CODEWITHSUNDEEP
phpajaxvariablesfirebug
stergosz
stergosz

Reputation: 5860

How to prevent users changing variables via firebug in ajax requests

I allow my users to favorite an update or a forum topic. So when a user tries to favorite one of these i will send via Ajax 2 things, the item_id(update or topic) as id(ex. 1321313213) and its type("update" or "topic") as string.

However lets say someones tries to favorite an update with the id untouched but the type is changed to "topic"(via firebug or whatever else)...

This should not procceed since this combination is not correct... how can i assure that the item_id being sent is an update or a topic since this ID might co-exist in both tables???

Current solution:

Create a hidden input element and add as value 5 random characters (a-zA-Z0-9) and md5 type name(update or topic)

like:

$random_str = $this->my_model->generateRandomString(5);
<input type="hidden" value="<?php echo $random_str.md5("update"); ?>" id="type" />

so when i try to validate the data to check if it is an update or topic i split the type on the first 5 characters and later and check if the later characters are md5 hashed are update or topic and continue validation

I would like some help in case this can be altered as well...

Upvotes: 3

Views: 1984

Answers (2)

petraszd
petraszd

Reputation: 4297

Your are looking at the problem from the wrong perspective. Especially from You server side (PHP) code.

Your server gets data. It gets data which is something like that: user (from session), id and type. Your server needs to ask a question: is it valid data? If it is -- save it to DB; If it is not -- do not save it to DB. It is that simple.

You can look from this perspective: Your client side code is just one way to communicate with Your server. Another way is using web browser + firebug. It is perfectly valid usage of Your server side application. And Your PHP code should not care how request reaches it.

So if Your current code does not allow You in Your PHP code feel comfortable and freely decide if is it update or topic creation than Your need to change Your server side code (and perhaps DB schema) as well.

Your current solution is not good, because if I know how to use firebug I would probably find out that "9d9b68ac2b1de18d3712096354b3c3a5" means "topic" and "3ac340832f29c11538fbe2d6f75e8bcc" means "update".

I think Your are trying to invent Your own CSRF protection. So go on Internet and read about it.

Upvotes: 0

rodneyrehm
rodneyrehm

Reputation: 13557

Your server side script (PHP) must always assume it's getting bogus data. Never rely solely on javascript to handle any sanitization / verification.

If your javascript can determine if the job should be "update" or "topic", I'm sure your PHP can do that as well. Probably using a few more DB queries or some such, but that's the price you've got to pay.

Upvotes: 2

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