entityMap|blocks|key|fngav|text|您可以尝试intval($pageNumber)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|bo5a1^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|D|9|@]|A|@]|B|$]]|$2|C|4|-4|6|7|8|E|9|@]|A|@]|B|$]]]]

you could try intval($pageNumber)

entityMap|blocks|key|56uns|text|$pageNumber+=+"1\'";
preg_match("%7C\d%2B%7C",+$pageNumber,+$number);
var_dump($number);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|f76m0|这将只提取字符串中的数值，并将其放入$number中。|unstyled|dhsrj^0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|I|9|@]|A|@]|B|$C|D]]|$2|E|4|F|6|G|8|J|9|@]|A|@]|B|$]]|$2|H|4|-4|6|G|8|K|9|@]|A|@]|B|$]]]]

<pre><code>$pageNumber = "1\'";
preg_match("|\d+|", $pageNumber, $number);
var_dump($number);
</code></pre>

This will just extract the numeric value in the string and put it inside $number.

entityMap|blocks|key|3efhn|text|从胡言乱语中解析一个数字有点棘手，因为如果文本中有其他数字，您就不知道要解析出哪个数字。(例如1\2，你会解析出哪个#？1还是2?)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|18ho1|最简单的解决方案是，如果无法解析页面，则将其缺省为整数。|5ghr2|例如：|5anfh|if+(!is_int($pageNum))
++++$pageNum+=+1;|code-block|syntax|javascript|fl53r|因此，如果有人篡改POST变量，他们不能做任何棘手的事情，因为它只会默认为1。|a602b^0|1B|3|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|S|9|@$A|T|B|U|C|D]]|E|@]|F|$]]|$2|G|4|H|6|7|8|V|9|@]|E|@]|F|$]]|$2|I|4|J|6|7|8|W|9|@]|E|@]|F|$]]|$2|K|4|L|6|M|8|X|9|@]|E|@]|F|$N|O]]|$2|P|4|Q|6|7|8|Y|9|@]|E|@]|F|$]]|$2|R|4|-4|6|7|8|Z|9|@]|E|@]|F|$]]]]

Parsing a number out of gibberish is a bit tricky because if there are other numbers in the text, you wouldn't know which number to parse out. (e.g. <code>1\2</code>, which # would you parse out? 1 or 2?)

The simplest solution would be to have the page default to an integer if you cannot parse it. 

For example:

<pre><code>if (!is_int($pageNum))
 $pageNum = 1;
</code></pre>

So if anyone is tampering with the POST variables, they cannot do anything tricky as it'll just default to 1.

entityMap|blocks|key|d8ane|text|$_POST的值总是字符串类型，所以is_int()不起作用。只需检查is_numeric()|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|745qg|if+(!is_numeric($pageNumber))+{++++//+$pageNumber+=+"1\'"+for+this+example
++++$pageNumber+=+'1';
}|code-block|syntax|javascript|50j7|但是，由于您知道它应该是一个整数，因此如果它以数字开头，则将获得正确的数字；如果不是，则将获得0：|401k5|$pageNumber+=+(int)$_POST['pageNum'];|5ac5g|你不能从胡言乱语中可靠地解析出正确的页码。另外，如果有人试图入侵你的服务器，你不应该在意给他们正确的页码。|biofb^0|0|6|I|8|Z|C|0|0|1B|1|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|S|9|@$A|T|B|U|C|D]|$A|V|B|W|C|D]|$A|X|B|Y|C|D]]|E|@]|F|$]]|$2|G|4|H|6|I|8|Z|9|@]|E|@]|F|$J|K]]|$2|L|4|M|6|7|8|10|9|@$A|11|B|12|C|D]]|E|@]|F|$]]|$2|N|4|O|6|I|8|13|9|@]|E|@]|F|$J|K]]|$2|P|4|Q|6|7|8|14|9|@]|E|@]|F|$]]|$2|R|4|-4|6|7|8|15|9|@]|E|@]|F|$]]]]

<code>$_POST</code> values are always type string so <code>is_int()</code> won't work. Just check <code>is_numeric()</code>:

<pre><code>if (!is_numeric($pageNumber)) { // $pageNumber = "1\'" for this example
 $pageNumber = '1';
}
</code></pre>

But since you know it should be an integer, this will get the proper number if it begins with the number or <code>0</code> if not:

<pre><code>$pageNumber = (int)$_POST['pageNum'];
</code></pre>

You can't reliably parse a correct page number out of gibberish. Also, if someone is trying to hack your server, you shouldn't care about giving them the correct page number.

entityMap|0|type|LINK|mutability|MUTABLE|data|url|http://php.net/manual/en/function.filter-var.php|blocks|key|840b|text|您可以使用filter_var函数。例如:+filter_var($your_integer_value，FILTER_VALIDATE_INT，$options)；|unstyled|depth|inlineStyleRanges|entityRanges|d16bh|查看此链接：http://php.net/manual/en/function.filter-var.php|offset|length|4591n^0|0|6|1C|0|0^^$0|$1|$2|3|4|5|6|$7|8]]]|9|@$A|B|C|D|2|E|F|N|G|@]|H|@]|6|$]]|$A|I|C|J|2|E|F|O|G|@]|H|@$K|P|L|Q|A|R]]|6|$]]|$A|M|C|-4|2|E|F|S|G|@]|H|@]|6|$]]]]

You can use the filter_var function. For example:
filter_var($your_integer_value, FILTER_VALIDATE_INT, $options);

See this link: <a href="http://php.net/manual/en/function.filter-var.php" rel="nofollow noreferrer">http://php.net/manual/en/function.filter-var.php</a>

I have a form with a hidden input that will always have an integer as a value automatically assigned in the code.

<pre><code>&lt;form action=""&gt;
 &lt;input name="pageNum" type="hidden" value="1"&gt;
&lt;/form&gt;
</code></pre>

How can I sanitize that value to always be the given integer? Even when someone alter the request like you would do with sql injection by adding a <code>'</code> after the value in the request. Because this was actually the case, somebody reported it as a sql injection security incident, after doing some testing. But the <code>1</code> just becomes <code>1\</code>'and breaks my code because <code>1\</code>'does not exist in an array where 1 does.

<pre><code>$pageNumber = $_POST['pageNum'];

if (!is_int($pageNumber)) { // $pageNumber = "1\'" for this example
 // what can I do to make it 1 again?
}
</code></pre>

Is there a way how I can accomplish this? Because this make that I get an internal server error further down in my code that makes people think it is a sql injection security vulnerability.

Sanitize a hidden input value to always be an integer

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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我有一个带有隐藏输入的表单，它将始终有一个整数作为代码中自动分配的值。 <form action=""> <input name="pageNum" type="hidden" value="1"></form> 如何才能将该值整理为始终为给定整数？即使有人修改了请求，您也可以通过在请求中的值后面添加'来处理s...

问清理隐藏的输入值，使其始终为整数
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