检查一个数字是否在PHP中浮动

问题描述

这真的很奇怪

  $ rewardAmt = $ amt; 
 if（is_float（$ rewardAmt））{
 print_r（is float）; die; 
} else {
 print_r（is not float）;死; 
} 
  

$ amt的值为0.01。但它正在进入其他条件。所以我做了一个$ amt var_dump。它说的字符串（4）
所以我决定改型$ amt

  $ rewardAmt =（float）$ amt; 
  

但是这个问题是即使$ amt的值是1，它仍然会被转换为浮动并进入条件，这不应该发生。有没有其他方法可以做到这一点？谢谢 > $ rewardAmt = $ amt + 0;


$ rewardAmt应该转换为数字。

This is really weird. I have this piece of code.

$rewardAmt = $amt;
if(is_float($rewardAmt)){
      print_r("is float");die;
} else {
      print_r("is not float"); die;
}

value of $amt is 0.01. But it is going into else condition. So I did a var_dump of $amt. it says string(4)So I decided to typecast $amt

   $rewardAmt = (float)$amt;

But the problem with this is even if the value of $amt is 1, it still gets typecast to float and goes into if condition, which shouldn't happen. Is there any other way to do this ? Thanks

解决方案

If you change the first line to

$rewardAmt = $amt+0;

$rewardAmt should be cast to a number.

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