User
WHERE Email = in_Email AND passhash = SHA2(CONCAT(salt, in_PassOld),512);
IF (KnowsOldPassword > 0) THEN
UPDATE User
SET Passhash = SHA2(CONCAT(salt, inPassNew),512)
WHERE Email = in_Email;
END IF;
END $$
DELIMITER ;
The salt
is an extra field in table user
that is more or less random, but does not need to be secret. It serves to defeat rainbow tables.
You can set salt to a short string char(10) or randomish data. e.g.
salt = ROUND(RAND(unix_timestamp(now())*9999999999);
You don't need to update the salt, just generate it once and then store it.
For more on this issue see:
Salting my hashes with PHP and MySQL
How should I ethically approach user password storage for later plaintext retrieval?
A comment on your code
IF(@PassOld == in_PassOld) THEN //incorrect
IF(@PassOld = in_PassOld) THEN //correct, SQL <> PHP :-)
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