php 浮点数后最多显示 2 位数字……仅当它是具有 2 个以上浮点数的浮点数时
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show 2 digits max after floating point ... only if it is a float number with more than 2 float digits
提问by max
in my app i do some math and the result can be float or int
在我的应用程序中，我做了一些数学运算，结果可以是 float 或 int
i want to show the final result with two digit after the decimal point max ... if result is a float number
我想显示小数点后两位最大的最终结果......如果结果是一个浮点数
there are two options to do this
有两种选择可以做到这一点
number_format($final ,2);
and
和
sprintf ("%.2f", $final );
but problem is ... if my final result is a int like 25i end up with
但问题是......如果我的最终结果是一个像25我一样的 int
25.00
or if final result is some thing like 12.3it gives me
或者如果最终结果是12.3它给我的东西
12.30
and i dont want that
我不想要那个
is there any way to format a number to show 2 digits after float point ONLY IF it's a float number with more than 2 digits after decimal point ? or should i do some checking before formatting my number ?
有没有办法格式化一个数字以在浮点数后显示 2 位数字，前提是它是一个小数点后超过 2 位数字的浮点数？或者我应该在格式化我的号码之前做一些检查？
回答by user4035
<?php
$number = 25;
print round($number, 2);
print "\n";
$number = 25.3;
print round($number, 2);
print "\n";
$number = 25.33;
print round($number, 2);
prints:
印刷：
25
25.3
25.33
回答by rnbguy
I think there is no such short bypass for it.
Manually check if it has 2 or more digits after decimal.
我认为没有这么短的旁路。
手动检查小数点后是否有 2 位或更多位。
How to know if it has less one or zero digits after decimal ? Just multiply with 10 and check if it is an integer. If it is, print the number as it is. If it's now use '%.2f' to print it.
如何知道小数点后是少一位还是零位？只需乘以 10 并检查它是否为整数。如果是，则按原样打印数字。如果现在使用 '%.2f' 打印它。
回答by SmootQ
You have to add an if condition, if it has more than 0 digits after the point. I don't see any other solution.
你必须添加一个 if 条件，如果它在点后有超过 0 个数字。我没有看到任何其他解决方案。
A simple and fast way to do this.
一个简单而快速的方法来做到这一点。
$final=3.40;
$decimalNbr= strlen(substr(strrchr($final, "."), 1));
$final = number_format($final,(is_float($final) ? (($decimalNbr>2) ? 2 : $decimalNbr) : 0));
echo $final;
Keep in mind too, that I added another decimal digits count before using the number_format.
还要记住，我在使用 number_format 之前添加了另一个十进制数字计数。
回答by bitWorking
I found another option. Cast to floatto strip the trailing zeros:
我找到了另一种选择。强制转换float为去除尾随零：
echo (float)sprintf("%.2f", $final);
// or
echo (float)number_format($final ,2);
But these functions seems to round the number just like round:
但这些函数似乎四舍五入数字就像round：
echo sprintf("%.2f", 12.556); // 12.56
echo number_format(12.556, 2); // 12.56
So if you don't want this behaviour use this:
因此，如果您不希望这种行为，请使用以下命令：
$final = 12.556;
echo (int)(($final*100))/100; // 12.55
echo (int)((12*100))/100; // 12
echo (int)((12.3*100))/100; // 12.3
echo (int)((12.34567*100))/100; // 12.34
回答by Santosh Kumar Singh
You can use this example
你可以使用这个例子
function num_format($number,$precision=0)
{
$precision = ($precision == 0 ? 1 : $precision);
$pow = pow(10, $precision);
$value = (int)((trim($number)*$pow))/$pow;
return number_format($value,$precision);
}
echo num_format(71730116.048758);
//Output
//71,730,116.04
回答by KRISHNA MK
Try below function to maintain a constant 2 digits after decimal point
尝试下面的函数来保持小数点后的 2 位数不变
in case you want to change the number of digits change AFTER_DESIMAL_DIGITS value
如果您想更改位数更改 AFTER_DESIMAL_DIGITS 值
function decimal_digit_maintainer($value){
define("AFTER_DESIMAL_DIGITS ",2);
$value_array = explode('.',$value);
if(isset($value_array[1])){
$value_array[1] = str_pad($value_array[1],AFTER_DESIMAL_DIGITS,'0',STR_PAD_RIGHT);
}
return implode('.',$value_array);
}
回答by Anthony Sterling
I don't like this.
我不喜欢这个。
<?php
function format($number) {
return preg_replace(
'~\.[0-9]0+$~',
null,
sprintf('%.2f', $number)
);
}
echo format(23), PHP_EOL; //23
echo format(23.3), PHP_EOL; //23
echo format(23.33), PHP_EOL; //23.33
回答by Ivan0x32
Yes, you should do some checking. For example check if ceil($number) > floor($number);If you need specificly two digits, that is going to take more effort.
是的，你应该做一些检查。例如检查ceil($number) > floor($number);如果您需要特定的两位数，那将需要更多的努力。
