php - 制作字符串 int 并检查字符串是否不是 int 的最佳实践

Question

我有以下代码：

<?php

function s2int($pinned_id) { 
$action="aaa";
if ( $action && is_numeric($pinned_id) && (float)$pinned_id==(int)$pinned_id) {
   /**
   * @param [string] $action is setted
   * @param [int/string as int] $pinned_id is setted
   */
echo "-chekpoint- $pinned_id\n";
    $pinned_id = (int)$pinned_id;
} 
else { echo "-passpoint- $pinned_id\n";}
return $pinned_id;
}

echo s2int("000010")."\n";
echo s2int(10.00001)."\n";
echo s2int(10)."\n";
echo s2int("10")."\n";
echo s2int("0")."\n";
echo s2int("a")."\n";
echo s2int("a10")."\n";
echo s2int("10a")."\n";
echo s2int("0x1A")."\n";
echo s2int("-100")."\n";

输出：

-chekpoint- 000010
10
-passpoint- 10.00001
10.00001
-chekpoint- 10
10
-chekpoint- 10
10
-chekpoint- 0
0
-passpoint- a
a
-passpoint- a10
a10
-passpoint- 10a
10a
-chekpoint- 0x1A
0
-chekpoint- -100
-100

预期输出：

-chekpoint- 000010
10
-passpoint- 10.00001
10.00001
-chekpoint- 10
10
-chekpoint- 10
10
-chekpoint- 0
0
-passpoint- a
a
-passpoint- a10
a10
-passpoint- 10a
10a
-passpoint- 0x1A
0x1A
-chekpoint- -100
-100

什么是使 s2int 返回正确的 (int) 变量并在变量无法转换为 (int) 时采取行动的最佳实践（如果输入是十六进制，您会看到结果意外？

http://codepad.org/lN84HKzV

Answer 1

我会为此使用filter_var()：

if (false === ($x = filter_var($pinned_id, FILTER_VALIDATE_INT))) {
    echo "Could not convert $pinned_id to an integer";
}
// $x is an integer

的情况000010不明确，因为它也可能表示八进制；但如果你想要一个 10 基数，你必须去掉任何前导零：

$num = preg_replace('/^0+(\d+)/', '\\1', $pinned_id);
if (false === ($x = filter_var($num, FILTER_VALIDATE_INT))) {
    echo "Could not convert $pinned_id to an integer";
}
// $x is an integer

如果您还想允许十六进制：

if (false === ($x = filter_var($num, FILTER_VALIDATE_INT, array(
    'flags' => FILTER_FLAG_ALLOW_HEX,
))) {
    echo "Could not convert $pinned_id to an integer";
}
// $x is an integer

编辑

您也可以选择以下preg_match()路线：

function s2int($pinned_id) { 
    echo "/$pinned_id/  ";
    if (!preg_match('/^-?\d+$/', $pinned_id)) {
        echo "Could not convert $pinned_id to an integer";
        return false;
    }
    return (int)$pinned_id;
}

由于某种原因，这不会在键盘上运行，但它应该在任何其他系统上运行

Answer 2

function s2int($pinned_id) {
// use regex (regular expression)
$pattern = "/[a-zA-Z.]/";
$bool = preg_match($pattern, $pinned_id);

// check if $pinned_id is without any character in the $pattern
if ( $bool == false ) {
    /**
     * @param [string] $action is setted
     * @param [int/string as int] $pinned_id is setted
     */
    echo "-chekpoint- $pinned_id<br />";
    $pinned_id = (int)$pinned_id;
} else { 
    echo "-passpoint- $pinned_id<br />";
}
    return $pinned_id;
}