用户输入的字符串验证 [英] String validation for user input

问题描述

我正在进行strin验证，当用户输入以下字符时我会抛出错误，就像无效的字符串一样。

/ \ [] :; | =，+ *？< >



但是我收到了错误

Microsoft C ++异常：std :: regex_error在内存位置0x000000000019A210。



我尝试过：



bool ValidateLoginid（CString inputString）

{std :: regex stringpattern（[！/ \ []：; | =，+ *<>] *）;



std :: string s_input（CW2A（inputString.GetString（）））;

if（std :: regex_match（s_input，stringpattern））

{

返回false;

}



返回true;





}



如果此处有任何问题，请指教我

解决方案

看看你的正则表达式：

 [！/ \ []：; | =，+ *<>] * 

它包含字符集开始和结束字符集中的此集中的任何字符，以及Escape字符！

试试这个：

 [！/ \\ [\]：; | =，+ *<>] * 

但是......那将不匹配特殊字符 - 试试这个：

 [！/ \\ [\]：; | =，+ *<>] + 

哦，您可能需要将C ++字符串中的每个'\'字符加倍，以免它被吞噬为C ++ Escape字符：

 std :: regex stringpattern（[！/ \\\\ [\\]：; | =，+ *<>] + ）; 

如果你想使用正则表达式，请获取 Expresso [ ^ ] - 它是免费的，它会检查并生成正则表达式。

y有一些问题我们的字符串：

• 你的角色类中的反斜杠不会被转义，所以 \ [被视为转义序列，但它不存在。如果你像 \\ 那样逃避反斜杠，那么它将最终成为一个反斜杠，正则表达式会将其解释为'下一个字符是意思是一个文字字符'，所以你必须添加另一个反斜杠，并且，在你的字符串中，这看起来像另一个双反斜杠。因此，对于正则表达式中的字面反斜杠，您的字符串中需要4个反斜杠。
  
  
• 字符中的] class结束了角色类，但是你希望它继续，所以你必须逃避] 。在正则表达式中，这将是 \] ，但因为你需要转义字符串中的反斜杠，它变为 \\] 。
  
  

因此它变为：

 std :: regex stringpattern（  [！/ \\\\ [\\]： ; | =，+ *<>] *）; 

但我们尚未完成！此代码不会执行您想要的操作：如果字符串包含一个无效字符，则返回false。原因是regex_match匹配完整字符串，而您想要搜索。而不是 regex_match ，使用 regex_search 。



And最后一件事：字符类意味着零或更多之后的 * 。这不是你想要的：你想要'一个或多个'。您可以用*替换*但是因为您正在搜索，您也可以删除 * ，因此您的代码变为：

< pre lang =c ++> std :: regex stringpattern（ [！/ \\\\ [ \\] :; | =，+ * LT;>]）;

// ...

< span class =code-keyword> if （std :: regex_search（s_input，stringpattern））
{
return < span class =code-keyword> false ;
}
返回 true ;

那是因为你忘了逃避字符组中的右方括号并且它们必须被双重转义（因此在字符串中放置一个反斜杠以供正则表达式解析器看到）： br />

错误：[！/ \ []：; | =，+ *<>] *
正确：[！/ \\ \\\ [\\]：; | =，+ *<>] *

您也不需要将输入转换为ANSI当使用Visual C ++提供的 wregex 类时：

 bool ValidateLoginid（const CString& inputString）
 { 
 std :: wregex stringpattern（L[！/ \\ [\\]：; | =，+ *<>] *）; 
 return！std :: regex_match（inputString.GetString（），stringpattern）; 
} 

另请注意，我已将参数更改为const引用，以避免传递值并指示它未更改。



如果您只想检查特定字符是否存在，也不需要使用正则表达式。然后只需使用 strstr 或 CString :: FindOneOf 。

i am doin strin validation ,when ever user give input as below characters i ahve throw error like not valid string.
" / \ [ ] : ; | = , + * ? < >

but i am getting error as
"Microsoft C++ exception: std::regex_error at memory location 0x000000000019A210."

What I have tried:

bool ValidateLoginid(CString inputString)
{std::regex stringpattern("[!/\[]:;|=,+*<>]*");

std::string s_input(CW2A(inputString.GetString()));
if (std::regex_match(s_input, stringpattern))
{
return false;
}

return true;


}

could you please suggest me if anything is wrong here

解决方案

Look at your regex:

[!/\[]:;|=,+*<>]*

It contains the characters to start and end the "any character in this set" within the set of characters, as well as the Escape character!
Try this:

[!/\\[\]:;|=,+*<>]*

But... that will match no special characters at all - try this:

[!/\\[\]:;|=,+*<>]+

Oh, and you probably need to double each '\' character in your C++ string, so that it doesn't get "swallowed" as the C++ Escape character:

std::regex stringpattern("[!/\\\\[\\]:;|=,+*<>]+");

If you want to use regular expressions, get a copy of Expresso[^] - it's free, and it examines and generates Regular expressions.

There are a few problems with your string:

• The backslash inside your character class isn't escaped, so \[ is being treated as escape sequence, but it doesn't exist. And if you'd escape the backslash like \\, so it would 'end up' being a single backslash, the regex will interpret that as 'the next character is meant to be a literal character', so you have to add another backslash, and, in your string, this will look like another double backslash. So for a literal backslash in the regex, you need 4 backslashes in your string.
  
• The ] in the character class ends the character class, but you want it to 'go on', so you have to escape the ]. In regular expression, that would be \], but because you need to escape the backslash in the string, it becomes \\].
  

So it becomes:

std::regex stringpattern("[!/\\\\[\\]:;|=,+*<>]*");

But we aren't done yet! This code won't do what you want: return false if the string contains one invalid character. The reason is that regex_match matches the full string, whereas you want to search. Instead of regex_match, use regex_search.

And one last thing: the * after the character class means 'zero or more'. That's not what you want: you want 'one or more'. You can replace the * by a + but because you're searching, you can also just remove the *, so your code becomes:

std::regex stringpattern("[!/\\\\[\\]:;|=,+*<>]");

// ...

if (std::regex_search(s_input, stringpattern))
{
    return false;
}
return true;

That is because you forgot to escape the closing square bracket in the character group and that they must be double escaped (so that a backslash is placed in the string to be seen by the regex parser):

wrong:   "[!/\[]:;|=,+*<>]*"
correct: "[!/\\[\\]:;|=,+*<>]*"

You also don't need to convert the input to ANSI when using the wregex class provided by Visual C++:

bool ValidateLoginid(const CString& inputString)
{
    std::wregex stringpattern(L"[!/\\[\\]:;|=,+*<>]*");
    return !std::regex_match(inputString.GetString(), stringpattern);
}

Note also that I have changed the parameter to be a const reference to avoid passing by value and indicate that it is not changed.

There is also no need to use a regular expression if you only want to check if specific characters are present or not. Then just use strstr or CString::FindOneOf.

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