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

使s2int返回正确（int）变量并在变量无法转换为（int）时执行操作的最佳做法是什么（如果输入为十六进制，您会看到结果意外？

http://codepad.org/lN84HKzV

Answer 1

我会使用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

000010的情况含糊不清，因为它也可能意味着八进制;但是如果你想要一个10个碱基的数字，你必须去除任何前导零：

$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

修改

您也可以选择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; }

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;
}

生成string int并检查string是否不是int的最佳实践

2 个答案: