blocks|key|4488112|text|试一试|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4488113|$level+=+min(max((int)(log($xp+/+100,+1.5)+%2B+1),+1),+$maxLevel);|code-block|syntax|javascript|4488114|这需要xp/100的对数底1.5+(即必须乘以100才能得到对数的次数)，然后加1(因为如果$x小于1.5，则$xp($x，1.5)小于1)。min(max(...,+minLevel),+maxLevel)构造允许您将级别限制在1和$maxLevel之间，还可以避免负级别的任何问题(如果$xp足够小于150)。|offset|length|style|CODE|4488115|entityMap^0|0|0|1Z|X|39|9|41|3|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]|$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|X|8|@]|9|@]|A|$]]]|N|$]]

Try

<pre><code>$level = min(max((int)(log($xp / 100, 1.5) + 1), 1), $maxLevel);
</code></pre>

That takes the logarithm base 1.5 of xp/100 (i.e. the number of times you'd have to multiply 100 by to get $xp), then adds one (since log($x, 1.5) is less than one if $x is less than 1.5). The <code>min(max(..., minLevel), maxLevel)</code> construct lets you clamp the level to lie between 1 and <code>$maxLevel</code>, also avoiding any issues with negative levels (if <code>$xp</code> is sufficiently less than 150).

blocks|key|2851913|text|我同意。对象或数组都是最好的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2851914|类似于：|2851915|$array+=+array(
++++array(
++++++++'xp_required'+=>+150,
++++++++'level'+=>+1
++++),
++++array(
++++++++'xp_required'+=>+225,
++++++++'level'+=>+2
++++),
++++array(
++++++++'xp_required'+=>+337.5,
++++++++'level'+=>+3
++++)
);
$current_xp+=+155;
foreach(+$array+as+$reqs+){
++++if(+$current_xp+>+$reqs['xp_required']+){
++++++++$level+=+$reqs['level'];
++++}
}
echo+$level';|code-block|syntax|javascript|2851916|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I'd agree. either objects or arrays would be best.

Something like:

<pre><code>$array = array(
 array(
 'xp_required' =&gt; 150,
 'level' =&gt; 1
 ),
 array(
 'xp_required' =&gt; 225,
 'level' =&gt; 2
 ),
 array(
 'xp_required' =&gt; 337.5,
 'level' =&gt; 3
 )
);
$current_xp = 155;
foreach( $array as $reqs ){
 if( $current_xp &gt; $reqs['xp_required'] ){
 $level = $reqs['level'];
 }
}
echo $level';
</code></pre>

blocks|key|2851933|text|目前，一旦获取用户exp，就可以将该代码块作为函数，不断检查exp级别的变化，但要使用ranged语句|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2851934|+if($xp+<+150){
+++++$level+=+"1";
++}elseif($xp+>+150+&&+$xp+<+225+){
++++$level+=+"2";
++}elseif($xp+>+225+&&+$xp+<+337.5){
++++$level+=+"3";
++}|code-block|syntax|javascript|2851935|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|$]]

well at the moment once you grab the users exp you can have that code block as a function to constantly check for variations in exp levels, but use a ranged statement

<pre><code> if($xp &lt; 150){
 $level = "1";
 }elseif($xp &gt; 150 &amp;&amp; $xp &lt; 225 ){
 $level = "2";
 }elseif($xp &gt; 225 &amp;&amp; $xp &lt; 337.5){
 $level = "3";
 }
</code></pre>

blocks|key|2851980|text|这是我如何完成我的关卡系统的。这是一个更高级的特点，有技能点数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2851981|++++<?php
++++$level_up+=+($level+%2B+1);
if+($exp+>=+$max_exp)
++++{
++++$sql+=+"UPDATE+users+SET+level=(level+%2B+1)+,+max_exp=(exp+*+1.05)+,+skill_points=(skill_points+%2B+3)+WHERE+id='".$id."'+LIMIT+1";
++++$res+=+mysql_query($sql);
if+($exp+>=+$max_exp)
++++++++echo+'<div+class="Leveled">'+.+'You+sucessfully+leveled+up+to+'+.+$level_up+.+'!'+.+'+As+a+reward+you+were+given+3+skill+points!'+.+'</div>';
++++}
++++else
++++{
++++}+
++++?>|code-block|syntax|javascript|2851982|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|$]]

Here's how I did my level system. It's a little more advanced featuring skill points..

<pre><code> &lt;?php
 $level_up = ($level + 1);
if ($exp &gt;= $max_exp)
 {
 $sql = "UPDATE users SET level=(level + 1) , max_exp=(exp * 1.05) , skill_points=(skill_points + 3) WHERE id='".$id."' LIMIT 1";
 $res = mysql_query($sql);
if ($exp &gt;= $max_exp)
 echo '&lt;div class="Leveled"&gt;' . 'You sucessfully leveled up to ' . $level_up . '!' . ' As a reward you were given 3 skill points!' . '&lt;/div&gt;';
 }
 else
 {
 } 
 ?&gt;
</code></pre>

I'm making a leveling system based on experience you have on the site. I already have all the experience stuff figured out, and how I want to do the leveling, but I need a more efficient way to do it. I know this would probably would be achieved using an array, but I don't really know how to go about doing that. Enough blabbering, though, this is what I'm trying to do...

Level 1 will be anything under 150 experience
Then I'm going to multiply that by 1.5, so
Level 2 will be anything under 225
Level 3 will be anything under 337.5 and so on. This is the inefficient way that I was going to do.

<pre><code>if($xp &lt; 150){
$level = "1";
}elseif($xp &lt; 225){
$level = "2";
}elseif($xp &lt; 337.5){
$level = "3";
}
</code></pre>

I could use a variable for the number and multiple by 1.5 (<code>$number*1.5</code>), but like I said before I don't really know how that'd work. 

*Some more info.. 
I have a session file included on every page and I have queries that would check every time there is new experience earned and this code would be used to update the level on the database automatically.

Efficient Leveling System

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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