blocks|key|4166221|text|只需检查mod是否在给定的数字范围内。也就是说。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4166222|function+mode_is_valid($mode)+{
++++if+(is_string($mode))+{
++++++++$mode+=+octdec($mode);
++++}
++++return+$mode+>=+0+&&+$mode+<=+0777;
}|code-block|syntax|javascript|4166223|如果您还希望关注特殊权限，则需要添加其他逻辑。|4166224|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Just check, if the mod is in the given number range. I.e.

<pre><code>function mode_is_valid($mode) {
 if (is_string($mode)) {
 $mode = octdec($mode);
 }
 return $mode &gt;= 0 &amp;&amp; $mode &lt;= 0777;
}
</code></pre>

If you want also to take care about special permissions, you need to add additional logic.

blocks|key|4166278|text|chmod遵循经典的Unix+/+Linux文件模式逻辑。该数字表示八进制的位模式!！符号。有9位用于读/写/执行，外加一些额外的标志用于特殊处理，如所有权继承(所谓的S位)……|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4166279|9位的顺序为：|4166280|3+bits+for+"owner"+of+the+file+/+link+/+directory
3+bits+for+"group"+of+the+file+/+link+/+directory
3+bits+for+the+rest+of+the+world|code-block|syntax|javascript|4166281|单个比特表示|4166282|READ+permission
WRITE+permission
EXECUTE+permission|4166283|示例|4166284|READ+(1)+WRITE+(1)+EXEC+(1)+=+binary+111+=+octal+7
READ+(1)+WRITE+(1)+EXEC+(0)+=+binary+110+=+octal+6
READ+(1)+WRITE+(0)+EXEC+(0)+=+binary+100+=+octal+4|4166285|以上三种组合是最常见的组合。READ+%2B+EXEC+=5也很常见。目录上需要EXEC才能列出目录。|4166286|在Windows上，这是在底层文件系统允许的情况下模拟的。|4166287|因此，请避免任何会剥夺所有者读取文件权限的操作。如果web服务器UID不是所有者，而是组，并且web服务器进程应该能够读取该文件，则还应避免删除第二个数字中的读取权限。|4166288|entityMap^0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|11|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|12|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|13|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|14|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|6|7|15|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|16|8|@]|9|@]|A|$]]|$1|U|3|V|5|6|7|17|8|@]|9|@]|A|$]]|$1|W|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|X|$]]

chmod follows the classic Unix / Linux file mode logic. The number represents a bit pattern in OCTAL !! notation. There are 9 bits for read / write / execute plus some additional flags for special treatment like inheritance of ownership (the so called S-Bit) ...

Order of the 9 bits is:

<pre><code>3 bits for "owner" of the file / link / directory
3 bits for "group" of the file / link / directory
3 bits for the rest of the world
</code></pre>

Individual bits represent

<pre><code>READ permission
WRITE permission
EXECUTE permission
</code></pre>

Example

<pre><code>READ (1) WRITE (1) EXEC (1) = binary 111 = octal 7
READ (1) WRITE (1) EXEC (0) = binary 110 = octal 6
READ (1) WRITE (0) EXEC (0) = binary 100 = octal 4
</code></pre>

The three combinations above are the most common ones. READ + EXEC = 5 is also common. The EXEC is needed on directories to allow directory listing.

On Windows this is emulated as the underlying file system permits.

So avoid anything that removes the owners permission to read the file. If the web-server UID is not the owner but the group and the web-server process should be able to read the file, also avoid removal of the read permission in the second digit.

I can set file permissions using <code>chmod</code>:

<pre><code>$file = 'somefile.txt';
$mode = 0755;

chmod($file, $mode);
</code></pre>

How do I check if the mode is valid? It seems entering a random mode will really mess up the file permissions.

Like:

<pre><code>if (mode_is_valid($mode)) {
 chmod($file, $mode);
}
</code></pre>

PHP: How to check if a valid CHMOD mode?

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我可以使用chmod设置文件权限$file = 'somefile.txt';$mode = 0755;chmod($file, $mode);如何检查该模式是否有效？似乎进入随机模式会使文件权限变得混乱。像这样：if (mode_is_valid($mode)) {  chmod($file, $mode);}

问PHP:如何检查CHMOD模式是否有效？
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