blocks|key|479116|text|$numbers+=+array(1,+9,+14,+53,+112,+725,+1001,+1200);
foreach($numbers+as+$number)+{
++++printf('%25d+=>+%25d'
++++++++++++,+$number
++++++++++++,+$number+-+$number+%25+pow(10,+floor(log10($number)))
++++++++++++);
++++echo+"\n";
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|479117|不幸的是，当$number为0时，这将非常失败，但它确实为正整数产生了预期的结果。而且这是一种仅限于数学的解决方案。|unstyled|479118|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>$numbers = array(1, 9, 14, 53, 112, 725, 1001, 1200);
foreach($numbers as $number) {
 printf('%d =&gt; %d'
 , $number
 , $number - $number % pow(10, floor(log10($number)))
 );
 echo "\n";
}
</code></pre>

Unfortunately this fails horribly when $number is 0, but it does produce the expected result for positive integers. And it is a math-only solution.

blocks|key|479154|text|这是一个纯粹的数学解决方案。如果您想要向上或向下取整，而不仅仅是向下取整，这也是一个更灵活的解决方案。它在0上有效:)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|479155|if($num+===+0)+return+0;
$digits+=+(int)(log10($num));
$num+=+(pow(10,+$digits))+*+floor($num/(pow(10,+$digits)));|code-block|syntax|javascript|479156|您可以将floor替换为round或ceil。实际上，如果您想四舍五入到最近的值，您可以进一步简化第三行。|offset|length|style|CODE|479157|$num+=+round($num,+-$digits);|479158|entityMap^0|0|0|4|5|C|5|I|4|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]|$I|V|J|W|K|L]|$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|P|$]]

Here's a pure math solution. This is also a more flexible solution if you ever wanted to round up or down, and not just down. And it works on 0 :)

<pre><code>if($num === 0) return 0;
$digits = (int)(log10($num));
$num = (pow(10, $digits)) * floor($num/(pow(10, $digits)));
</code></pre>

You could replace <code>floor</code> with <code>round</code> or <code>ceil</code>. Actually, if you wanted to round to the nearest, you could simplify the third line even more.

<pre><code>$num = round($num, -$digits);
</code></pre>

blocks|key|4172436|text|如果你真的想要一个数学解决方案，试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4172437|function+floorToFirst($int)+{
++++if+(0+===+$int)+return+0;

++++$nearest+=+pow(10,+floor(log($int,+10)));
++++return+floor($int+/+$nearest)+*+$nearest;
}|code-block|syntax|javascript|4172438|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If you do want to have a mathy solution, try this:

<pre><code>function floorToFirst($int) {
 if (0 === $int) return 0;

 $nearest = pow(10, floor(log($int, 10)));
 return floor($int / $nearest) * $nearest;
}
</code></pre>

blocks|key|2394834|text|如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2394835|$str+=+(string)$value;
echo+(int)($str[0]+.+str_repeat('0',+strlen($str)+-+1));|code-block|syntax|javascript|2394836|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Something like this:

<pre><code>$str = (string)$value;
echo (int)($str[0] . str_repeat('0', strlen($str) - 1));
</code></pre>

blocks|key|2098441|text|这完全不是数学问题，但我只会利用刺的长度……可能有一种更平滑的方法来处理它，但你可以用以下方法来完成它|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2098442|function+significant($number){
++++$digits+=+count($number);
++++if($digits+>=+2){
++++++++$newNumber+=+substr($number,0,1);
++++++++$digits--;
++++++++for($i+=+0;+$i+<+$digits;+$i%2B%2B){
++++++++++++$newNumber+=+$newNumber+.+"0";
++++++++}
++++}
++++return+$newNumber;
}|code-block|syntax|javascript|2098443|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

It's totally non-mathy, but I would just do this utilizing sting length... there's probably a smoother way to handle it but you could acomplish it with

<pre><code>function significant($number){
 $digits = count($number);
 if($digits &gt;= 2){
 $newNumber = substr($number,0,1);
 $digits--;
 for($i = 0; $i &lt; $digits; $i++){
 $newNumber = $newNumber . "0";
 }
 }
 return $newNumber;
}
</code></pre>

blocks|key|2394918|text|一个基于数学的替代方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2394919|$mod+=+pow(10,+intval(round(log10($value)+-+0.5)));+
$answer+=+((int)($value+/+$mod))+*+$mod;|code-block|syntax|javascript|2394920|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

