PHP - floatval 削减小数点后的位置？

发布于 2024-12-07 01:11:17 字数 453 浏览 1 评论 0原文

我有这个 PHP 代码：

<?php
    $float = "1,99";
    echo "<p>$float<br>";
    $float = floatval($float); // Without this line number_format throws a notice "A non well formed numeric value encountered" / (float) $float leads to the same output
    $val = number_format($float, 2,'.', ',');
    echo "$float</p>";
?>

为什么它返回 1？别明白。

并且：是的，将 1,99 转换为 1,99 是有意义的；-)

感谢您的建议...

原文

I have this PHP code:

<?php
    $float = "1,99";
    echo "<p>$float<br>";
    $float = floatval($float); // Without this line number_format throws a notice "A non well formed numeric value encountered" / (float) $float leads to the same output
    $val = number_format($float, 2,'.', ',');
    echo "$float</p>";
?>

Why does it return 1? Don't get that.

And: yes, there is a sense in converting 1,99 to 1,99 ;-)

Thanks for advise...

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一页 2024-12-14 01:11:17

问题是，PHP 无法将 1,99 中的 , 识别为小数点分隔符。 float 类型定义为具有以下正式定义：

LNUM          [0-9]+
DNUM          ([0-9]*[\.]{LNUM}) | ({LNUM}[\.][0-9]*)
EXPONENT_DNUM [+-]?(({LNUM} | {DNUM}) [eE][+-]? {LNUM})

这意味着它只接受.作为小数点分隔符。事实上，这与 number_format 对无效数据类型发出警告的原因相同，因为它无法在内部将 1,99 转换为 float。

以下应该有效：

$float = "1,99";
echo "<p>$float<br>";
$val = number_format(str_replace(',', '.', $float), 2,'.', ',');
echo "$float</p>";

The problem is, that PHP does not recognize the , in 1,99 as a decimal separator. The float type is defined as having the following formal definition:

LNUM          [0-9]+
DNUM          ([0-9]*[\.]{LNUM}) | ({LNUM}[\.][0-9]*)
EXPONENT_DNUM [+-]?(({LNUM} | {DNUM}) [eE][+-]? {LNUM})

That means it'll only accept . as a decimal separator. That's in fact the same reason why number_format throws a warning on an invalid datatype because it cannot convert 1,99 to a float internally.

The following should work:

$float = "1,99";
echo "<p>$float<br>";
$val = number_format(str_replace(',', '.', $float), 2,'.', ',');
echo "$float</p>";

回复收藏 0 原文

阳光①夏 2024-12-14 01:11:17

floatval 将逗号 (,) 识别为字符而不是数字，因此它会截断其后面的所有内容。在本例中，即 99。请使用点 (.) 而不是逗号 (,)，这样可能会起作用。

floatval 示例（来源：http://php.net/manual/en/function.floatval。 php):

<?php
$var = '122.34343The';
$float_value_of_var = floatval($var);
echo $float_value_of_var; // 122.34343
?>

floatval recognizes the comma (,) as a character and not as a number, so it cuts off everything that comes after it. In this case, that's the 99. Please use a dot (.) instead of a comma (,) and it will probably work.

Example floatval (source: http://php.net/manual/en/function.floatval.php):

<?php
$var = '122.34343The';
$float_value_of_var = floatval($var);
echo $float_value_of_var; // 122.34343
?>

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二货你真萌 2024-12-14 01:11:17

1,99 不是有效的 php 浮点数。问题是你的逗号（，）。在 PHP 中，必须使用点 (.) 作为浮点分隔符。

<?php
    $float = "1.99";
    echo "<p>$float<br>";
    $float = (float)$float; 
    $val = number_format($float, 2,'.', ',');
    echo "$float</p>";
?>

1,99 is not a valid php float. The issue is your comma (,). In PHP you have to use dot (.) as floating point separator.

<?php
    $float = "1.99";
    echo "<p>$float<br>";
    $float = (float)$float; 
    $val = number_format($float, 2,'.', ',');
    echo "$float</p>";
?>

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