blocks|key|2506675|text|也许：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2506676|$num+=+245354;
$numlength+=+strlen((string)$num);|code-block|syntax|javascript|2506677|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|$]]

Maybe:

<pre><code>$num = 245354;
$numlength = strlen((string)$num);
</code></pre>

blocks|key|37817|text|被接受的答案不适用于大数字。要计算任意数字的长度，更好的方法是通过检查0来调用floor(log10($num)+%2B+1)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|37818|$num+=+12357;
echo+$num+!==+0+?+floor(log10($num)+%2B+1)+:+1;+//+prints+5|code-block|syntax|javascript|37819|它有多个优点。它更快，你不需要进行类型转换，它适用于大数字，它适用于不同的数字系统，如bin，hex，oct。|37820|该方程以10为底进行对数运算，然后将其作为底数并加1。|37821|这个解决方案可以独立地在底上工作，所以如果你想计算二进制或十六进制的长度，只需更改对数的底。|37822|Working+fiddle|37823|entityMap|0|LINK|mutability|MUTABLE|url|http://sandbox.onlinephpfunctions.com/code/767bea2461fe3cbbbef3ca4886f1f2dcc919653b^0|Z|1|13|M|0|0|0|0|0|0|E|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|16|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|17|8|@]|D|@]|E|$]]|$1|O|3|P|5|6|7|18|8|@]|D|@]|E|$]]|$1|Q|3|R|5|6|7|19|8|@]|D|@$9|1A|A|1B|1|1C]]|E|$]]|$1|S|3|-4|5|6|7|1D|8|@]|D|@]|E|$]]]|T|$U|$5|V|W|X|E|$Y|Z]]]]

Accepted answer won't work with the big numbers. The better way to calculate the length of any number is to invoke <code>floor(log10($num) + 1)</code> with a check for <code>0</code>. 

<pre><code>$num = 12357;
echo $num !== 0 ? floor(log10($num) + 1) : 1; // prints 5
</code></pre>

It has multiple advantages. It's faster, you don't do the casting of types, it works on big numbers, it works with different number systems like bin, hex, oct.

The equation does the logarithm with base 10 then makes the floor of it and adds 1. 

This solution can work independently on the base, so if you want to calculate the length of binary or hex just change the base of the logarithm.

<a href="http://sandbox.onlinephpfunctions.com/code/767bea2461fe3cbbbef3ca4886f1f2dcc919653b" rel="noreferrer">Working fiddle</a>

blocks|key|2717213|text|可接受的解决方案在评估负数时会出现问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2717214|它与正数一起工作：|2717215|$num+=+245354;
$numlength+=+strlen((string)$num);
//+Result:+6|code-block|syntax|javascript|2717216|但如果是负数，则将(-)添加到计数：|2717217|$num+=+-245354;
$numlength+=+strlen((string)$num);
//+Result:+7|2717218|快速解决方法：|2717219|$num+=+-245354;
$numlength+=+strlen((string)abs($num));
//+Result:+6|2717220|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|V|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|W|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

The accepted solution presents a problem when evaluating negative numbers.

It works with a positive number:

<pre><code>$num = 245354;
$numlength = strlen((string)$num);
// Result: 6
</code></pre>

But with a negative number, the (-) is added to the count:

<pre><code>$num = -245354;
$numlength = strlen((string)$num);
// Result: 7
</code></pre>

Quick workaround:

<pre><code>$num = -245354;
$numlength = strlen((string)abs($num));
// Result: 6
</code></pre>

blocks|key|37691|text|更优雅的方式:)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|37692|ceil(log10($num));|code-block|syntax|javascript|37693|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|$]]

More elegant way :) 

<pre><code>ceil(log10($num));
</code></pre>

blocks|key|38771|text|你也可以使用一些基本的数学！|type|unstyled|depth|inlineStyleRanges|entityRanges|data|38772|$digits+=+(int)(log($num,10)%2B1)+

<?php
++$num+=+123;
++$num2+=+1234;
++$num3+=+12345;

++function+digits($num){
++++return+(int)+(log($num,+10)+%2B+1);
++}

++echo+"\n+$num:+"+.+digits($num);++//+123:+3
++echo+"\n+$num2:"+.+digits($num2);+//+1234:+4
++echo+"\n+$num3:"+.+digits($num3);+//+12345:+5+
++echo+"\n";|code-block|syntax|javascript|38773|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|$]]

You could also use some basic math!