A math based alternative:

<pre><code>$mod = pow(10, intval(round(log10($value) - 0.5))); 
$answer = ((int)($value / $mod)) * $mod;
</code></pre>

blocks|key|479215|text|我知道这是一个古老的帖子，但我在寻找如何解决这个问题的灵感时读到了它。这是我想出来的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|479216|class+Math
{
++++public+static+function+round($number,+$numberOfSigFigs+=+1)
++++{
++++++++//+If+the+number+is+0+return+0
++++++++if+($number+==+0)+{
++++++++++++return+0;
++++++++}

++++++++//+Deal+with+negative+numbers
++++++++if+($number+<+0)+{
++++++++++++$number+=+-$number;
++++++++++++return+-Math::sigFigRound($number,+$numberOfSigFigs);
++++++++}

++++++++return+Math::sigFigRound($number,+$numberOfSigFigs);
++++}

++++private+static+function+sigFigRound($number,+$numberOfSigFigs)
++++{
++++++++//+Log+the+number+passed
++++++++$log+=+log10($number);

++++++++//+Round+$log+down+to+determine+the+integer+part+of+the+log
++++++++$logIntegerPart+=+floor($log);

++++++++//+Subtract+the+integer+part+from+the+log+itself+to+determine+the+fractional+part+of+the+log
++++++++$logFractionalPart+=+$log+-+$logIntegerPart;

++++++++//+Calculate+the+value+of+10+raised+to+the+power+of+$logFractionalPart
++++++++$value+=+pow(10,+$logFractionalPart);

++++++++//+Round+$value+to+specified+number+of+significant+figures
++++++++$value+=+round($value,+$numberOfSigFigs+-+1);

++++++++//+Return+the+correct+value
++++++++return+$value+*+pow(10,+$logIntegerPart);+
++++}
}|code-block|syntax|javascript|479217|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I know this is an old thread but I read it when looking for inspiration on how to solve this problem. Here's what I came up with:

<pre><code>class Math
{
 public static function round($number, $numberOfSigFigs = 1)
 {
 // If the number is 0 return 0
 if ($number == 0) {
 return 0;
 }

 // Deal with negative numbers
 if ($number &lt; 0) {
 $number = -$number;
 return -Math::sigFigRound($number, $numberOfSigFigs);
 }

 return Math::sigFigRound($number, $numberOfSigFigs);
 }

 private static function sigFigRound($number, $numberOfSigFigs)
 {
 // Log the number passed
 $log = log10($number);

 // Round $log down to determine the integer part of the log
 $logIntegerPart = floor($log);

 // Subtract the integer part from the log itself to determine the fractional part of the log
 $logFractionalPart = $log - $logIntegerPart;

 // Calculate the value of 10 raised to the power of $logFractionalPart
 $value = pow(10, $logFractionalPart);

 // Round $value to specified number of significant figures
 $value = round($value, $numberOfSigFigs - 1);

 // Return the correct value
 return $value * pow(10, $logIntegerPart); 
 }
}
</code></pre>

