如何使number_format（）不圓人數達到

Question

我有這個數：使用number_format($double, 2, ',', ' ')

$double = '21.188624'; 

後我得到：

21,19 

但我想要的是：

21,18 

任何ideea我該如何做這項工作？

謝謝。

Answer 1

number_format總是會做到這一點，你唯一的辦法就是給它不同的東西：

$number = intval(($number*100))/100; 

或者：

$number = floor(($number*100))/100; 

Answer 2

floor($double*100)/100 

Answer 3

使用此功能：

function number_format_unlimited_precision($number,$decimal = '.') 
{ 
    $broken_number = explode($decimal,$number); 
    return number_format($broken_number[0]).$decimal.$broken_number[1]); 
} 

Answer 4

$finalCommishParts = explode('.',$commission); 
$commisshSuffix = (isset($finalCommishParts[1])?substr($finalCommishParts[1],0,2):'00'); 
$finalCommish = $finalCommishParts[0].'.'.$commisshSuffix; 

Answer 5

public function numberFormatPrecision($number, $separator = '.', $format = 2){ 

    $response = ''; 
    $brokenNumber = explode($separator, $number); 
    $response = $brokenNumber[0] . $separator; 
    $brokenBackNumber = str_split($brokenNumber[1]); 

    if($format < count($brokenBackNumber)){ 

     for($i = 1; $i <= $format; $i++) 
      $response .= $brokenBackNumber[$i]; 
    } 

    return $response; 
} 

Answer 6

如果你不在乎什麼來小數點後，您可以施放float爲int避免四捨五入：

$float = 2.8; 
echo (int) $float; // outputs '2' 

Answer 7

我用這個函數：

function cutNum($num, $precision = 2){ 
    return floor($num).substr($num-floor($num),1,$precision+1); 
} 

floor($num)是整數部分，$num-floor($num)是小數（如0.2345）。

所以substr($num-floor($num),1,$precision+1)得到「.23」（注意這一點也取決於）。

使用示例：

cutNum(1.2345) //returns 1.23 
cutNum(1.2345,3) //returns 1.234 
cutNum(1.2345,10) //returns 1.2345 
cutNum(1.2345,0) //returns 1. (!don't use with zero as second argument: use floor instead!) 

Answer 8

**Number without round**   

    $double = '21.188624'; 
    echo intval($double).'.'.substr(end(explode('.',$double)),0,2); 

**Output** 21.18 

Answer 9

我知道，這一個老問題，但它仍然實際:)。

這個功能怎麼樣？

function numberFormatPrecision($number, $precision = 2, $separator = '.') 
{ 
    $numberParts = explode($separator, $number); 
    $response = $numberParts[0]; 
    if(count($numberParts)>1){ 
     $response .= $separator; 
     $response .= substr($numberParts[1], 0, $precision); 
    } 
    return $response; 
} 

用法：

// numbers test 
numberFormatPrecision(19, 2, '.'); // expected 19 return 19 
numberFormatPrecision(19.1, 2, '.'); //expected 19.1 return 19.1 
numberFormatPrecision(19.123456, 2, '.'); //expected 19.12 return 19.12 

// negative numbers test 
numberFormatPrecision(-19, 2, '.'); // expected -19 return -19 
numberFormatPrecision(-19.1, 2, '.'); //expected -19.1 return -19.1 
numberFormatPrecision(-19.123456, 2, '.'); //expected -19.12 return -19.12 

// precision test 
numberFormatPrecision(-19.123456, 4, '.'); //expected -19.1234 return -19.1234 

// separator test 
numberFormatPrecision('-19,123456', 3, ','); //expected -19,123 return -19,123 -- comma separator 

Answer 10

如果你有小float值，您可以使用number_format功能這種方式。

$number = 21.23; 

echo number_format($number, 2, '.', ',')); // 21.23 

如果你有你有長十進制數，然後也將格式號這樣

$number = 201541.23; 

echo number_format($number, 2, '.', ',')); // 201,541.23 

Answer 11

使用PHP的本地函數bcdiv。

function numberFormat($number, $decimals = 2, $sep = ".", $k = ","){ 
    $number = bcdiv($number, 1, $decimals); // Truncate decimals without rounding 
    return number_format($number, $decimals, $sep, $k); // Format the number 
} 

詳情請參閱this answer。