PHP json_encode(); exclude array

Question

In my array, I have very long variables that are used as a security token to prevent people who don't have authority to the PHP server out. However, I want to be able to log the data to review for bad actions if the variables passed the security wall, but I can't seem to find a way to exclude some variables to prevent getting my log file too big and rotating frequently.

logissue("User ".$authname." is now preforming ".json_encode($data));

returns and logs this huge hunk of data..

66.***.***.**  2015/08/03 06:28:52  User Nickoplier is now preforming {"Validate":"n@xrbycrzF6*au8c8.....","Action":"message","auth":":ZzWh[a....","Parameter1":"2933***","Parameter2":"hello :D","Parameter3":"just a test"}

Is it possible to exclude logging 'Validate' and 'auth'?

M1ke · Accepted Answer · 2015-08-03 11:09:33Z

3

Yes, simply add:

unset($data['Validate'], $data['auth']);

Before the line converting the array into JSON. You can replicate this with as many array keys as you'd like. You can even exclude sub-keys of child arrays: unset($data['stuff']['more-stuff']['field'])

This will delete that key, so a better way would be to wrap the whole component inside a function and pass $data in.

edited Aug 3, 2015 at 11:09

answered Aug 3, 2015 at 10:46

M1ke

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1 Comment

jmattheis Over a year ago

Note: unset deletes that entry so be sure that you don't need them later

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