php number_format() 导致错误“遇到格式不正确的数值”
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number_format() causes error "A non well formed numeric value encountered"
提问by Tu Hoang
I am using number_format to round floats to only 2 decimal digits. The problem is that some of my inputs don't have more than 2 decimals digits to begin with. So the code:
我正在使用 number_format 将浮点数四舍五入到只有 2 个十进制数字。问题是我的一些输入开始时没有超过 2 个小数位。所以代码：
number_format($value, 2)
Instead of peacefully adding 0 in case it doesn't have enough decimal digits, it raises errors inside Apache log and that's not desirable.
如果没有足够的十进制数字，它不是和平地添加 0，而是在 Apache 日志中引发错误，这是不可取的。
So number_format(2.1, 2)or number_format(0, 2)will raise error in Apache log.
所以number_format(2.1, 2)ornumber_format(0, 2)会在 Apache 日志中引发错误。
[Thu Jun 30 17:18:04 2011] [error] [client 127.0.0.1] PHP Notice: A non well formed numeric value encountered in /home/tahoang/Desktop/Projects/weatherData/weatherData.php on line 41
[Thu Jun 30 17:18:04 2011] [error] [client 127.0.0.1] PHP 注意：在第 41 行的 /home/tahoang/Desktop/Projects/weatherData/weatherData.php 中遇到格式不正确的数值
How to fix this?
如何解决这个问题？
回答by Emre Yazici
Try type casting first parameter of number_format()to float:
尝试将number_format() 的第一个参数转换为浮点数：
$format = number_format((float)0, 2);
or
或者
$format = number_format(floatval(0), 2);
回答by user2659841
Try to replace decimal point and after that cast to float.
尝试替换小数点，然后转换为浮点数。
var_dump((float)number_format((float)str_replace(",", ".", "20,5"), 2, ".", ""));
result: float(20.5);
Without replacing:
无需更换：
var_dump((float)number_format(floatval("20,5"), 2, ".", ""));
result: float(20);
var_dump((float)number_format((float) "20,5", 2, ".", ""));
result: float(20);
回答by Leo Chashchin
When doing calculations, use
进行计算时，使用
number_format($value, 2, ".", "")
number_format($value, 2, ".", "")
And if you would like to display numbers with .00 at the end (like "50.50" not "50.5") then you would use
如果您想在末尾显示 .00 的数字（例如“50.50”而不是“50.5”），那么您可以使用
numer_format($value, 2)
numer_format($value, 2)
回答by Matheno
I used this:
我用过这个：
str_replace(array(".", ","), array(",", "."), $value)
Maybe it'll help someone out.
也许它会帮助某人。
回答by Elyor
Function is recognizing it as a string.
函数将其识别为字符串。
Convert to number by adding zero:
通过添加零转换为数字：
number_format($NUMBER+0)
