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php圆形浮点数

[英]php round float number

I have read the documentation and have clear idea of round but i didn't find useful information to solve the problem. 我已经阅读了文档并且对解决方法有清楚的了解,但是我没有找到解决问题的有用信息。

The problem is i have float number which is let say 1.09 and i want to display it 2 instead of 1. if we use round function it display 1. Help me solve this problem. 问题是我有浮点数,可以说是1.09,我想显示2而不是1。如果我们使用取整功能,则显示1。请帮助我解决此问题。

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$TotalPaidRemaining=1090;
$monthly_installments=1000;
$MakingNumberOfMonths=$TotalPaidRemaining/$monthly_installments;
echo round($MakingNumberOfMonths, 0, PHP_ROUND_HALF_UP);// it display 1. i want it display 2..

What i want is if the value after decimal point is greater than 0. For example 0.01. 我想要的是如果小数点后的值大于0。例如0.01。 I want to consider it as 1. 我想将其视为1。

Hope i am clear at my question. 希望我清楚我的问题。

Use the ceil() function instead. 请改用ceil()函数。

$number = ceil(0.1); // $number will be 1

From the documentation : 从文档中:

Returns the next highest integer value by rounding up value if necessary. 必要时通过舍入值返回下一个最大的整数值。

You can use the ceil() php function instead of round(). 您可以使用ceil()php函数代替round()。 It will round up your values. 它将丰富您的价值观。 Docs: http://php.net/manual/en/function.ceil.php Example: ceil(1.09); // return 2 文件: http : ceil(1.09); // return 2范例: ceil(1.09); // return 2 ceil(1.09); // return 2

You could use ceil($yourNumber) , which will round the number to its next higher integer. 您可以使用ceil($yourNumber) ,它将数字四舍五入到下一个更高的整数。

Or you could use round($yourNumber + 0.499999999999999) . 或者您可以使用round($yourNumber + 0.499999999999999)

Or you could use floor($yourNumber + 1) , which rounds the number to its previous highest integer. 或者,您可以使用floor($yourNumber + 1) ,将数字四舍五入到其先前的最高整数。

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