php - 使字符串为 int 并检查字符串是否为 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

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

http://codepad.org/lN84HKzV

Answer 1

我会使用 filter_var()为此:

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

000010 的大小写不明确，因为它也可能表示八进制；但是如果你想要一个 10 基数，你必须去除所有前导零:

$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

如果你也想允许十六进制:

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

你也可以选择 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;
}

出于某种原因，这不会在键盘上运行，但它应该在任何其他系统上运行