blocks|key|479337|text|虽然这里的函数工作正常，但我需要对非常小的数字使用有效数字(将低价值加密货币与比特币进行比较)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|479338|在Format+number+to+N+significant+digits+in+PHP上，答案在某种程度上是有效的，尽管PHP用科学记数法显示的数字非常少，这使得一些人很难阅读它们。|offset|length|479339|​|479340|📷|atomic|479341|479342|我试过使用number_format，尽管它需要在小数后面加上特定的位数，这会破坏数字的“有效”部分(如果输入了一个集合数字)，有时还会返回0(对于小于集合数字的数字)。|479343|解决方案是修改函数以识别非常小的数字，然后对它们使用number_format+-将科学记数法数字的数量作为number_format的数字数量：|479344|function+roundRate($rate,+$digits)
{
++++$mod+=+pow(10,+intval(round(log10($rate))));
++++$mod+=+$mod+/+pow(10,+$digits);
++++$answer+=+((int)($rate+/+$mod))+*+$mod;
++++$small+=+strstr($answer,"-");
++++if($small)
++++{
++++++++$answer+=+number_format($answer,str_replace("-","",$small));
++++}
++++return+$answer;
}|code-block|syntax|javascript|479345|此函数保留有效数字，并以易于阅读的格式向每个人显示数字。(我知道，这对科学工作者来说不是最好的，甚至不是最一致的长度“漂亮”的数字，但总的来说，它是我们所需的最佳解决方案。)|479346|479347|479348|479349|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/37618679/format-number-to-n-significant-digits-in-php|1|IMAGE|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/bfd19eb801bbc8dce041186bf5590418.png|imageAlt|2|https://i.stack.imgur.com/OA0VL.png^0|0|1|18|0|0|0|0|1|1|0|0|0|0|0|0|0|0|1|2|0|0^^$0|@$1|2|3|4|5|6|7|1F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1G|8|@]|9|@$D|1H|E|1I|1|1J]]|A|$]]|$1|F|3|G|5|6|7|1K|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|1L|8|@]|9|@$D|1M|E|1N|1|1O]]|A|$]]|$1|K|3|G|5|6|7|1P|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|1Q|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|1R|8|@]|9|@]|A|$]]|$1|P|3|Q|5|R|7|1S|8|@]|9|@]|A|$S|T]]|$1|U|3|V|5|6|7|1T|8|@]|9|@]|A|$]]|$1|W|3|G|5|6|7|1U|8|@]|9|@]|A|$]]|$1|X|3|I|5|J|7|1V|8|@]|9|@$D|1W|E|1X|1|1Y]]|A|$]]|$1|Y|3|G|5|6|7|1Z|8|@]|9|@]|A|$]]|$1|Z|3|-4|5|6|7|20|8|@]|9|@]|A|$]]]|10|$11|$5|12|13|14|A|$15|16]]|17|$5|18|13|19|A|$1A|1B|1C|-4]]|1D|$5|18|13|19|A|$1A|1E|1C|-4]]]]

While the functions here worked, I needed significant digits for very small numbers (comparing low-value cryptocurrency to bitcoin).

The answer at <a href="https://stackoverflow.com/questions/37618679/format-number-to-n-significant-digits-in-php">Format number to N significant digits in PHP</a> worked, somewhat, though very small numbers are displayed by PHP in scientific notation, which makes them hard for some people to read.

<a href="https://i.stack.imgur.com/Kl2cY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Kl2cY.png" alt="enter image description here"></a>

I tried using number_format, though that needs a specific number of digits after the decimal, which broke the 'significant' part of the number (if a set number is entered) and sometimes returned 0 (for numbers smaller than the set number). 

The solution was to modify the function to identify really small numbers and then use number_format on them - taking the number of scientific notation digits as the number of digits for number_format:

<pre><code>function roundRate($rate, $digits)
{
 $mod = pow(10, intval(round(log10($rate))));
 $mod = $mod / pow(10, $digits);
 $answer = ((int)($rate / $mod)) * $mod;
 $small = strstr($answer,"-");
 if($small)
 {
 $answer = number_format($answer,str_replace("-","",$small));
 }
 return $answer;
}
</code></pre>

This function retains the significant digits as well as presents the numbers in easy-to-read format for everyone. (I know, it is not the best for scientific people nor even the most consistently length 'pretty' looking numbers, but it is overall the best solution for what we needed.)

<a href="https://i.stack.imgur.com/OA0VL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OA0VL.png" alt="enter image description here"></a>

Is there any slick way to round down to the nearest significant figure in php?

So:

<pre><code>0-&gt;0
9-&gt;9
10-&gt;10
17-&gt;10
77-&gt;70
114-&gt;100
745-&gt;700
1200-&gt;1000
</code></pre>

?

How to round down to the nearest significant figure in php

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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在php中有没有什么灵巧的方法可以四舍五入到最接近的有效数字？所以：0->09->910->1017->1077->70114->100745->7001200->1000

问如何在php中向下舍入到最接近的有效数字
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