将所有 XML 特殊字符转换回常规字符(在 SQL 中)

时间:2023-02-08
本文介绍了将所有 XML 特殊字符转换回常规字符(在 SQL 中)的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着跟版网的小编来一起学习吧!

问题描述

如何将 XML 中的所有特殊字符转换为 ASCII 值?

How do I convert all Special Characters in XML to the ASCII Value?

例如

DECLARE @xml XML = (SELECT 'abc & xyz><' FOR XML PATH(''))
SELECT @xml --@xml is now 'abc &amp; xyz &gt;&lt;'

我希望转换回 ASCII varchar 值(即abc & xyz><").我发现的唯一方法是手动替换所有特殊的 XML 字符,即

I wish to convert back to the ASCII varchar value (i.e. 'abc & xyz><'). The only way I have found is to manually replace all special XML Characters i.e.

SELECT REPLACE(REPLACE(REPLACE(CAST(@xml AS VARCHAR(MAX)),'&amp;','&'),'&gt;','>'),'&lt;','<');
--RETURNS 'abc & xyz><'

然而,这是一个非常不雅的解决方案,并不能处理所有的 XML 字符转换.是否有任何内置的 SQL Server 函数可以执行此操作?

However, this is a very inelegant solution, and does not handle all XML Character conversions. Is there any built-in SQL Server function to do this?

推荐答案

更新:将我之前的解决方案保留在下方,但根据 Jeremy 发布的内容提出了更好的解决方案.

Update: leaving my previous solution available below but came up with a better one based on what Jeremy Posted.

新解决方案:

DECLARE @xml XML = 'abc &amp; xyz &gt;&lt;';

SELECT newstring = ((SELECT @xml FOR XML PATH(''), TYPE).value('.', 'varchar(8000)'));

退货:

abc & xyz ><

旧解决方案(仍然可行):

对于这种类型的事情,我有几个函数.首先你需要 rangeAB 和 CharMapAB

I have a couple functions for this type of thing. First you need rangeAB and CharMapAB

RangeAB

CREATE FUNCTION dbo.rangeAB
(
  @low  bigint, 
  @high bigint, 
  @gap  bigint,
  @row1 bit
)
/****************************************************************************************
[Purpose]:
 Creates up to 531,441,000,000 sequentia integers numbers beginning with @low and ending 
 with @high. Used to replace iterative methods such as loops, cursors and recursive CTEs 
 to solve SQL problems. Based on Itzik Ben-Gan's getnums function with some tweeks and 
 enhancements and added functionality. The logic for getting rn to begin at 0 or 1 is 
 based comes from Jeff Moden's fnTally function. 

 The name range because it's similar to clojure's range function. The name "rangeAB" as 
 used because "range" is a reserved SQL keyword.

[Author]: Alan Burstein

[Compatibility]: 
 SQL Server 2008+ and Azure SQL Database

[Syntax]:
 SELECT r.RN, r.OP, r.N1, r.N2
 FROM dbo.rangeAB(@low,@high,@gap,@row1) AS r;

[Parameters]:
 @low  = a bigint that represents the lowest value for n1.
 @high = a bigint that represents the highest value for n1.
 @gap  = a bigint that represents how much n1 and n2 will increase each row; @gap also
         represents the difference between n1 and n2.
 @row1 = a bit that represents the first value of rn. When @row = 0 then rn begins
         at 0, when @row = 1 then rn will begin at 1.

[Returns]:
 Inline Table Valued Function returns:
 rn = bigint; a row number that works just like T-SQL ROW_NUMBER() except that it can 
      start at 0 or 1 which is dictated by @row1.
 op = bigint; returns the "opposite number that relates to rn. When rn begins with 0 and
      ends with 10 then 10 is the opposite of 0, 9 the opposite of 1, etc. When rn begins
      with 1 and ends with 5 then 1 is the opposite of 5, 2 the opposite of 4, etc...
 n1 = bigint; a sequential number starting at the value of @low and incrimentingby the
      value of @gap until it is less than or equal to the value of @high.
 n2 = bigint; a sequential number starting at the value of @low+@gap and  incrimenting 
      by the value of @gap.

