正则表达式删除单行SQL注释（ - ） [英] Regex to remove single-line SQL comments (--)

问题描述

问：

任何人都可以给我一个工作正则表达式（C＃/ VB.NET），可以消除一行从SQL语句评论

？

我的意思是这些注释：

   - 这是一个注释
  

而不是那些

  / *这是一个注释* / 
  

因为我已经可以处理恒星意见。

我有一个作了少许解析器当他们在该行的开始，消除这些意见，但它们也可以是代码或更糟后的某处，在SQL串'你好 - 测试 - 世界
那些意见也应该被删除（除非那些理所当然的SQL字符串 - 如果可能的话）。

令人惊讶的我没有得到正则表达式的工作。我会认为明星的意见是比较困难的，但实际上，他们不是

根据要求，在这里我的代码删除/ ** / - 风格评论
（为了让它忽略SQL风格的字符串，你有一个唯一标识符，以替补多串（我用4 concated），然后应用注释去除，再申请字符串回代。

 静态字符串RemoveCstyleComments（字符串strInput）
 {
串strPattern = @/ [*] [\w\ d\s] + [*] /; 
 // strPattern = @/ \ * * * \ /。？; //不行的
 // strPattern = /\\*.*?\\*/; //不行的
 // strPattern = @/ \ *（[^ *] | [\r\\\
 ] |（\ * +（[^ * /] | [\r\\\
]）））* \ * + /; //不行的
 // strPattern = @/ \ *（[^ *] | [\r\\\
] |（\ * +（[^ * /] | [\r\\\
]）））* \ * + /; / /不工作
 
 // http://stackoverflow.com/questions/462843/improving-fixing-a-regex-for-c-style-block-comments 
 strPattern = @/ \ *（？>（？：？！（大于[^ *] +）| \ *（/））*）\ * /; //作品！ 
 
串strOutput = System.Text.RegularExpressions.Regex.Replace（strInput，strPattern，的String.Empty，System.Text.RegularExpressions.RegexOptions.Multiline）; 
 Console.WriteLine（strOutput）; 
返回strOutput; 
} //端功能RemoveCstyleComments 
  

解决方案

我会辜负大家。这不能用正则表达式来完成。当然，这是很容易找到不是在一个字符串评论（连OP可以做），真正的交易是在一个字符串意见。还有就是看变通的一点希望，但是这仍然不够。告诉你有一个报价之前在一条线上将不能保证什么。什么保证你的时候，唯一的事情是报价的怪胎。有些事情，你不能用正则表达式找到。所以只是简单地去与非正则表达式的方法

编辑：
这里的C＃代码：

 字符串SQL = - 这是一个test\r\\\
select的东西在那里substaff像 - 这评论应该留 - 这应removed\r\\\
 
的char [] =报价{'\''，''}; 
 INT newCommentLiteral，lastCommentLiteral = 0; 
，而（（newCommentLiteral = sql.IndexOf（ - ， lastCommentLiteral））= -1）
 {
 INT countQuotes = sql.Substring（lastCommentLiteral，newCommentLiteral  -  lastCommentLiteral）.Split（引号）。长度 -  1;！
如果（countQuotes％2 = = 0）//这是一条评论，因为有偶数个前
 {
 INT EOL = sql.IndexOf（\r\\\
）+ 2报价; 
如果（EOL == -1）
 EOL = sql.Length; //不换行，这意味着字符串
 SQL = sql.Remove（newCommentLiteral，EOL  -  newCommentLiteral）结束; 
 lastCommentLiteral = newCommentLiteral; 
} 
，否则//这是一个字符串中，找到字符串结束，并移动到它
 {
 INT singleQuote = sql.IndexOf（'，newCommentLiteral ）; 
如果（singleQuote == -1）
 singleQuote = sql.Length; 
 INT双引号= sql.IndexOf（''，newCommentLiteral）; 
如果（双引号== -1）
双引号= sql.Length; 
 
 lastCommentLiteral = Math.Min（singleQuote，双引号）+ 1; 
 
 //而不是找到字符串，你可以简单地做+ = 2，但该方案将成为稍慢
} $ b的结束$ b} 
 
 Console.WriteLine（SQL）; 
  

这能做什么：找到每一个注释文字对于每一个，检查它是否是一个注释中与否，通过计算当前的匹配，最后一个的报价数量如果这个数字是偶数，那么它的评论，从而将其删除（查找第一个。 。行尾和删除之间什么），如果是奇数，所以这是一个字符串中，找到字符串的结尾，并移动到它RGIS片段是基于一个奇怪的SQL招：这是一个有效的字符串。即使寿2报价有所不同。如果这不是你的SQL语言真的，你应该尝试一个完全不同的方法。我会写一个程序的太多，如果是这样的话，但是这一次的更快，更简单。

Question:

Can anybody give me a working regex expression (C#/VB.NET) that can remove single line comments from a SQL statement ?

