将所有 XML 特殊字符转换回常规字符(在 SQL 中)
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 & xyz ><'
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)),'&','&'),'>','>'),'<','<');
--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 & xyz ><';
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 ">".
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  0x0100
2 "" 0 1 0  0x0200
....
32 " " 0 1 0   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 & 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&123<xxx>'
SELECT REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(@yourstring,
'	', CHAR(9)),
'
', CHAR(10)),
'
', CHAR(13)),
' ', CHAR(32)),
'&', CHAR(38)),
'<', CHAR(60)),
'>', 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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