php字符串到整数转换和比较

Question

I'm somewhat confused and was hoping you guys could help. I'm writing something simple that takes a value from the posted name field and returns it, checking to see beforehand that its a string(correct) or an integer(in which case it'll say 'a name is not a number').

<?php

$number = array("1","2","3","4","5","6","7","8","9","0");

if(!isset($_POST['submit'])){
?>
<p>Enter your name</p>
<form method="post" action="action4.php">
    <input type="text" name="name">
    <input type="submit" name="submit">
</form>
<?php
}else{
$name = $_POST['name'];
foreach($number as $v){
$int = (int)$v;
}
if($name == $int){
    print "This is a number and not a name.";
}else{
            print $name;
   }
}
?>

Given that every number entered into the name field was seen as a string, I cycled through the array and converted every digit to an integer. When entering an integer into the name field however, rather than getting "this is a number and not a name" I get $name. Reciprocally if I enter in a string, I get "this is a number and not a name."

var_dumping both name and int in the if statement returns as expected: $name as string and $int as integer. So why are they comparing as true?

Answer 1

Because you are doing nothing but assigning a variable in your foreach loop the final value of $v is 0. When PHP compares strings to integers it converts the string to an integer to do the comparison (see Type Juggling). So $name becomes 0. 0 equals 0 so your if statement is always true.

To overcome this you need to compare variable type and value (and do your comparison inside your foreach loop). This means using the === comparison operator which compares both type and value as opposed to just value which is what == does.

Reference: Comparison operators, Type juggling