Trouble getting XML converted into variable in C# - c#

I am trying to get an XML from Swagger. Let's just say the XML is this:
<note>
<to>Tove</to>
<from>Jani</from>
<heading>Reminder</heading>
<body>Don't forget me this weekend!</body>
</note>
(HOwever, in reality my XML is MUCH larger). So if the above is being returned from swagger, how would I store these exact results in a variable in C#? Let me know if you need more detail. In theory. I would love to just be able to capture all this data and store it in a string variable or XMLDocument variable. Thanks!
I have tried doing storing it in a XML document and even a string, but the part I am having trouble with is capturing the XML from Swagger.

Related

how to skip invalid xml nodes when reading with xmlreader? [duplicate]

Currently, I'm working on a feature that involves parsing XML that we receive from another product. I decided to run some tests against some actual customer data, and it looks like the other product is allowing input from users that should be considered invalid. Anyways, I still have to try and figure out a way to parse it. We're using javax.xml.parsers.DocumentBuilder and I'm getting an error on input that looks like the following.
<xml>
...
<description>Example:Description:<THIS-IS-PART-OF-DESCRIPTION></description>
...
</xml>
As you can tell, the description has what appears to be an invalid tag inside of it (<THIS-IS-PART-OF-DESCRIPTION>). Now, this description tag is known to be a leaf tag and shouldn't have any nested tags inside of it. Regardless, this is still an issue and yields an exception on DocumentBuilder.parse(...)
I know this is invalid XML, but it's predictably invalid. Any ideas on a way to parse such input?
That "XML" is worse than invalid – it's not well-formed; see Well Formed vs Valid XML.
An informal assessment of the predictability of the transgressions does not help. That textual data is not XML. No conformant XML tools or libraries can help you process it.
Options, most desirable first:
Have the provider fix the problem on their end. Demand well-formed XML. (Technically the phrase well-formed XML is redundant but may be useful for emphasis.)
Use a tolerant markup parser to cleanup the problem ahead of parsing as XML:
Standalone: xmlstarlet has robust recovering and repair capabilities credit: RomanPerekhrest
xmlstarlet fo -o -R -H -D bad.xml 2>/dev/null
Standalone and C/C++: HTML Tidy works with XML too. Taggle is a port of TagSoup to C++.
Python: Beautiful Soup is Python-based. See notes in the Differences between parsers section. See also answers to this question for more
suggestions for dealing with not-well-formed markup in Python,
including especially lxml's recover=True option.
See also this answer for how to use codecs.EncodedFile() to cleanup illegal characters.
Java: TagSoup and JSoup focus on HTML. FilterInputStream can be used for preprocessing cleanup.
.NET:
XmlReaderSettings.CheckCharacters can
be disabled to get past illegal XML character problems.
#jdweng notes that XmlReaderSettings.ConformanceLevel can be set to
ConformanceLevel.Fragment so that XmlReader can read XML Well-Formed Parsed Entities lacking a root element.
#jdweng also reports that XmlReader.ReadToFollowing() can sometimes
be used to work-around XML syntactical issues, but note
rule-breaking warning in #3 below.
Microsoft.Language.Xml.XMLParser is said to be “error-tolerant”.
Go: Set Decoder.Strict to false as shown in this example by #chuckx.
PHP: See DOMDocument::$recover and libxml_use_internal_errors(true). See nice example here.
Ruby: Nokogiri supports “Gentle Well-Formedness”.
R: See htmlTreeParse() for fault-tolerant markup parsing in R.
Perl: See XML::Liberal, a "super liberal XML parser that parses broken XML."
Process the data as text manually using a text editor or
