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Homephp教程php手册如何实现一个php框架系列文章【5】安全处理输入,

如何实现一个php框架系列文章【5】安全处理输入,

Jun 13, 2016 am 08:46 AM
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如何实现一个php框架系列文章【5】安全处理输入,

所有的外部输入参数都应该检查合法性。

未正确处理输入数据将可能导致sql注入等漏洞。

 

框架提供系列函数来取$_REQUEST中的值

requestInt

requestString

requestFloat

requestBool

 

ps:注意$_REQUEST中变量类型可能会是数组

如请求为 ?i[]=1,那么$_REQUEST['i'] 的值为array(1)

做校验的时候要考虑全面以防止php warning信息泄露

 

 

另外再介绍一下kv json格式的数据校验。

有时为了在项目中保留一定扩展性,会使用json格式的数据,这种数据又该如何校验呢。

 

//校验键值形式{k1:v1, k2:v2, k3:v3 ...}的json数据,可以对每一对kv进行校验

requestKvJson

 

 

 

 

部分实现代码

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 //校验整数,失败返回$default function checkInt($var$default = 0) {     return  is_numeric($var) ? intval($var, (strncasecmp($var'0x', 2) == 0 || strncasecmp($var'-0x', 3) == 0) ? 16 : 10) : $default; }   //校验字符串 $check为正则表达式 function checkString($var$check ''$default '') {     if (!is_string($var)) {         if(is_numeric($var)) {             $var = (string)$var;         }         else {             return $default;         }     }     if ($check) {          return (preg_match($check$var$ret) ? $ret[1] : $default);     }       return $var; }   /*     校验kv json,     如果想要一个这样的数据{id:1, 'type':'single_text', 'required': true, 'desc':'this is a text'}     那么$desc可以这样写     array(    array('id', 'Int'),    array('type', 'string', PATTERN_NORMAL_STRING),    array('required', 'Bool', false),    array('desc', 'string', PATTERN_NORMAL_STRING), )) */ function checkKvJson($var$desc array()) {     if(is_string($var)) {         $var = json_decode($var, true);     }     if(!$var || !is_array($var)) {         return array();     }       if($desc)     foreach($desc as $d) {           if(!isset($var[$d[0]])) {             return array();         }           $ps array_slice($d, 2);         array_unshift($ps$var[$d[0]]);         $var[$d[0]] = call_user_func_array('check'.$d[1], $ps);         if($var[$d[0]] === false && strcasecmp($d[1], 'Bool')) {             return array();         }     }       return $var; }
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