An if judge, find a day can not find errors, only.

The code is as follows:

if ($webstate!== "Y" or $webstate!== "N") {echo ""; exit;}

My idea is: $ only two cases, one is Y, the other is n, if there is a third case, the error.

I now if only one condition, it is normal, if plus or more one condition, it is not normal (whether it is equal to Y or equal to n will error. ）

Reply to discussion (solution)

My idea is: $ only two cases, one is Y, the other is n, if there is a third case, the error.

According to you, the third case is not Y and not n, it should be with and not with OR.

Wrong with the logical relationship.

if ($webstate!== "Y" | | $webstate!== "N") {
//
}else{
//
}

if ($webstate!== "Y" and $webstate!== "N")
Or
if (! ($webstate = = "Y" or $webstate = = "N"))

My idea is: $ only two cases, one is Y, the other is n, if there is a third case, the error.

According to you, the third case is not Y and not n, it should be with and not with OR.

I walked into a dead end, thank you.

