The following will introduce to you two methods of using PHP to determine whether a variable is an integer. I hope this article will be helpful to you.
Method 1: You can round or round the number, and then compare it with the original number. For example, the result of floor(3.1) should be 3. In this case, obviously 3!=3.1, or you can use the ceil() function, which can also be judged. Whether the output is an integer.
Method 2: Use the function is_int() that comes with PHP to easily determine whether the number is an integer.
Example:
$a = 3.3;
//Method 1
The code is as follows
代码如下 |
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if(floor($a)==$a){
echo "$a 是整数!";
}else{
echo "$a 不是整数!";
}
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代码如下 |
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if(is_int($a)){
echo "$a 是整数!";
}else{
echo "$a 不是整数!";
}
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if(floor($a)==$a){
echo "$a is an integer!";
}else{
代码如下 |
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function str_is_int($str)
{
return 0 === strcmp($str , (int)$str);
}
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echo "$a is not an integer!"; |
}
Method 2,
The code is as follows
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if(is_int($a)){
echo "$a is an integer!";
}else{
echo "$a is not an integer!";
}
Note: is_int() and floor check the type of the variable, not the content in the variable. When judging a string, you can use the following instead:
The code is as follows
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function str_is_int($str)
{
return 0 === strcmp($str , (int)$str);
}
http://www.bkjia.com/PHPjc/632620.htmlwww.bkjia.comtruehttp: //www.bkjia.com/PHPjc/632620.htmlTechArticleThe following will introduce two methods for you to determine whether a variable is an integer in PHP. I hope this article will be helpful to you. Classmates will help. Method 1: You can round or round the number,...
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