[Dependencies]:
N/A

[Developer Notes]:

 1. The lowest and highest possible numbers returned are whatever is allowable by a 
    bigint. The function, however, returns no more than 531,441,000,000 rows (8100^3). 
 2. @gap does not affect rn, rn will begin at @row1 and increase by 1 until the last row
    unless its used in a query where a filter is applied to rn.
 3. @gap must be greater than 0 or the function will not return any rows.
 4. Keep in mind that when @row1 is 0 then the highest row-number will be the number of
    rows returned minus 1
 5. If you only need is a sequential set beginning at 0 or 1 then, for best performance
    use the RN column. Use N1 and/or N2 when you need to begin your sequence at any 
    number other than 0 or 1 or if you need a gap between your sequence of numbers. 
 6. Although @gap is a bigint it must be a positive integer or the function will
    not return any rows.
 7. The function will not return any rows when one of the following conditions are true:
      * any of the input parameters are NULL
      * @high is less than @low 
      * @gap is not greater than 0
    To force the function to return all NULLs instead of not returning anything you can
    add the following code to the end of the query:

      UNION ALL 
      SELECT NULL, NULL, NULL, NULL
      WHERE NOT (@high&@low&@gap&@row1 IS NOT NULL AND @high >= @low AND @gap > 0)

    This code was excluded as it adds a ~5% performance penalty.
 8. There is no performance penalty for sorting by rn ASC; there is a large performance 
    penalty for sorting in descending order WHEN @row1 = 1; WHEN @row1 = 0
    If you need a descending sort the use op in place of rn then sort by rn ASC. 

Best Practices:
--===== 1. Using RN (rownumber)
 -- (1.1) The best way to get the numbers 1,2,3...@high (e.g. 1 to 5):
 SELECT RN FROM dbo.rangeAB(1,5,1,1);
 -- (1.2) The best way to get the numbers 0,1,2...@high-1 (e.g. 0 to 5):
 SELECT RN FROM dbo.rangeAB(0,5,1,0);

--===== 2. Using OP for descending sorts without a performance penalty
 -- (2.1) The best way to get the numbers 5,4,3...@high (e.g. 5 to 1):
 SELECT op FROM dbo.rangeAB(1,5,1,1) ORDER BY rn ASC;
 -- (2.2) The best way to get the numbers 0,1,2...@high-1 (e.g. 5 to 0):
 SELECT op FROM dbo.rangeAB(1,6,1,0) ORDER BY rn ASC;

--===== 3. Using N1
 -- (3.1) To begin with numbers other than 0 or 1 use N1 (e.g. -3 to 3):
 SELECT N1 FROM dbo.rangeAB(-3,3,1,1);
 -- (3.2) ROW_NUMBER() is built in. If you want a ROW_NUMBER() include RN:
 SELECT RN, N1 FROM dbo.rangeAB(-3,3,1,1);
 -- (3.3) If you wanted a ROW_NUMBER() that started at 0 you would do this:
 SELECT RN, N1 FROM dbo.rangeAB(-3,3,1,0);

--===== 4. Using N2 and @gap
 -- (4.1) To get 0,10,20,30...100, set @low to 0, @high to 100 and @gap to 10:
 SELECT N1 FROM dbo.rangeAB(0,100,10,1);
 -- (4.2) Note that N2=N1+@gap; this allows you to create a sequence of ranges.
 --       For example, to get (0,10),(10,20),(20,30).... (90,100):
 SELECT N1, N2 FROM dbo.rangeAB(0,90,10,1);
 -- (4.3) Remember that a rownumber is included and it can begin at 0 or 1:
 SELECT RN, N1, N2 FROM dbo.rangeAB(0,90,10,1);

[Examples]:
--===== 1. Generating Sample data (using rangeAB to create "dummy rows")
 -- The query below will generate 10,000 ids and random numbers between 50,000 and 500,000
 SELECT
   someId    = r.rn,
   someNumer = ABS(CHECKSUM(NEWID())%450000)+50001 
 FROM rangeAB(1,10000,1,1) r;

--===== 2. Create a series of dates; rn is 0 to include the first date in the series
 DECLARE @startdate DATE = '20180101', @enddate DATE = '20180131';

 SELECT r.rn, calDate = DATEADD(dd, r.rn, @startdate)
 FROM dbo.rangeAB(1, DATEDIFF(dd,@startdate,@enddate),1,0) r;
 GO

--===== 3. Splitting (tokenizing) a string with fixed sized items
 -- given a delimited string of identifiers that are always 7 characters long
 DECLARE @string VARCHAR(1000) = 'A601225,B435223,G008081,R678567';

 SELECT
   itemNumber = r.rn, -- item's ordinal position 
   itemIndex  = r.n1, -- item's position in the string (it's CHARINDEX value)
   item       = SUBSTRING(@string, r.n1, 7) -- item (token)
 FROM dbo.rangeAB(1, LEN(@string), 8,1)  r;
 GO

--===== 4. Splitting (tokenizing) a string with random delimiters
 DECLARE @string VARCHAR(1000) = 'ABC123,999F,XX,9994443335';

 SELECT
   itemNumber = ROW_NUMBER() OVER (ORDER BY r.rn), -- item's ordinal position 
   itemIndex  = r.n1+1, -- item's position in the string (it's CHARINDEX value)
   item       = SUBSTRING
               (
                 @string,
                 r.n1+1,
                 ISNULL(NULLIF(CHARINDEX(',',@string,r.n1+1),0)-r.n1-1, 8000)
               ) -- item (token)
 FROM dbo.rangeAB(0,DATALENGTH(@string),1,1) r
 WHERE SUBSTRING(@string,r.n1,1) = ',' OR r.n1 = 0;
 -- logic borrowed from: http://www.sqlservercentral.com/articles/Tally+Table/72993/

--===== 5. Grouping by a weekly intervals
 -- 5.1. how to create a series of start/end dates between @startDate & @endDate
 DECLARE @startDate DATE = '1/1/2015', @endDate DATE = '2/1/2015';
 SELECT 
   WeekNbr   = r.RN,
   WeekStart = DATEADD(DAY,r.N1,@StartDate), 
   WeekEnd   = DATEADD(DAY,r.N2-1,@StartDate)
 FROM dbo.rangeAB(0,datediff(DAY,@StartDate,@EndDate),7,1) r;
 GO

 -- 5.2. LEFT JOIN to the weekly interval table
 BEGIN
  DECLARE @startDate datetime = '1/1/2015', @endDate datetime = '2/1/2015';
  -- sample data 
  DECLARE @loans TABLE (loID INT, lockDate DATE);
  INSERT @loans SELECT r.rn, DATEADD(dd, ABS(CHECKSUM(NEWID())%32), @startDate)
  FROM dbo.rangeAB(1,50,1,1) r;

  -- solution 
  SELECT 
    WeekNbr   = r.RN,
    WeekStart = dt.WeekStart, 
    WeekEnd   = dt.WeekEnd,
    total     = COUNT(l.lockDate)
  FROM dbo.rangeAB(0,datediff(DAY,@StartDate,@EndDate),7,1) r
  CROSS APPLY (VALUES (
    CAST(DATEADD(DAY,r.N1,@StartDate) AS DATE), 
    CAST(DATEADD(DAY,r.N2-1,@StartDate) AS DATE))) dt(WeekStart,WeekEnd)
  LEFT JOIN @loans l ON l.lockDate BETWEEN  dt.WeekStart AND dt.WeekEnd
  GROUP BY r.RN, dt.WeekStart, dt.WeekEnd ;
 END;

--===== 6. Identify the first vowel and last vowel in a along with their positions
 DECLARE @string VARCHAR(200) = 'This string has vowels';

 SELECT TOP(1) position = r.rn, letter = SUBSTRING(@string,r.rn,1)
 FROM dbo.rangeAB(1,LEN(@string),1,1) r
 WHERE SUBSTRING(@string,r.rn,1) LIKE '%[aeiou]%'
 ORDER BY r.rn;

