Asked  7 Months ago    Answers:  5   Viewed   37 times

I have the following code

    $page = $_GET['p'];

    if($page == "")
    {
        $page = 1;
    }
    if(is_int($page) == false)
    {
        setcookie("error", "Invalid page.", time()+3600);
        header("location:somethingwentwrong.php");
        die();
    }
    //else continue with code

which I am going to use for looking at different "pages" of a database (results 1-10, 11-20, etc). I can't seem to get the is_int() function to work correctly, however. Putting "1" into the url (noobs.php?p=1) gives me the invalid page error, as well as something like "asdf".

 Answers

77

Using is_numeric() for checking if a variable is an integer is a bad idea. This function will return TRUE for 3.14 for example. It's not the expected behavior.

To do this correctly, you can use one of these options:

Considering this variables array :

$variables = [
    "TEST 0" => 0,
    "TEST 1" => 42,
    "TEST 2" => 4.2,
    "TEST 3" => .42,
    "TEST 4" => 42.,
    "TEST 5" => "42",
    "TEST 6" => "a42",
    "TEST 7" => "42a",
    "TEST 8" => 0x24,
    "TEST 9" => 1337e0
];

The first option (FILTER_VALIDATE_INT way) :

# Check if your variable is an integer
if ( filter_var($variable, FILTER_VALIDATE_INT) === false ) {
  echo "Your variable is not an integer";
}

Output :

TEST 0 : 0 (type:integer) is an integer ?
TEST 1 : 42 (type:integer) is an integer ?
TEST 2 : 4.2 (type:double) is not an integer ?
TEST 3 : 0.42 (type:double) is not an integer ?
TEST 4 : 42 (type:double) is an integer ?
TEST 5 : 42 (type:string) is an integer ?
TEST 6 : a42 (type:string) is not an integer ?
TEST 7 : 42a (type:string) is not an integer ?
TEST 8 : 36 (type:integer) is an integer ?
TEST 9 : 1337 (type:double) is an integer ?

The second option (CASTING COMPARISON way) :

# Check if your variable is an integer
if ( strval($variable) !== strval(intval($variable)) ) {
  echo "Your variable is not an integer";
}

Output :

TEST 0 : 0 (type:integer) is an integer ?
TEST 1 : 42 (type:integer) is an integer ?
TEST 2 : 4.2 (type:double) is not an integer ?
TEST 3 : 0.42 (type:double) is not an integer ?
TEST 4 : 42 (type:double) is an integer ?
TEST 5 : 42 (type:string) is an integer ?
TEST 6 : a42 (type:string) is not an integer ?
TEST 7 : 42a (type:string) is not an integer ?
TEST 8 : 36 (type:integer) is an integer ?
TEST 9 : 1337 (type:double) is an integer ?

The third option (CTYPE_DIGIT way) :

# Check if your variable is an integer
if ( ! ctype_digit(strval($variable)) ) {
  echo "Your variable is not an integer";
}

Output :

TEST 0 : 0 (type:integer) is an integer ?
TEST 1 : 42 (type:integer) is an integer ?
TEST 2 : 4.2 (type:double) is not an integer ?
TEST 3 : 0.42 (type:double) is not an integer ?
TEST 4 : 42 (type:double) is an integer ?
TEST 5 : 42 (type:string) is an integer ?
TEST 6 : a42 (type:string) is not an integer ?
TEST 7 : 42a (type:string) is not an integer ?
TEST 8 : 36 (type:integer) is an integer ?
TEST 9 : 1337 (type:double) is an integer ?

The fourth option (REGEX way) :

# Check if your variable is an integer
if ( ! preg_match('/^d+$/', $variable) ) {
  echo "Your variable is not an integer";
}

Output :

TEST 0 : 0 (type:integer) is an integer ?
TEST 1 : 42 (type:integer) is an integer ?
TEST 2 : 4.2 (type:double) is not an integer ?
TEST 3 : 0.42 (type:double) is not an integer ?
TEST 4 : 42 (type:double) is an integer ?
TEST 5 : 42 (type:string) is an integer ?
TEST 6 : a42 (type:string) is not an integer ?
TEST 7 : 42a (type:string) is not an integer ?
TEST 8 : 36 (type:integer) is an integer ?
TEST 9 : 1337 (type:double) is an integer ?
Wednesday, March 31, 2021
 
Juicy
answered 7 Months ago
86

You should point to your vendor/autoload.php at Settings | PHP | PHPUnit when using PHPUnit via Composer.

This blog post has all the details (with pictures) to successfully configure IDE for such scenario: http://confluence.jetbrains.com/display/PhpStorm/PHPUnit+Installation+via+Composer+in+PhpStorm

Related usability ticket: http://youtrack.jetbrains.com/issue/WI-18388

P.S. The WI-18388 ticket is already fixed in v8.0

Wednesday, March 31, 2021
 
ojrac
answered 7 Months ago
79

On Mac OS X environment variables available in Terminal and for the normal applications can be different, check the related question for the solution how to make them similar.

Note that this solution will not work on Mountain Lion (10.8).

Saturday, May 29, 2021
 
Nate
answered 5 Months ago
40

If you need to do this, do

isinstance(<var>, int)

unless you are in Python 2.x in which case you want

isinstance(<var>, (int, long))

Do not use type. It is almost never the right answer in Python, since it blocks all the flexibility of polymorphism. For instance, if you subclass int, your new class should register as an int, which type will not do:

class Spam(int): pass
x = Spam(0)
type(x) == int # False
isinstance(x, int) # True

This adheres to Python's strong polymorphism: you should allow any object that behaves like an int, instead of mandating that it be one.

BUT

The classical Python mentality, though, is that it's easier to ask forgiveness than permission. In other words, don't check whether x is an integer; assume that it is and catch the exception results if it isn't:

try:
    x += 1
except TypeError:
    ...

This mentality is slowly being overtaken by the use of abstract base classes, which let you register exactly what properties your object should have (adding? multiplying? doubling?) by making it inherit from a specially-constructed class. That would be the best solution, since it will permit exactly those objects with the necessary and sufficient attributes, but you will have to read the docs on how to use it.

Tuesday, June 1, 2021
 
SheppardDigital
answered 5 Months ago
30

You can use the is_a? method

>> 1.is_a? Integer
=> true
>> "dadadad@asdasd.net".is_a? Integer
=> false
>> nil.is_a? Integer
=> false
Tuesday, July 27, 2021
 
superfell
answered 3 Months ago
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