converting types

[irishgirl2004]

Hi,
I have a variable which is is a varchar in a mysql database, can php convert this to an integer or a float so i can perform calculations on it?

thanks

[stolzyboy]

PHP doesn't really care whether or not it's a varchar in a mysql database.

As long as it's a number like: 5.2334324 and not like five-thousand, it will do the calculations.

For example:

<?php
$start = "5.3454";
$add = 199;
$total = $start + $add;
echo $total;
?>

$start is a "string" whereas $add is a number, but it will give a total of 204.3454

[NogDog]

PHP is a weakly typed language, and generally speaking you do not need to worry about what type a variable is, as PHP will automatically convert it as needed for different types of operations. For instance:

$a = "1";  // $a is set to the string "1"
$b = $a + 2;  // $b will now be 3

If for some reason you need/want to reference a variable as a specific type, you can typecast it:

$b = (integer) $a + 2;
$c = "This is test number " . (string) $a;

You can also use the [man]settype/man function.

But to reiterate, the vast majority of the time you can probably let PHP do its own automatic typecasting and not worry about it.

[irishgirl2004]

this is odd, the following calculation is coming up with a zero, here goes....

			$i = $myrow1[3];
			$j = i / 1200 ;
			$n = 10 * 12 * -1;

both $j and $n are 0 when i print them? have i done something really dumb?

[irishgirl2004]

forgot to say, $myrow1[3] is 4.6900000000000004

[laserlight]

Looks like a typo error: you mean to use $i, not i.

[irishgirl2004]

deserves a great big smack across the head

sorry!!!!

[NogDog]

A good reason to develop with error_reporting(E_ALL | E_STRICT) so that you would have gotten a notice about using the undefined constant "i".

[bradgrafelman]

Don't forget to mark this thread resolved (if it is).