Get variable type from object

Question

Im trying to get the variable type but I dont know why, always says that the variable is a string.

Here is my object:

$question

stdClass Object
        (
            [reply_id] => 8
            [q_key] => 35BBs2xy1346230680
            [answer] => 1
        )


echo $question->answer." is a ".gettype($question->answer);

The output is:

1 is a string

But 1 is a integer...

What am I doing wrong ?

Do you want to know if the type is technically an integer or if the scalar value only contains digits? Also, please print the object using `var_dump()`. — Daniel M, Sep 05 '12 at 15:50
If it says it's a string, then it's a string. Where does the data come from for you to think it's not a string? — deceze, Sep 05 '12 at 15:56

score 5 · Answer 1 · edited May 23 '17 at 11:52

If the value was passed from a $_GET or $_POST then it was automatically treated as a string. If it was returned from a mysql query it is being returned as a string.

At best, you can use settype to set it to an integer. Having said that, PHP is pretty liberal when it comes to types and their values.

Example of settype:

<?php
$foo = "5bar"; // string
$bar = true;   // boolean

settype($foo, "integer"); // $foo is now 5   (integer)
settype($bar, "string");  // $bar is now "1" (string)
?>

Notes from Types:

PHP does not require (or support) explicit type definition in variable declaration; a variable's type is determined by the context in which the variable is used. That is to say, if a string value is assigned to variable $var, $var becomes a string. If an integer value is then assigned to $var, it becomes an integer.

An example of PHP's automatic type conversion is the addition operator '+'. If either operand is a float, then both operands are evaluated as floats, and the result will be a float. Otherwise, the operands will be interpreted as integers, and the result will also be an integer. Note that this does not change the types of the operands themselves; the only change is in how the operands are evaluated and what the type of the expression itself is.

score 1 · Answer 2 · answered Sep 05 '12 at 16:15

1

Im using EZsql to get the object... I'm not defining the variable. I think is_numeric() function will do the job :)

Thanks!

answered Sep 05 '12 at 16:15

Klian

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score 0 · Answer 3 · answered Sep 05 '12 at 15:53

0

This is because of the way you assign this variable. Here's the example:

$a = 1;
var_dump($a);
$a = "1";
var_dump($a);

which would produce:

int(1)
string(1) "1"

answered Sep 05 '12 at 15:53

Marcin Orlowski

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score 0 · Answer 4 · edited Sep 05 '12 at 15:58

0

You did not define it as a string, rather than define it this way:

$question->answer = 1;

You need to define it thusly:

$question->answer = "1";

PHP also works with implicit variables, so it may automatically declare a numeric answer as an integer. To remedy this, if $var = "1", define it this way:

$question->answer = strval($var);

edited Sep 05 '12 at 15:58

Fluffeh

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answered Sep 05 '12 at 15:55

Sommer

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Get variable type from object

4 Answers4