What does $var = value actually return?

This seems like a very mysterious part of PHP to me, and I was wondering if someone could clarify because the manual doesn't seem to include this (or I just couldn't find it anywhere).

What would some of these things return?

if($c = mysql_connect($host, $user, $pass)){
 echo 'Success';
}else{
 echo 'Failure';
}

Woul开发者_如何学Cd this always echo 'Success' because $c is successfully assigned to true or false? I'm wondering if I can do this or if I have to define $c on the previous line.

Thanks.

It is mentioned in Assignment Operators:

The value of an assignment expression is the value assigned. That is, the value of "$a = 3" is 3.

So $c = ... will result in true if true is assigned to $c and in false if false is assigned.

That is also the reason why iterating over query results

while(($row = mysql_fetch_array(...)))

works.

When in doubt, build a test-case!

Build a test-case:

if ( $c = true ) {
  echo 'Success';
} else {
  echo 'No Success';
}

Run this online: http://codepad.org/jxylNzOu

You'll note that the first block will only be executed if the conditions turn up true (or what PHP evaluates as true when asked), so it's not asking if a value was successfully applied to $c, but rather whether the value applied to $c is true.

Regarding your specific example...

If your case, where you're attempting to open a connection to MySQL, $c will be true if the connection is made, resulting in the first block of the if-statement being ran. Otherwise, if no connection is made, $c will be false, making the condition of the if-statement false, resulting in the running of the else block.

According to the documentation for mysql_connect(), one of two things can be returned from this function.

1. A MySQL Link Identifier (if a connection is made)
2. FALSE, indicating a failure to make a connection.

So there's no difference between the following code examples

$conn = mysql_connect( $host, $user, $pass );
if ( $conn ) { /*...*/ }

And

if ( $conn = mysql_connect( $host, $user, $pass ) {
  /*...*/
}

It does not always echo 'Success'. PHP first assigns the result of mysql_connect to $to then evaluates boolean value of $to. But it's better to use this way to ensure the understandability:

$c = mysql_connect($host, $user, $pass)
if($to) {
    echo 'Success';
} else{
    echo 'Failure';
}

PHP is a "weakly typed" language which means php doesn't require (nor support it for that matter) explicit type declaration of variables.

Pay attention to consufe or evaluate 0 1 as true/false (boolean value)

Take this case as sample:

$s = "0"; //String s = '0'
$res = strstr($s,'0'); //Search the character zero into the string $s
if ($res){
  echo "Zero found!";
}else{
  echo "Zero not found!"
}
//Hey!! Whats up!!?? Zero is not found!

This is because zero, which is the returned value of the function strstr, is evaluated as FALSE producing in some cases unexpected results.

The correct way is to use the Not Identical operator !== where value and type is compared

The previous example should be:

$s = "0"; //String s = '0'
$res = strstr($s,'0'); //Search the character zero into the string $s
if ($res !== FALSE){//Check for value AND type
  echo "Zero found!";
}else{
  echo "Zero not found!"
}
//yeah now it works!

So in your case I would write the if statement as:

if(($c = mysql_connect($host, $user, $pass)) !== FALSE){
 echo 'Success';
}else{
 echo 'Failure';
}