programmatically using character/string functions. Doing this
programmatically can range from tricky to impossible as
what appears to be
predictable often is not -- rule breaking is rarely bound by rules.
For invalid character errors, use regex to remove/replace invalid characters:
PHP: preg_replace('/[^\x{0009}\x{000a}\x{000d}\x{0020}-\x{D7FF}\x{E000}-\x{FFFD}]+/u', ' ', $s);
Ruby: string.tr("^\u{0009}\u{000a}\u{000d}\u{0020}-\u{D7FF}\u{E000‌​}-\u{FFFD}", ' ')
JavaScript: inputStr.replace(/[^\x09\x0A\x0D\x20-\xFF\x85\xA0-\uD7FF\uE000-\uFDCF\uFDE0-\uFFFD]/gm, '')
For ampersands, use regex to replace matches with &: credit: blhsin, demo
&(?!(?:#\d+|#x[0-9a-f]+|\w+);)
Note that the above regular expressions won't take comments or CDATA
sections into account.
A standard XML parser will NEVER accept invalid XML, by design.
Your only option is to pre-process the input to remove the "predictably invalid" content, or wrap it in CDATA, prior to parsing it.
The accepted answer is good advice, and contains very useful links.
I'd like to add that this, and many other cases of not-wellformed and/or DTD-invalid XML can be repaired using SGML, the ISO-standardized superset of HTML and XML. In your case, what works is to declare the bogus THIS-IS-PART-OF-DESCRIPTION element as SGML empty element and then use eg. the osx program (part of the OpenSP/OpenJade SGML package) to convert it to XML. For example, if you supply the following to osx
<!DOCTYPE xml [
<!ELEMENT xml - - ANY>
<!ELEMENT description - - ANY>
<!ELEMENT THIS-IS-PART-OF-DESCRIPTION - - EMPTY>
]>
<xml>
<description>blah blah
<THIS-IS-PART-OF-DESCRIPTION>
</description>
</xml>
it will output well-formed XML for further processing with the XML tools of your choice.
Note, however, that your example snippet has another problem in that element names starting with the letters xml or XML or Xml etc. are reserved in XML, and won't be accepted by conforming XML parsers.
IMO these cases should be solved by using JSoup.
Below is a not-really answer for this specific case, but found this on the web (thanks to inuyasha82 on Coderwall). This code bit did inspire me for another similar problem while dealing with malformed XMLs, so I share it here.
Please do not edit what is below, as it is as it on the original website.
The XML format, requires to be valid a unique root element declared in the document.
So for example a valid xml is:
<root>
<element>...</element>
<element>...</element>
</root>
But if you have a document like:
<element>...</element>
<element>...</element>
<element>...</element>
<element>...</element>
This will be considered a malformed XML, so many xml parsers just throw an Exception complaining about no root element. Etc.
In this example there is a solution on how to solve that problem and succesfully parse the malformed xml above.
Basically what we will do is to add programmatically a root element.
So first of all you have to open the resource that contains your "malformed" xml (i. e. a file):
File file = new File(pathtofile);
Then open a FileInputStream:
FileInputStream fis = new FileInputStream(file);
If we try to parse this stream with any XML library at that point we will raise the malformed document Exception.
Now we create a list of InputStream objects with three lements:
A ByteIputStream element that contains the string: <root>
Our FileInputStream
A ByteInputStream with the string: </root>
So the code is:
List<InputStream> streams =
Arrays.asList(
new ByteArrayInputStream("<root>".getBytes()),
fis,
new ByteArrayInputStream("</root>".getBytes()));
Now using a SequenceInputStream, we create a container for the List created above:
InputStream cntr =
new SequenceInputStream(Collections.enumeration(str));
Now we can use any XML Parser library, on the cntr, and it will be parsed without any problem. (Checked with Stax library);

