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$salt2I am working on a login system and am having a problem with updating my salt tables. I am NOT an expert in sql BUT I know my way around.

My query to update is:

UPDATE hashTable SET `salt1` = 'salt1here' AND `salt2` = 'salt2here' WHERE `userID` = userID

userID is an integer value so I don't need to quote that.

When I update that table salt1 is set to the value of 0. I am using php to create my sql query and that looks like:

UPDATE hashTable SET `salt1` = '$salt1' AND `$salt2` = 'salt2here' WHERE `userID` = $userID

SIDE NOTE: I know about sql injection and I do have protection against that in my code. In this case I do not need this because the salt values are being generated by the script and user id is a value returned by a function. Any place where I do have user input I strip slashes and have ways to prevent injection.

To me my sql query seems correct and I know that my values are correct because this is what a dynamically created query looks like:

UPDATE hashTable SET `salt1` = '9d6db1743e5e0cf1bb0e8cd799c0640231a10ec21e1612a6ed46e8ea16862835' AND `salt2` = '0824b2aac446ccfbd719645f84b13443cbcf59ee4e6dabace8c421ff6a8c6688' WHERE `userID` = 1374770432

I have even entered that in directly to phpMyAdmin and it says 0 rows affected but still changes my salt1 row to 0.

I am somewhat baffled because it seems like I'm doing everything correctly but obviously I am not.

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您的 SQL 查询错误；

UPDATE hashTable 
SET `salt1` = '$salt1' AND `$salt2` = 'salt2here' 
WHERE `userID` = $userID

...应该...

UPDATE hashTable 
SET `salt1` = '$salt1', `$salt2` = 'salt2here' 
WHERE `userID` = $userID

目前，您正在和之间进行操作AND（在这种情况下似乎返回 0）并将其存储在.$salt1$salt2 = 'salt2here'salt1

于 2013-07-25T17:10:25.030 回答