<pre><code>$digits = (int)(log($num,10)+1) 

&lt;?php
 $num = 123;
 $num2 = 1234;
 $num3 = 12345;

 function digits($num){
 return (int) (log($num, 10) + 1);
 }

 echo "\n $num: " . digits($num); // 123: 3
 echo "\n $num2:" . digits($num2); // 1234: 4
 echo "\n $num3:" . digits($num3); // 12345: 5 
 echo "\n";
</code></pre>

blocks|key|2717260|text|另一种计算数字长度的方法是将数字的整数部分除以10，直到它变成0。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2717261|示例：|2717262|2021/10+=+202.1+
202/10+=+20.2+
20/10+=+2+
2/10+=+0.2|code-block|syntax|javascript|2717263|代码：|2717264|function+numberGetLength($number)+{
+++++$count+=+0;
+++++while+(intval($number)+>+0)+{
++++++++$number+=+intval($number)+/+10;
++++++++$count+%2B=+1;
+++++}
+++++return+$count
}|2717265|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Another way to find out the length of a number in digits would be to divide the integer part of the number to 10 until it becomes 0.
Example:
<pre><code>2021/10 = 202.1 
202/10 = 20.2 
20/10 = 2 
2/10 = 0.2
</code></pre>
Code:
<pre><code>function numberGetLength($number) {
 $count = 0;
 while (intval($number) &gt; 0) {
 $number = intval($number) / 10;
 $count += 1;
 }
 return $count
}
</code></pre>

blocks|key|2506783|text|仅仅使用某些版本的(int)(log($num,10)%2B1)会失败10,100,1000等等。它将数字10算作1位数，100算作两位数，等等。它也会在0或任何负数时失败。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2506784|如果必须使用数学(且数字为非负数)，请使用：|2506785|$numlength+=+(int)(log($num%2B1,+10)%2B1);|2506786|或者用于计算正数或负数的位数的数学解决方案：|2506787|$numlength+=+($num>=0)+?+(int)(log($num%2B1,+10)%2B1)+:+(int)(log(1-$num,+10)%2B1);|2506788|但strlen解决方案在PHP中的速度也差不多。|2506789|entityMap^0|9|L|0|0|0|12|0|0|0|25|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|U|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|Y|8|@]|D|@]|E|$]]|$1|L|3|M|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|N|3|O|5|6|7|12|8|@]|D|@]|E|$]]|$1|P|3|-4|5|6|7|13|8|@]|D|@]|E|$]]]|Q|$]]

Just using some version of <code>(int)(log($num,10)+1)</code> fails for 10, 100, 1000, etc. It counts the number 10 as 1 digit, 100 as two digits, etc. It also fails with 0 or any negative number.
 If you must use math (and the number is non-negative), use:
 <code>$numlength = (int)(log($num+1, 10)+1);</code>
 Or for a math solution that counts the digits in positive OR negative numbers:
 <code>$numlength = ($num&gt;=0) ? (int)(log($num+1, 10)+1) : (int)(log(1-$num, 10)+1);</code>
 But the strlen solution is just about as fast in PHP.

blocks|key|2717182|text|以下函数适用于整数或浮点数(适用于PHP7%2B)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2717183|function+digitsCount($number):+int
{
++++$number+=+abs($number);
++++$numberParts+=+explode(".",+$number);

++++return
++++++++strlen($numberParts[0])+%2B
++++++++(strlen($numberParts[1]+??+0));
}|code-block|syntax|javascript|2717184|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|$]]

The following function work for either integers or floats (works with PHP7+):
<pre><code>function digitsCount($number): int
{
 $number = abs($number);
 $numberParts = explode(&quot;.&quot;, $number);

 return
 strlen($numberParts[0]) +
 (strlen($numberParts[1] ?? 0));
}
</code></pre>

blocks|key|2717199|text|在PHP中，类型是松散设置和猜测的，如果你想看到一个字符串，如果它是一个整数，浮点数和(我没有尝试过)+bool，那么@Gorjunav是最正确的答案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2717200|将变量重置为字符串|2717201|$stringNum+=+(string)+$num;|code-block|syntax|javascript|2717202|然后你可以用它去任何你想要的字符串！反之亦然，用于将字符串更改为int|2717203|$number+=+(int)+$stringNum;|2717204|以此类推。|2717205|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

In PHP types are loosely set and guessed, if you want to see something as a string if it is an integer, float, and (i have not tried this) bool then @Gorjunav is the most correct answer.

