Question

我认为这非常简单，但是在PHP中，要确定字符串是否仅包含整数或浮点数（整数或半数）是很棘手的。

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

很难（仅）检测字符串中的浮点数：
- is_float()显然拒绝字符串（例如is_float("2.5") = false）
- 通过类型转换为浮点数可以使字符串包含字母（例如is_float((float)"wat2.5") = true）
确定浮点数是完整还是一半也很奇怪：
- 模运算符（%）由于某种原因会自动转换为int，因此您需要使用fmod()（例如，PHP中的10.33 % 0.5返回0 5.4）

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：

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

为什么不仅仅使用正则表达式，例如

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

此数字将允许在数字后跟0，因此4.50和4.5都会通过。如果您不希望这样做，只需删除其中的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

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？

2 个答案: