How to detect whether a character belongs to a Rig

问题:

What is a good way to tell whether a string contains text in a Right To Left language.

I have found this question which suggests the following approach:

public bool IsArabic(string strCompare)
{
  char[] chars = strCompare.ToCharArray();
  foreach (char ch in chars)
    if (ch >= '\u0627' && ch <= '\u0649') return true;
  return false;
}

While this may work for Arabic this doesn't seem to cover other RTL languages such as Hebrew. Is there a generic way to know that a particular character belongs to a RTL language?

回答1:

Unicode characters have different properties associated with them. These properties cannot be derived from the code point; you need a table that tells you if a character has a certain property or not.

You are interested in characters with bidirectional property "R" or "AL" (RandALCat).

A RandALCat character is a character with unambiguously right-to-left directionality.

Here's the complete list as of Unicode 3.2 (from RFC 3454):

D. Bidirectional tables

D.1 Characters with bidirectional property "R" or "AL"

----- Start Table D.1 -----
05BE
05C0
05C3
05D0-05EA
05F0-05F4
061B
061F
0621-063A
0640-064A
066D-066F
0671-06D5
06DD
06E5-06E6
06FA-06FE
0700-070D
0710
0712-072C
0780-07A5
07B1
200F
FB1D
FB1F-FB28
FB2A-FB36
FB38-FB3C
FB3E
FB40-FB41
FB43-FB44
FB46-FBB1
FBD3-FD3D
FD50-FD8F
FD92-FDC7
FDF0-FDFC
FE70-FE74
FE76-FEFC
----- End Table D.1 -----

Here's some code to get the complete list as of Unicode 6.0:

var url = "http://www.unicode.org/Public/6.0.0/ucd/UnicodeData.txt";

var query = from record in new WebClient().DownloadString(url).Split('\n')
            where !string.IsNullOrEmpty(record)
            let properties = record.Split(';')
            where properties[4] == "R" || properties[4] == "AL"
            select int.Parse(properties[0], NumberStyles.AllowHexSpecifier);

foreach (var codepoint in query)
{
    Console.WriteLine(codepoint.ToString("X4"));
}

Note that these values are Unicode code points. Strings in C#/.NET are UTF-16 encoded and need to be converted to Unicode code points first (see Char.ConvertToUtf32). Here's a method that checks if a string contains at least one RandALCat character:

static void IsAnyCharacterRightToLeft(string s)
{
    for (var i = 0; i < s.Length; i += char.IsSurrogatePair(s, i) ? 2 : 1)
    {
        var codepoint = char.ConvertToUtf32(s, i);
        if (IsRandALCat(codepoint))
        {
            return true;
        }
    }
    return false;
}

回答2:

You can try using "named blocks" in regular expressions. Just pick out the blocks that are right to left, and form the regex. For example:

\p{IsArabic}|\p{IsHebrew}

If that regex returns true, then there was at least one hebrew or arabic character in the string.

回答3:

All "AL" or "R" of Unicode 6.0 (from http://www.unicode.org/Public/6.0.0/ucd/UnicodeData.txt)

bool hasRandALCat = 0;
if(c >= 0x5BE && c <= 0x10B7F)
{
    if(c <= 0x85E)
    {
        if(c == 0x5BE)                        hasRandALCat = 1;
        else if(c == 0x5C0)                   hasRandALCat = 1;
        else if(c == 0x5C3)                   hasRandALCat = 1;
        else if(c == 0x5C6)                   hasRandALCat = 1;
        else if(0x5D0 <= c && c <= 0x5EA)     hasRandALCat = 1;
        else if(0x5F0 <= c && c <= 0x5F4)     hasRandALCat = 1;
        else if(c == 0x608)                   hasRandALCat = 1;
        else if(c == 0x60B)                   hasRandALCat = 1;
        else if(c == 0x60D)                   hasRandALCat = 1;
        else if(c == 0x61B)                   hasRandALCat = 1;
        else if(0x61E <= c && c <= 0x64A)     hasRandALCat = 1;
        else if(0x66D <= c && c <= 0x66F)     hasRandALCat = 1;
        else if(0x671 <= c && c <= 0x6D5)     hasRandALCat = 1;
        else if(0x6E5 <= c && c <= 0x6E6)     hasRandALCat = 1;
        else if(0x6EE <= c && c <= 0x6EF)     hasRandALCat = 1;
        else if(0x6FA <= c && c <= 0x70D)     hasRandALCat = 1;
        else if(c == 0x710)                   hasRandALCat = 1;
        else if(0x712 <= c && c <= 0x72F)     hasRandALCat = 1;
        else if(0x74D <= c && c <= 0x7A5)     hasRandALCat = 1;
        else if(c == 0x7B1)                   hasRandALCat = 1;
        else if(0x7C0 <= c && c <= 0x7EA)     hasRandALCat = 1;
        else if(0x7F4 <= c && c <= 0x7F5)     hasRandALCat = 1;
        else if(c == 0x7FA)                   hasRandALCat = 1;
        else if(0x800 <= c && c <= 0x815)     hasRandALCat = 1;
        else if(c == 0x81A)                   hasRandALCat = 1;
        else if(c == 0x824)                   hasRandALCat = 1;
        else if(c == 0x828)                   hasRandALCat = 1;
        else if(0x830 <= c && c <= 0x83E)     hasRandALCat = 1;
        else if(0x840 <= c && c <= 0x858)     hasRandALCat = 1;
        else if(c == 0x85E)                   hasRandALCat = 1;
    }
    else if(c == 0x200F)                      hasRandALCat = 1;
    else if(c >= 0xFB1D)
    {
        if(c == 0xFB1D)                       hasRandALCat = 1;
        else if(0xFB1F <= c && c <= 0xFB28)   hasRandALCat = 1;
        else if(0xFB2A <= c && c <= 0xFB36)   hasRandALCat = 1;
        else if(0xFB38 <= c && c <= 0xFB3C)   hasRandALCat = 1;
        else if(c == 0xFB3E)                  hasRandALCat = 1;
        else if(0xFB40 <= c && c <= 0xFB41)   hasRandALCat = 1;
        else if(0xFB43 <= c && c <= 0xFB44)   hasRandALCat = 1;
        else if(0xFB46 <= c && c <= 0xFBC1)   hasRandALCat = 1;
        else if(0xFBD3 <= c && c <= 0xFD3D)   hasRandALCat = 1;
        else if(0xFD50 <= c && c <= 0xFD8F)   hasRandALCat = 1;
        else if(0xFD92 <= c && c <= 0xFDC7)   hasRandALCat = 1;
        else if(0xFDF0 <= c && c <= 0xFDFC)   hasRandALCat = 1;
        else if(0xFE70 <= c && c <= 0xFE74)   hasRandALCat = 1;
        else if(0xFE76 <= c && c <= 0xFEFC)   hasRandALCat = 1;
        else if(0x10800 <= c && c <= 0x10805) hasRandALCat = 1;
        else if(c == 0x10808)                 hasRandALCat = 1;
        else if(0x1080A <= c && c <= 0x10835) hasRandALCat = 1;
        else if(0x10837 <= c && c <= 0x10838) hasRandALCat = 1;
        else if(c == 0x1083C)                 hasRandALCat = 1;
        else if(0x1083F <= c && c <= 0x10855) hasRandALCat = 1;
        else if(0x10857 <= c && c <= 0x1085F) hasRandALCat = 1;
        else if(0x10900 <= c && c <= 0x1091B) hasRandALCat = 1;
        else if(0x10920 <= c && c <= 0x10939) hasRandALCat = 1;
        else if(c == 0x1093F)                 hasRandALCat = 1;
        else if(c == 0x10A00)                 hasRandALCat = 1;
        else if(0x10A10 <= c && c <= 0x10A13) hasRandALCat = 1;
        else if(0x10A15 <= c && c <= 0x10A17) hasRandALCat = 1;
        else if(0x10A19 <= c && c <= 0x10A33) hasRandALCat = 1;
        else if(0x10A40 <= c && c <= 0x10A47) hasRandALCat = 1;
        else if(0x10A50 <= c && c <= 0x10A58) hasRandALCat = 1;
        else if(0x10A60 <= c && c <= 0x10A7F) hasRandALCat = 1;
        else if(0x10B00 <= c && c <= 0x10B35) hasRandALCat = 1;
        else if(0x10B40 <= c && c <= 0x10B55) hasRandALCat = 1;
        else if(0x10B58 <= c && c <= 0x10B72) hasRandALCat = 1;
        else if(0x10B78 <= c && c <= 0x10B7F) hasRandALCat = 1;
    }
}

回答4:

EDIT:

This is what I use now, it includes the Vowelization chars and everything in Hebrew and Arabic:

[\u0591-\u07FF]

OLD ANSWER:

If you need to detect RTL language in a sentence, this simplified RegEx will probably be enough:

[א-ת؀-ۿ]

If one wants to write something in Hebrew it will have to use one of these characters, and the case is similar with Arabic.

It does not include vowelization characters, so if you need to catch all whole words or absolutely all RTL chars you better use one of the other answers. Vowelization chars in Hebrew are very rare in non-poetry texts. I don't know about Arabic texts.

回答5:

On my implementation of regex I could not use neither \u, \x, nor {} language named groups.

So I built my own pattern programatically based on all "R" and "AL" (RandALCat) bidirectional characters as listed in UnicodeData.txt.

[־׀׃א-״؛-ي٭-ە‏ײַ-ﳝﶈ-ﷺﺂ-ﻼ]

This should be decently comprehensive and I've tested it on Arabic and Hebrew text so far.