 -- To avoid a sort in the execution plan we'll use op instead of rn
 SELECT TOP(1) position = r.op, letter = SUBSTRING(@string,r.op,1)
 FROM dbo.rangeAB(1,LEN(@string),1,1) r
 WHERE SUBSTRING(@string,r.rn,1) LIKE '%[aeiou]%'
 ORDER BY r.rn;

---------------------------------------------------------------------------------------
[Revision History]:
 Rev 00 - 20140518 - Initial Development - Alan Burstein
 Rev 01 - 20151029 - Added 65 rows to make L1=465; 465^3=100.5M. Updated comment section
                   - Alan Burstein
 Rev 02 - 20180613 - Complete re-design including opposite number column (op)
 Rev 03 - 20180920 - Added additional CROSS JOIN to L2 for 530B rows max - Alan Burstein
****************************************************************************************/
RETURNS TABLE WITH SCHEMABINDING AS RETURN
WITH L1(N) AS 
(
  SELECT 1
  FROM (VALUES
   (0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),
   (0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),
   (0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),
   (0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),
   (0),(0)) T(N) -- 90 values 
),
L2(N)  AS (SELECT 1 FROM L1 a CROSS JOIN L1 b CROSS JOIN L1 c),
iTally AS (SELECT rn = ROW_NUMBER() OVER (ORDER BY (SELECT 1)) FROM L2 a CROSS JOIN L2 b)
SELECT  
  r.RN,
  r.OP,
  r.N1,
  r.N2
FROM
(
  SELECT
    RN = 0,
    OP = (@high-@low)/@gap,
    N1 = @low,
    N2 = @gap+@low
  WHERE @row1 = 0
  UNION ALL -- COALESCE required in the TOP statement below for error handling purposes
  SELECT TOP (ABS((COALESCE(@high,0)-COALESCE(@low,0))/COALESCE(@gap,0)+COALESCE(@row1,1)))
    RN = i.rn,
    OP = (@high-@low)/@gap+(2*@row1)-i.rn,
    N1 = (i.rn-@row1)*@gap+@low,
    N2 = (i.rn-(@row1-1))*@gap+@low
  FROM iTally AS i
  ORDER BY rn
) AS r
WHERE @high&@low&@gap&@row1 IS NOT NULL AND @high >= @low AND @gap > 0;

CharMapAB

CREATE FUNCTION dbo.charmapAB
(
  @asciiOnly BIT,
  @xmlCheck  BIT
) 
/*****************************************************************************************
[Purpose]:
 Generates a table containing the numbers 1 through 65535 along with the
 corrsponding CHAR(N) value (e.g. CHAR(65) = "A") and/or UNICODE value (e.g. 
 NCHAR(324) = "ń", aka the Latin minuscule: ń. 

 The ascii_xml_special and unicode_xml_special columns at bits that indicate if 
 the character is an ASCII or UNICODE Reserved XML character. The ascii_xml and 
 unicode_xml columns show what will be displayed when the character is output as
 in XML format (e.g. SELECT CAST('>' AS XML) will return "&gt;". 

 is_ascii_whitespace indicates if the character is a "whitespace character" (such
 as CHAR(9), CHAR(32) and CHAR(160)). abin is the character's ascii binary value 
 and ubin is the characters unicode binary value. 

[Developer Notes]:
 1. Have not determined UNICODE whitespace characters. 

[Examples]:
--===== Get a list of ASCII whitespace characters
  SELECT cm.* -- WhiteSpaceCharacters = 'CHAR('+CAST(n AS varchar(3))+')'
  FROM   dbo.CharmapAB(0,0) AS cm;

  SELECT cm.* -- WhiteSpaceCharacters = 'CHAR('+CAST(n AS varchar(3))+')'
  FROM   dbo.CharmapAB(1,1) AS cm;

