PHP Two ways to judge whether a variable is an integer _php tutorial

Here are the students to introduce two about the PHP judgment variable is a whole number of two methods, I hope this article will help you.

Method One: Can be used four or five into the number, and then compared with the original number, such as floor (3.1) The result should be 3, at this time obviously 3!=3.1, or with ceil () function also line, this can also determine whether it is an integer.

Method two: Using PHP's own function is_int () can easily determine whether the number is an integer.

To illustrate:

$a = 3.3;

Method One

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if (floor ($a) = = $a) { echo "$a is an integer!"; }else{ echo "$a not an integer!"; }

Method Two,

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if (Is_int ($a)) { echo "$a is an integer!"; }else{ echo "$a not an integer!"; }

Note:is_int () and floor check the type of the variable, not the contents of the variable, when judging the string, you can use the following to replace:

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function Str_is_int ($STR) { return 0 = = = strcmp ($str, (int) $str); }

http://www.bkjia.com/PHPjc/632620.html www.bkjia.com true http://www.bkjia.com/PHPjc/632620.html techarticle Here are the students to introduce two about the PHP judgment variable is a whole number of two methods, I hope this article will help you. Method One: You can use four or five into the number, ...

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