Reset the variable as a string

<pre><code>$stringNum = (string) $num;
</code></pre>

Then you can go anything string related you want with it! And vice-versa for changing a string to an int 

<pre><code>$number = (int) $stringNum;
</code></pre>

and so on...

blocks|key|37884|text|仅计算整数值|type|unstyled|depth|inlineStyleRanges|entityRanges|data|37885|++`<?php
++++++++$n1+=12345;
++++++++$n2+=123454.55;
++++++++$n3+=12345564.557;
++++++++echo+"The+Number+you+Type:+".$n1." ";
++++++++$count+=+0;+
++++++++while+($n1+!=+0)++
++++++++{+
++++++++++++$n1+=+$n1+/+10;
++++++++++++$n1+=+intval($n1);
++++++++++++%2B%2B$count;+
++++++++}+
++++++++echo+"The+Digit+in+a+Number:+".$count;
++++}
++++?>`|code-block|syntax|javascript|37886|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|$]]

count only integer value

<pre><code> `&lt;?php
 $n1 =12345;
 $n2 =123454.55;
 $n3 =12345564.557;
 echo "The Number you Type: ".$n1."&lt;br&gt;";
 $count = 0; 
 while ($n1 != 0) 
 { 
 $n1 = $n1 / 10;
 $n1 = intval($n1);
 ++$count; 
 } 
 echo "The Digit in a Number: ".$count;
 }
 ?&gt;`
</code></pre>

blocks|key|37418|text|echo+strlen((string)+abs($num));+//+using+**abs**+it'll+work+with+negative+integers+as+well++|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|37419|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>echo strlen((string) abs($num)); // using **abs** it'll work with negative integers as well 
</code></pre>

blocks|key|2506872|text|在PHP+4.4.9+-+8.0.0中测试|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2506873|$array+=+array(-1,+0,+-0,+1,+4,+9,+10,+-10,+20,+-20,+100,+-100);
foreach(+$array+as+$key+=>+$num+){
++++echo+$key."\t{$num}\t=>\t".($num+!==+0+?+floor(log10(abs($num))+%2B+1)+:+1)."\n";
}
/*+Output:
0+++-1++=>++1
1+++0+++=>++1
2+++0+++=>++1
3+++1+++=>++1
4+++4+++=>++1
5+++9+++=>++1
6+++10++=>++2
7+++-10+=>++2
8+++20++=>++2
9+++-20+=>++2
10++100+=>++3
11++-100++++=>++3
*/|code-block|syntax|javascript|2506874|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|$]]

Tested in PHP 4.4.9 - 8.0.0
<pre><code>$array = array(-1, 0, -0, 1, 4, 9, 10, -10, 20, -20, 100, -100);
foreach( $array as $key =&gt; $num ){
 echo $key.&quot;\t{$num}\t=&gt;\t&quot;.($num !== 0 ? floor(log10(abs($num)) + 1) : 1).&quot;\n&quot;;
}
/* Output:
0 -1 =&gt; 1
1 0 =&gt; 1
2 0 =&gt; 1
3 1 =&gt; 1
4 4 =&gt; 1
5 9 =&gt; 1
6 10 =&gt; 2
7 -10 =&gt; 2
8 20 =&gt; 2
9 -20 =&gt; 2
10 100 =&gt; 3
11 -100 =&gt; 3
*/
</code></pre>

I would like to ask how I can get the length of digits in an Integer. For example:

<pre><code>$num = 245354;
$numlength = mb_strlen($num);
</code></pre>

<code>$numlength</code> should be 6 in this example. Somehow I can't manage it to work?

Thanks

EDIT: The example code above --^ and its respective method <code>mb_strlen();</code> works just fine.

PHP - Get length of digits in a number

我想问一下怎样才能得到Integer中的数字长度。例如：$num = 245354;$numlength = mb_strlen($num);在本例中，$numlength应为6。不知何故我不能让它工作？谢谢编辑：上面的示例代码--^及其各自的方法mb_strlen();工作得很好。

问PHP -获取数字中的位数长度
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回答 12

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问PHP -获取数字中的位数长度EN

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