  SELECT cm.* -- WhiteSpaceCharacters = 'CHAR('+CAST(n AS varchar(3))+')'
  FROM  dbo.CharmapAB(0,1) AS cm
  WHERE cm.char_nbr IN (9,10,13,32,38,60,62);
-----------------------------------------------------------------------------------------
[Revision History]:
 Rev 00 - May 2015 - Initial Development - Alan Burstein
 Rev 01 - 20150819 changed whitespace val, column names, added quoted_val
        - Alan Burstein
*****************************************************************************************/
RETURNS TABLE WITH SCHEMABINDING AS RETURN
WITH rowz(N) AS (SELECT CASE @asciiOnly WHEN 0 THEN 255 ELSE 65535 END)
SELECT
char_nbr        = i.RN, 
ascii_val       = CHAR(cs.RN),
unicode_val     = u.unicode_val,
quoted_val      = uq.quoted_val,
is_unicode_only = SIGN(i.RN&256),
is_acsii_ws     = CASE WHEN cs.RN IN ((2),(9),(10),(13),(32),(160)) THEN 1 ELSE 0 END,
is_ascii_blank  = CASE WHEN cs.RN BETWEEN 28  AND 31 
                         OR cs.RN BETWEEN 129 AND 159 THEN 1 ELSE 0 END,
unicode_xml_val = x.unicode_xml_val,
bin             = CAST(NCHAR(cs.RN) AS varbinary)
FROM rowz
CROSS APPLY dbo.rangeAB(1,rowz.N,1,1)       AS i
CROSS APPLY (VALUES(CHECKSUM(i.RN)))        AS cs(RN)
CROSS APPLY (SELECT TOP (@xmlCheck*1) NCHAR(cs.RN) 
             WHERE @xmlCheck = 1 
             FOR XML PATH(''))              AS x(unicode_xml_val)
CROSS APPLY (VALUES(NCHAR(cs.RN)))          AS u(unicode_val)  
CROSS APPLY (VALUES('"'+u.unicode_val+'"')) AS uq(quoted_val);

CharmapAB 将帮助您识别哪些字符是 XML:

CharmapAB will help you identify which characters are XML:

如果您运行此查询,您可以识别哪些 ASCII 字符是XML 保护的"

If you run this query you can identify which ASCII characters are "XML Protected"

SELECT cm.*
FROM  dbo.CharmapAB(0,1) AS cm;

返回(为简洁起见被截断)

char_nbr  ascii_val unicode_val quoted_val is_unicode_only      is_acsii_ws is_ascii_blank unicode_xml_val      bin
--------- --------- ----------- ---------- -------------------- ----------- -------------- -------------------- ------
1                             ""        0                    0           0              &#x01;               0x0100
2                             ""        0                    1           0              &#x02;               0x0200
....
32                              " "        0                    1           0              &#x20;               0x2000
33        !         !           "!"        0                    0           0              !                    0x2100
34        "         "           """        0                    0           0              "                    0x2200
35        #         #           "#"        0                    0           0              #                    0x2300
36        $         $           "$"        0                    0           0              $                    0x2400
37        %         %           "%"        0                    0           0              %                    0x2500
38        &         &           "&"        0                    0           0              &amp;                0x2600
39        '         '           "'"        0                    0           0              '                    0x2700
...

我的经验是前 31 个字符从不使用,除了 char(9)、char(10) 和 char(13)(制表符回车和换行).以及 char(32)、char(38)、char(60) 和 char(62),它们是:空格、与号 (&),然后大于和小于(<"和>").此查询可能足以为您提供所需的字符:

My experience has been that the first 31 characters are never used except char(9),char(10) and char(13) (tab carriage return and line returns). As well as char(32),char(38),char(60) and char(62) which are: space, ampersand (&), then greater than and less than ("<" and ">"). This query will likely be enough to get you the characters you need:

DECLARE @yourstring VARCHAR(8000) = 'ABC&amp;123&lt;xxx&gt;'

SELECT REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(@yourstring,
  '&#x09;', CHAR(9)),
  '&#x0A;', CHAR(10)),
  '&#x0D;', CHAR(13)),
  '&#x20;', CHAR(32)),
  '&amp;', CHAR(38)),
  '&lt;', CHAR(60)),
  '&gt;', CHAR(62));

退货:ABC&123

您可以根据需要使用 CharMapAB 进行更新.

You can use CharMapAB to update this as needed.

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