Convert string to valid XML [duplicate]

Currently, I'm working on a feature that involves parsing XML that we receive from another product. I decided to run some tests against some actual customer data, and it looks like the other product is allowing input from users that should be considered invalid. Anyways, I still have to try and figure out a way to parse it. We're using javax.xml.parsers.DocumentBuilder and I'm getting an error on input that looks like the following.
<xml>
...
<description>Example:Description:<THIS-IS-PART-OF-DESCRIPTION></description>
...
</xml>
As you can tell, the description has what appears to be an invalid tag inside of it (<THIS-IS-PART-OF-DESCRIPTION>). Now, this description tag is known to be a leaf tag and shouldn't have any nested tags inside of it. Regardless, this is still an issue and yields an exception on DocumentBuilder.parse(...)
I know this is invalid XML, but it's predictably invalid. Any ideas on a way to parse such input?
That "XML" is worse than invalid – it's not well-formed; see Well Formed vs Valid XML.
An informal assessment of the predictability of the transgressions does not help. That textual data is not XML. No conformant XML tools or libraries can help you process it.
Options, most desirable first:
Have the provider fix the problem on their end. Demand well-formed XML. (Technically the phrase well-formed XML is redundant but may be useful for emphasis.)
Use a tolerant markup parser to cleanup the problem ahead of parsing as XML:
Standalone: xmlstarlet has robust recovering and repair capabilities credit: RomanPerekhrest
xmlstarlet fo -o -R -H -D bad.xml 2>/dev/null
Standalone and C/C++: HTML Tidy works with XML too. Taggle is a port of TagSoup to C++.
Python: Beautiful Soup is Python-based. See notes in the Differences between parsers section. See also answers to this question for more
suggestions for dealing with not-well-formed markup in Python,
including especially lxml's recover=True option.
See also this answer for how to use codecs.EncodedFile() to cleanup illegal characters.
Java: TagSoup and JSoup focus on HTML. FilterInputStream can be used for preprocessing cleanup.
.NET:
XmlReaderSettings.CheckCharacters can
be disabled to get past illegal XML character problems.
#jdweng notes that XmlReaderSettings.ConformanceLevel can be set to
ConformanceLevel.Fragment so that XmlReader can read XML Well-Formed Parsed Entities lacking a root element.
#jdweng also reports that XmlReader.ReadToFollowing() can sometimes
be used to work-around XML syntactical issues, but note
rule-breaking warning in #3 below.
Microsoft.Language.Xml.XMLParser is said to be “error-tolerant”.
Go: Set Decoder.Strict to false as shown in this example by #chuckx.
PHP: See DOMDocument::$recover and libxml_use_internal_errors(true). See nice example here.
Ruby: Nokogiri supports “Gentle Well-Formedness”.
R: See htmlTreeParse() for fault-tolerant markup parsing in R.
Perl: See XML::Liberal, a "super liberal XML parser that parses broken XML."
Process the data as text manually using a text editor or
programmatically using character/string functions. Doing this
programmatically can range from tricky to impossible as
what appears to be
predictable often is not -- rule breaking is rarely bound by rules.
For invalid character errors, use regex to remove/replace invalid characters:
PHP: preg_replace('/[^\x{0009}\x{000a}\x{000d}\x{0020}-\x{D7FF}\x{E000}-\x{FFFD}]+/u', ' ', $s);
Ruby: string.tr("^\u{0009}\u{000a}\u{000d}\u{0020}-\u{D7FF}\u{E000‌​}-\u{FFFD}", ' ')
JavaScript: inputStr.replace(/[^\x09\x0A\x0D\x20-\xFF\x85\xA0-\uD7FF\uE000-\uFDCF\uFDE0-\uFFFD]/gm, '')
For ampersands, use regex to replace matches with &: credit: blhsin, demo
&(?!(?:#\d+|#x[0-9a-f]+|\w+);)
Note that the above regular expressions won't take comments or CDATA
sections into account.
A standard XML parser will NEVER accept invalid XML, by design.
Your only option is to pre-process the input to remove the "predictably invalid" content, or wrap it in CDATA, prior to parsing it.
The accepted answer is good advice, and contains very useful links.
I'd like to add that this, and many other cases of not-wellformed and/or DTD-invalid XML can be repaired using SGML, the ISO-standardized superset of HTML and XML. In your case, what works is to declare the bogus THIS-IS-PART-OF-DESCRIPTION element as SGML empty element and then use eg. the osx program (part of the OpenSP/OpenJade SGML package) to convert it to XML. For example, if you supply the following to osx
<!DOCTYPE xml [
<!ELEMENT xml - - ANY>
<!ELEMENT description - - ANY>
<!ELEMENT THIS-IS-PART-OF-DESCRIPTION - - EMPTY>
]>
<xml>
<description>blah blah
<THIS-IS-PART-OF-DESCRIPTION>
</description>
</xml>
it will output well-formed XML for further processing with the XML tools of your choice.
Note, however, that your example snippet has another problem in that element names starting with the letters xml or XML or Xml etc. are reserved in XML, and won't be accepted by conforming XML parsers.
IMO these cases should be solved by using JSoup.
Below is a not-really answer for this specific case, but found this on the web (thanks to inuyasha82 on Coderwall). This code bit did inspire me for another similar problem while dealing with malformed XMLs, so I share it here.
Please do not edit what is below, as it is as it on the original website.
The XML format, requires to be valid a unique root element declared in the document.
So for example a valid xml is:
<root>
<element>...</element>
<element>...</element>
</root>
But if you have a document like:
<element>...</element>
<element>...</element>
<element>...</element>
<element>...</element>
This will be considered a malformed XML, so many xml parsers just throw an Exception complaining about no root element. Etc.
In this example there is a solution on how to solve that problem and succesfully parse the malformed xml above.
Basically what we will do is to add programmatically a root element.
So first of all you have to open the resource that contains your "malformed" xml (i. e. a file):
File file = new File(pathtofile);
Then open a FileInputStream:
FileInputStream fis = new FileInputStream(file);
If we try to parse this stream with any XML library at that point we will raise the malformed document Exception.
Now we create a list of InputStream objects with three lements:
A ByteIputStream element that contains the string: <root>
Our FileInputStream
A ByteInputStream with the string: </root>
So the code is:
List<InputStream> streams =
Arrays.asList(
new ByteArrayInputStream("<root>".getBytes()),
fis,
new ByteArrayInputStream("</root>".getBytes()));
Now using a SequenceInputStream, we create a container for the List created above:
InputStream cntr =
new SequenceInputStream(Collections.enumeration(str));
Now we can use any XML Parser library, on the cntr, and it will be parsed without any problem. (Checked with Stax library);

