When does a php if statment not give an output either ways?

Why do you expect for a "true" or "false" be ing printed?

First, echoing an expression of boolean in PHP will never result with a "true" or "false", if the expression is equal to "true", 1 will be printed otherwise nothing will be printed. Secondly, your code shows me that one of "pass" and "fail" should be surely printed. But nothing?

That's a bit strange. Except that your calls to mysql functions failed and threw some exception. Try using echo mysql_error() to know what it is. You can also want to try making your code simpler by writing it step by step. My experience tells me that one of your commands failed, otherwise "pass" or "fail" should be printed!

1. Add

error-reporting(E_ALL);

at the very beginning of your file. That will tell you all the mistakes you made.

2. I assume that you have open the DB and selected the db, of course

3. Do not mix your queries into one: bad habit! If something goes wrong, you can't find out why.

Rewrite (and take care of the syntax below!)

// define your query in a string:

$sql = "select * from `situation` where `situationID`='" . $id . "' ";

// Place the query:

$res = mysql_query($sql) or die ("Error: " . mysql_error() . "<br>SQL: " . $sql);

// this will report the syntax errors if you have any

// Read the results:

$row = mysql_fetch_array($res); // this assumes that it can only be ONE ID, otherwise, you need to use "while"

// now, you can display the results:

foreach ($row as $key => $value)

echo $key . " => " . $value . "<br>";

// don't forget to release the result!

mysql_free_result($res);

echo "asdf <hr>"; // testing php is not having parse errors

$sql = "SELECT * FROM situation WHERE situationID= '{$id}' "; // some sql

$go = mysql_query($sql); // run the query

$row = mysql_fetch_array($go); // fetch the array

print_r($row); // print out the results so you can see whats up.