php syntax - using 'or' and '&&' in statement

Question

This is for a text field password input. Here is my code:

$vaar is what is entered into that field.

if(($vaar !='pass01' or 'pass02') && (!empty($vaar))) {

I can't figure out the correct syntax. I've tried ||, or, xor. I've also tried placing 'pass01' and 'pass02' in their own ( )'s.

What I want it to do is this:

If $vaar isn't 'pass01' or 'pass02' and $vaar is not empty then do this:

just a syntax error, but I can't figure it out!

Besides the good answers below, I want to add something general: PHP checks the expressions from left to right, if one fails completely (without an alternative expression evaluateing true by OR) it won't check the following anymore — Daniel W.
– Daniel W., Commented Jul 11, 2013 at 19:41

nickb · Accepted Answer · 2013-07-11 19:31:52Z

4

You're looking for:

if( !empty($vaar) && ($vaar != 'pass01' && $vaar != 'pass02')) {

Note that I've put the empty() check first, which will short-circuit the evaluation should $vaar in fact be empty.

Edit: To better reflect the OP's wording (and logic), this is identical to the following, because of Demorgan's law.):

if( !empty($vaar) && !($vaar == 'pass01' || $vaar == 'pass02')) {

Reads: If $vaar is not empty and $vaar is not pass01 or pass02

edited Jul 11, 2013 at 19:31

answered Jul 11, 2013 at 19:26

nickb

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5 Comments

Niet the Dark Absol Over a year ago

x != y || x != z for differing y and z always returns true. -1

nickb Over a year ago

@Kolink - You're absolutely right, I had originally written it with && and then changed it from reading the OP, but the correct logic expression is with &&. Thanks.

nickb Over a year ago

@Downvoters - Care to explain? I would like to improve my answer if it is inadequate.

rink.attendant.6 Over a year ago

@nickb I'm getting downvoted too but I think your answer is adequate.

Ghost Echo Over a year ago

both of you two ended up with the same answer, and it works. Points go here b/c he posted the empty() in the beginning, and also for the DeMorgan's law reference :-)

rink.attendant.6 · Accepted Answer · 2013-07-11 19:32:39Z

1

if(($vaar !== 'pass01' and $vaar !== 'pass02') and (!empty($vaar))) {
   // ...
}

Alternatively:

if(($vaar !== 'pass01' && $vaar !== 'pass02') && (!empty($vaar))) {
   // ...
}

What you actually mean is:

If $vaar is not equal to "pass01" AND $vaar is not equal to "pass02" AND $vaar is not empty, then ...

Note that I have kept your original parentheses in there, but since they are all AND, you may remove them, like this:

if(!empty($vaar) && $vaar !== 'pass01' && $vaar !== 'pass02') {
   // ...
}

Here, I have also moved the empty language construct to the beginning of the conditional, as if the variable is empty then the conditional will evaluate the false immediately, preventing an E_NOTICE error for Undefined variable $vaar if the variable is undefined.

edited Jul 11, 2013 at 19:32

answered Jul 11, 2013 at 19:25

rink.attendant.6

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2 Comments

Niet the Dark Absol Over a year ago

Suggest moving the empty check first, since that will short-circut the other two conditions (as it is more broad) but will also prevent notices in case of an undefined variable.

rink.attendant.6 Over a year ago

@Kolink Good idea, answer now improved.

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