如果PHP中的或语句无法正确评估

Question

I have this PHP statement:

if (($row['rest'] != "") or ($row['rest'] != "Select An Option")) {
    $rest = "<b>Rest Stops:</b> {$row['rest']},";
}
else {
    $rest = "";
}

which is not evaluating properly and I can't figure out why. What I want the statement to do is if the field 'rest' is blank or "Select An Option" then the variable $rest should evaluate to "Rest Stops:" followed by the data. My data is "Select An Option" and I get "Rest Stops: Select An Option" as the output. I did some testing of this statement and I figured out PHP is assigning the variable $row['rest'] as not equal to "" instead of evaluating the 'or' statement. What would be the correct syntax?

Answer 1

What I want the statement to do is if the field 'rest' is blank or Select An Option than the variable $rest should evaluate to Rest Stops: followed by the data.

Your logic is incorrect. You need to check if they are equal to , so use == instead of != .

if ($row['rest'] == ""  || $row['rest'] == "Select An Option") {
^--------------------^     ^--------------------------------^
if field 'rest' blank        if field is 'Select An Option'

This can be be improved by using empty() to perform the "is empty" check:

if (empty($row['test']) || $row['rest'] == "Select An Option") {

Answer 2

"If the field 'rest' is blank ...." - if that's true, then your logic is backwards:

if ($row['rest'] == '') || ($row['rest'] == 'Select an option') {
    $rest = 'rest stops';
} else {
    $rest = '';
}

Note the use of == for equality, rather than != for inequality.