on 2014 Jun 05 7:05 AM
So I've inherited this database with some invalid xml data in one table.
I know for a fact that the character is a control char and I can find it in the column (which is of type XML):
SELECT CHARINDEX(char(26), BLA_COLUMN) FROM BLA_TABLE where ID_RECORD = 1234
Returns a valid index.
Now, when I try to replace that char with something else:
UPDATE BLA_TABLE SET BLA_COLUMN = REPLACE(BLA_COLUMN, char(26), char(32)) WHERE ID_RECORD = 1234
The statement fails with message: "XML parser error: character: 603, line:1, column:603 Illegal control character SQLCODE=-888, ODBC 3 State="HY000"
Any ideas how to get rid of that character? (Running SQLA 12)
Thank you!
EDIT
Looks like Volker's suggestion to first update the BLA_COLUMN to a Null value then to update to a valid XML, replacing char(26) with char (32), provides a workaround. Thanks Volker!
Request clarification before answering.
Is the same true if you try to set the column to null (which would look like the previous contents is parsed, too)...?
EDIT: As this seems to have done the trick, it shows that
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Setting the column to Null first, then updating it with a valid XML (ie char 26 replaced with char 32) seems to do the trick.
I had a go reproducing the problem (on 10.0.1 and 16.0) but couldn't - I could select and update the bad XML without an issue - so I was wondering what was causing the problem. Anyway, you have a work around so that's great.
create table TestXML(PK int default autoincrement, XMLStuff xml, primary key(PK));
insert into TestXML(XMLStuff) values ('<Bad XML>');
insert into TestXML(XMLStuff) values ('<GoodXML/>');
insert into TestXML(XMLStuff) values (string('<ReallyBadXML',char(26),'>'));
insert into TestXML(XMLStuff) values (string('<EvenWorseXML',char(26)));
select * from TestXML where PK = 1;
-- works OK
update TestXML set XMLStuff = '<BetterXML/>' where PK = 1;
-- works OK
select * from TestXML where PK = 3;
-- works OK
update TestXML set XMLStuff = '<BetterXML/>' where PK = 3;
-- works OK
select * from TestXML where PK = 4;
-- works OK
update TestXML set XMLStuff = '<BetterXML/>' where PK = 4;
-- works OK
drop table TestXML;
12.0.1.4085 shows the same success.
Also trying to turn a valid XML into an invalid one doesn't raise an error, such as:
update TestXML set XMLStuff =
replace(XMLStuff, char(47), char(26)) where PK = 2;
-- works OK
SQL Anywhere only validates XML when parsing it (for example, using OPENXML). It is not validated when inserting or updating a value, or casting to the XML type. I wonder if the original database has a trigger, computed column, or check constraint that ends up parsing the XML.
You guys are right. Having looked more carefully at the table, I've notice the presence of a trigger that indeed does some parsing on update (disabling the trigger allowed the updates of fixing the xml just fine). Can't believe that didn't cross my mind! So there you go, case closed. Thank you!
I think you need to convert the xml data to something that SQLA doesn't try to interpret.
eg (untested)
create variable mytext long varchar; set mytext = (SELECT (BLA_COLUMN) FROM BLA_TABLE where ID_RECORD = 1234); set mytext= REPLACE(mytext, char(26), char(32)); update BLA_TABLE set BLA_COLUMN = mytext where ID_RECORD = 1234; commit;
If you have to do this a lot you could write a user defined function. Also if you have characters which have problems because of collation tables eg
update person set notes=replace(notes,'ú','£')and every u in the field will also be changed as the collation sequence helpfully reckons that 'ú' is the same as 'u', you can use this approach:
create function BinaryReplace(in x long varchar,in targetascii smallint, in replacementascii smallint) returns long varchar deterministic begin declare rv long varchar; declare l integer; declare i integer; declare c char(1); set l=length(x); set i=0; set rv=''; while i < l loop set i=i+1; set c=substr(x,i,1); if ascii(c) = targetascii then set c=char(replacementascii) end if; set rv=rv+c end loop; return rv end;
not fast, but quicker than typing!
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