PHP - 强制整数转换浮点三位小数 [英] PHP - Force integer conversion to float with three decimals

问题描述

我正在尝试将一大串数字转换为+/- NNN.NNN的特定格式。
所以，如果我有这些数字：

$ $ p1 $ array1 = array（-1.23,0.3222,10,5.54） ;

我希望他们的最终格式为

  $ numbers2 = array（-001.023，+000.322，+0.10.000，+005.054）; 
  

我试图这样做：

  foreach（$ numbers1 as $ n）{
 $ fnum = abs（number_format（（float）$ n，3，'。'，''））; 
 if（$ fnum> 0&& $ fnum< 10）{
 $ fnum ='00'。$ fnum; 
} else if（$ fnum> = 10&& $ fnum< 100）{
 $ fnum ='0'。$ fnum; 
} 
 if（$ n> 0）$ fnum ='+'。$ fnum; 
 if（$ n <0）$ fnum =' - '。$ fnum; 
 if（$ n == 0）$ fnum ='+000.000'; 
 $ numbers2 [] = $ fnum; 
  

}

这是错的，我只是不知道要用什么来实现它。 解决方案

尝试使用 sprintf（）

  $ numbers2 [] = sprintf（％+ 07.3f，$ fnum）; 
  

I am trying to convert a big array of numbers to a specific format which is +/-NNN.NNN. So, if I have these numbers:

$numbers1 = array(-1.23, 0.3222, 10, 5.54);

I want their final format to be

$numbers2 = array(-001.023, +000.322, +010.000, +005.054);

I am trying to do it like this:

foreach ($numbers1 as $n) {
 $fnum = abs(number_format((float)$n, 3, '.', ''));
 if ($fnum>0 && $fnum<10) { 
     $fnum = '00'.$fnum;
 } else if ($fnum >= 10 && $fnum<100) {
     $fnum = '0'.$fnum;
 }
 if ($n>0) $fnum = '+'.$fnum;
 if ($n<0) $fnum = '-'.$fnum;
 if ($n == 0) $fnum = '+000.000';
 $numbers2[] = $fnum;

}

This is wrong and I just don't know what to use in order to achieve it.

解决方案

Try using sprintf()

$numbers2[] = sprintf("%+07.3f", $fnum);

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