I mean these comments:

-- This is a comment

not those

/* this is a comment */

because I already can handle the star comments.

I have a made a little parser that removes those comments when they are at the start of the line, but they can also be somewhere after code or worse, in a SQL-string 'hello --Test -- World' Those comments should also be removed (except those in a SQL string of course - if possible).

Surprisingly I didn't got the regex working. I would have assumed the star comments to be more difficult, but actually, they aren't.

As per request, here my code to remove /**/-style comments (In order to have it ignore SQL-Style strings, you have to subsitute strings with a uniqueidentifier (i used 4 concated), then apply the comment-removal, then apply string-backsubstitution.

    static string RemoveCstyleComments(string strInput) 
    { 
        string strPattern = @"/[*][\w\d\s]+[*]/"; 
        //strPattern = @"/\*.*?\*/"; // Doesn't work 
        //strPattern = "/\\*.*?\\*/"; // Doesn't work 
        //strPattern = @"/\*([^*]|[\r\n]|(\*+([^*/]|[\r\n])))*\*+/ "; // Doesn't work 
        //strPattern = @"/\*([^*]|[\r\n]|(\*+([^*/]|[\r\n])))*\*+/ "; // Doesn't work 

        // http://stackoverflow.com/questions/462843/improving-fixing-a-regex-for-c-style-block-comments 
        strPattern = @"/\*(?>(?:(?>[^*]+)|\*(?!/))*)\*/";  // Works ! 

        string strOutput = System.Text.RegularExpressions.Regex.Replace(strInput, strPattern, string.Empty, System.Text.RegularExpressions.RegexOptions.Multiline); 
        Console.WriteLine(strOutput); 
        return strOutput; 
    } // End Function RemoveCstyleComments 

解决方案

I will disappoint all of you. This can't be done with regular expressions. Sure, it's easy to find comments not in a string (that even the OP could do), the real deal is comments in a string. There is a little hope of the look arounds, but that's still not enough. By telling that you have a preceding quote in a line won't guarantee anything. The only thing what guarantees you something is the oddity of quotes. Something you can't find with regular expression. So just simply go with non-regular-expression approach.

EDIT: Here's the c# code:

        String sql = "--this is a test\r\nselect stuff where substaff like '--this comment should stay' --this should be removed\r\n";
        char[] quotes = { '\'', '"'};
        int newCommentLiteral, lastCommentLiteral = 0;
        while ((newCommentLiteral = sql.IndexOf("--", lastCommentLiteral)) != -1)
        {
            int countQuotes = sql.Substring(lastCommentLiteral, newCommentLiteral - lastCommentLiteral).Split(quotes).Length - 1;
            if (countQuotes % 2 == 0) //this is a comment, since there's an even number of quotes preceding
            {
                int eol = sql.IndexOf("\r\n") + 2;
                if (eol == -1)
                    eol = sql.Length; //no more newline, meaning end of the string
                sql = sql.Remove(newCommentLiteral, eol - newCommentLiteral);
                lastCommentLiteral = newCommentLiteral;
            }
            else //this is within a string, find string ending and moving to it
            {
                int singleQuote = sql.IndexOf("'", newCommentLiteral);
                if (singleQuote == -1)
                    singleQuote = sql.Length;
                int doubleQuote = sql.IndexOf('"', newCommentLiteral);
                if (doubleQuote == -1)
                    doubleQuote = sql.Length;

                lastCommentLiteral = Math.Min(singleQuote, doubleQuote) + 1;

                //instead of finding the end of the string you could simply do += 2 but the program will become slightly slower
            }
        }

        Console.WriteLine(sql);

What this does: find every comment literal. For each, check if it's within a comment or not, by counting the number of quotes between the current match and the last one. If this number is even, then it's a comment, thus remove it (find first end of line and remove whats between). If it's odd, this is within a string, find the end of the string and move to it. Rgis snippet is based on a wierd SQL trick: 'this" is a valid string. Even tho the 2 quotes differ. If it's not true for your SQL language, you should try a completely different approach. I'll write a program to that too if that's the case, but this one's faster and more straightforward.

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