How to parse an xml that has non-xml data in it

I am working with some xml in C# and am having some issues parsing an xml file due to the format it is in. It has non xml data in the file and I have no control over the format of this file. The file is "test.xml"(see below). I am only concerned with the xml portion of the data, but am unsure the best way to go about accessing it. Any thoughts or recommendations would be greatly appreciated.
Test data -1
Smith, 2234
##*j
Random--
#<?xml version="1.0" encoding="utf-16"?>
<ConfigMessage xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns="http://www.Test.com/schemas/Test.test.Config">
<Config>
<Version>10</Version>
<Build>00520</Build>
<EnableV>false</EnableV>
<BuildL>22</BuildL>
<BuildP>\\testpath\test</BuildP>
</Config>
</ConfigMessage>
#
Put the whole file into a string that contains anything within the first '<' and the last '>' characters detected on the file. Then you can treat it as normal XML from there. If there's random non-XML elements throughout it though you will need to add additional logic to detect starting/stopping XML "blocks".
I can suggest you such solution: open your pseudo-xml like simple text-file, read whole text, after that, with using regex you ought to take xml document (part of primordial document that is able to be converted to XML [|startTag|any symbols|/endTag|]), put it into XDocument (in memory) and now parse it like XML-file.

Is there a XmlReader method similar to the XmlWriter.WriteRaw method?

XmlWriter.WriteRaw will preserve &apos; and not send an actual apostrophe. Is there a method to read in &apos; and keep it as such?
You need to encode it properly. Let's take for example the following XML:
<root>&apos;</root>
The value of the <root> node is ' no matter which XML parser you use to read this XML.
On the other hand if you have the following XML:
<root>&apos;</root>
the value of the <root> node is &apos;.
So in both cases we have properly encoded XML so that when a standard compliant parser reads it, it is able to correctly retrieve the value.
So be very careful when using the WriteRaw method when generating the XML. Since it properly encode the argument it is now your responsibility to ensure that you are passing correct data to it.

Parsing .plist Files to plain XML C#

I'm trying to read my Apple Safari history with c#, which is stored in a plist file, however I always get an error and I'm not sure what the correct way is to do it.
The code I tried to execute is this:
XmlDocument xmd = new XmlDocument();
xmd.LoadXml(#"C:\Users\Oran\AppData\Roaming\AppleComputer\Safari\History.plist");
and I always get the following error:
"Data at the root level is invalid. Line 1, position 1."
Does anyone know whats wrong with this code and recommend what is the best way to read plist files?
It looks like that Apple Safari history.plist is binary plist. I've found a great project:
https://github.com/animetrics/PlistCS
From the readme:
This is a C# Property List (plist) serialization library (MIT
license). It supports both XML and binary versions of the plist
format.
try this and everyhing should be fine ;-)
xmd.Load(...)
The one you have used loads the xml data from a string not from a file.
A plist doesn't have to be XML. There are four different serialization methods — old-style (for NeXT; no longer used), XML, binary and JSON (new in 10.7). Safari's History.plist is most likely binary, for efficiency reasons.
If I'm not mistaken, Safari for Windows does ship with plutil.exe in Common Files\Apple Application Support. You can use that like plutil -convert xml1 SOME_FILE.plist to convert your file.
The problem is with the second line, saying
<!DOCTYPE plist PUBLIC "-//Apple//DTD PLIST 1.0//EN" "http://www.apple.com/DTDs/PropertyList-1.0.dtd">
Option 1. Remove it before parsing.
Option 2. Read the MSDN on
"XmlDocument.XmlResolver Property" and figure out how to make the
XmlDocument download, parse and use the DTD from the URI specified in the XML.

Categories

Resources