PHP-检查字符串是否仅包含整个或半个十进制数字的最优雅的方法？

Question

我以为这将非常简单，但是在PHP中，要确定字符串是仅包含整数还是包含整数或半数的浮点数将非常棘手。

例如，我想要：

// Approved  
"0.5"
"12"  
"22.5"  
"6.0"
"1"
"0"

// Rejected  
"0.65"  
"foo19bar"  
"ten"  
"39.4"
"s12"
"0x600"

棘手的事情：

• 检测（仅）字符串中的浮点数很难：

  • is_float()显然拒绝字符串（例如is_float("2.5") = false ）
  • 浮点型转换允许字符串包含字母（例如is_float((float)"wat2.5") = true ）

• 检测浮点是整个还是一半也是很奇怪的：

  • 模数运算符（ % ）由于某种原因会自动转换为int ，因此您需要改用fmod() （例如10.33 % 0.5在PHP 5.4中返回0 ）

到目前为止，我最简洁的解决方案是这个，但是我觉得应该有一个更好的方法来做到这一点：

if (ctype_digit($rating) || (is_float($rating + 0) && fmod($rating, 0.5) == 0)) {
    if (is_numeric($rating)) {
        echo "Approved";
    }
}

Answer 1

<?php

$test = [
    "0.5",
    "12",  
    "22.5",  
    "6.0",
    "1",
    "0",
    "0.65",
    "foo19bar",
    "ten",
    "39.4",
    "s12",
    "0x600",
];

foreach ($test as $number)
    if ((string)((int)((double)$number * 2) / 2) == $number)
        echo "Approved $number\n";
    else
        echo "Rejected $number\n";

输出：

Approved 0.5
Approved 12
Approved 22.5
Approved 6.0
Approved 1
Approved 0
Rejected 0.65
Rejected foo19bar
Rejected ten
Rejected 39.4
Rejected s12
Rejected 0x600

Answer 2

为什么不只使用正则表达式，例如

if (preg_match('/^\d+(\.[05]0*)?$/', $rating)) echo "\"$rating\": Approved\n";

这个将允许在数字后跟随0，因此4.50和4.5都会通过。 如果您不希望这样，只需删除它的0*部分即可。

例如

$ratings = array(
"0.5",
"12"  ,
"22.5" , 
"6.0",
"1",
"0",
"4.5",
"3.500",
"1.00",
"0.65"  ,
"foo19bar"  ,
"ten"  ,
"39.4",
"s12",
"0x600"
    );
foreach ($ratings as $rating) {
    echo "\"$rating\": " . (preg_match('/^\d+(\.[05]0*)?$/', $rating) ? "Approved\n" : "Rejected\n");
}

输出：

"0.5": Approved
"12": Approved
"22.5": Approved
"6.0": Approved
"1": Approved
"0": Approved
"4.5": Approved
"3.500": Approved
"1.00": Approved
"0.65": Rejected
"foo19bar": Rejected
"ten": Rejected
"39.4": Rejected
"s12": Rejected
"0x600": Rejected