超越PHP - Exceed PHP: is_int

描述

bool is_int ( mixed var)

如果 var 是 integer 则返回 TRUE，否则返回 FALSE。

注: 若想测试一个变量是否是数字或数字字符串（如表单输入，它们通常为字符串），必须使用 is_numeric()。

参见 is_bool()、is_float()、is_integer()、is_numeric()、is_string()、is_array() 和 is_object()。

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is_int

logan at logannet dot net
11-Feb-2003 07:42

[[Editors note: Or you can simply use is_numeric()]]



Some people
have offered their ways to find out if a string from a form is an integer
or not, here's my way:



if(ereg("^[0-9]+$",
$_POST["number"])) $_POST["number"] =
(int)$_POST["number"];



In psuedo code:

if you are a
string full of numbers then convert yourself to an integer



So
instead of just checking if its a string full of numbers you check and then
convert it, which means you can use the standard is_int. You can also
do:



if(ereg("^[0-9]+$", $_POST["number"]))
$_POST["number"] += 0;



I think the first way i mentioned
is better because your coding what you want to do, rather than the second
way that uses a side effect of adding 0 to convert the string.



The
first way also may make your code ever so slightly faster (nothing
noticeable) as php does not need to add 0 to the number after it converts
it.



Also note an integer is full numbers (1, 2, 3 etc) not decimal
numbers (1.1, 2.4, 3.7 etc), to convert decimal numbers you could use
something like:



if(ereg("^[.0-9]+$",
$_POST["number"])) $_POST["number"] =
(float)$_POST["number"];



OR



if(ereg("^[.0-9]+$",
$_POST["number"])) $_POST["number"] += 0;



But
note that these would not work with is_int(), because they are not
integers.

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