[RESOLVED] get part of a integer

nunomira

hi,

just like you have substr() to get part of a string, is there a way to get part of an integer?

If I have something like:

$myVar = 20060103142005;
// and want
$myVar = 20060103;

what's the best way to do it?

$myVar = 20060103142005;
$myVar = substr ($myVar, 0, 8);
echo $myVar;

doesn't work.
Trying to convert the number to a string doesn't work either:

$myVar = 20060103142005;
settype( $myVar, 'string');
$myVar = substr ($myVar, 0, 8);

Installer

As long as the left digit isn't a zero, your first code should work. Otherwise, is this an actual number that you do calculations with? If not, just initialize and use it as a string, then you should get what you're after.

bradgrafelman

Yep, both code examples produced the same result for me: 20060103

So... if you're having problems with such an implementation, you'll have to show us the actual code you're using.

nunomira

hi,

Thank you both.

It seems it has to do with the version of PHP I'm using
I get 2.006010 for both pieces of code, testing with PHP 4.4.2 or 5.x.

But it works as expected with PHP 4.3.x.

It seems that these more recent versions of PHP handle numbers differently:

$myVar = 20060103142005;
echo $myVar; // outputs 2.0060103142E+013

and

$myVar = 20060103142005;
echo (int) $myVar; // outputs -1689097611

Is there a function that converts to a base 10 integer?
I found intval():

$myVar = 20060103142005;
echo intval( $myVar, 10 ); // also outputs -1689097611

😕

laserlight

If you read the PHP Manual on integers, you will see that:

The size of an integer is platform-dependent, although a maximum value of about two billion is the usual value (that's 32 bits signed).

and

If you specify a number beyond the bounds of the integer type, it will be interpreted as a float instead. Also, if you perform an operation that results in a number beyond the bounds of the integer type, a float will be returned instead.

What is happening here is that 20060103142005 is so large that it is beyond the bounds of the integer type, at least on some servers. As such, it is interpreted as a floating point value.

But how did you get that value in the first place? If you are writing it, your problems are solved simply by declaring:

$myVar = "20060103142005";

nunomira

hi laserlight,

That explains it!
And using a string solves it. Besides, that 'number' is a date, and after all date() returns a string.
Your comment opened my eyes to this little detail.
:rolleyes:

Thank you.