TRUENEUTRAL
asked on
XMLException
Why am I getting this? It's maddening.
System.Xml.XmlException: The root element is missing.
at System.Xml.XmlTextReader.R
at System.Xml.XmlReader.MoveT
at System.Xml.XmlReader.IsSta
at System.Xml.Serialization.T
at System.Xml.Serialization.X
at eConnect80_Serializer.Seri
The program that is generating this is very complicated, so I put together a simple sample program to demonstrate the error.
Here is my function:
Public Function My_Deserialize(ByVal xmltext As String) As String
Try
Dim myobj As New MyType
Dim ser As New Xml.Serialization.XmlSeria
Dim st As New System.IO.MemoryStream
Dim writer As New Xml.XmlTextWriter(st, New System.Text.UTF8Encoding)
writer.WriteString(xmltext
st.Position = 0
Dim reader As New Xml.XmlTextReader(st)
reader.Normalization = True
st.Position = 0
If ser.CanDeserialize(reader)
myobj = ser.Deserialize(reader)
Else
MY_Deserialize = "Can't deserialize"
End If
Catch ex As Exception
MY_Deserialize = ex.ToString
End Try
End Function
'Here is my classes for the serializer
'the following classes should define a document to look like
'<MyType><InnerClass><TEST
<System.Xml.Serialization.
Public Class MyType
<System.Xml.Serialization.
Public InnerClass() As InnerClassType
End Class
Public Class InnerClassType
<System.Xml.Serialization.
Public TEST As String
End Class
When I pass it the string:
<MyType><InnerClass><TEST>
I get the "root element missing" error. Please correct me if I am wrong, but <MyType> *is* the root element here. Does anyone now what this error really means?
There is no inner exception on this one.
System.Xml.XmlException: The root element is missing.
at System.Xml.XmlTextReader.R
at System.Xml.XmlReader.MoveT
at System.Xml.XmlReader.IsSta
at System.Xml.Serialization.T
at System.Xml.Serialization.X
at eConnect80_Serializer.Seri
The program that is generating this is very complicated, so I put together a simple sample program to demonstrate the error.
Here is my function:
Public Function My_Deserialize(ByVal xmltext As String) As String
Try
Dim myobj As New MyType
Dim ser As New Xml.Serialization.XmlSeria
Dim st As New System.IO.MemoryStream
Dim writer As New Xml.XmlTextWriter(st, New System.Text.UTF8Encoding)
writer.WriteString(xmltext
st.Position = 0
Dim reader As New Xml.XmlTextReader(st)
reader.Normalization = True
st.Position = 0
If ser.CanDeserialize(reader)
myobj = ser.Deserialize(reader)
Else
MY_Deserialize = "Can't deserialize"
End If
Catch ex As Exception
MY_Deserialize = ex.ToString
End Try
End Function
'Here is my classes for the serializer
'the following classes should define a document to look like
'<MyType><InnerClass><TEST
<System.Xml.Serialization.
Public Class MyType
<System.Xml.Serialization.
Public InnerClass() As InnerClassType
End Class
Public Class InnerClassType
<System.Xml.Serialization.
Public TEST As String
End Class
When I pass it the string:
<MyType><InnerClass><TEST>
I get the "root element missing" error. Please correct me if I am wrong, but <MyType> *is* the root element here. Does anyone now what this error really means?
There is no inner exception on this one.
ASKER
Hmmm...
As you can see from my example, the XML is very simple (and valid)
Regardless of whether I use this
<?xml version='1.0'?><MyType><In
or this
<?xml version="1.0" encoding="ISO-8859-1"?><My
I get the same exact error
As you can see from my example, the XML is very simple (and valid)
Regardless of whether I use this
<?xml version='1.0'?><MyType><In
or this
<?xml version="1.0" encoding="ISO-8859-1"?><My
I get the same exact error
Which line is 58? My guess is myobj = ser.Deserialize(reader). I am going to assume also that you are deserializing an object that you have previously serialized.
What would be interesting to know is how important is this information to the deserialization process:
xmlns:xsd="http://www.w3.org/2001/XMLSchema"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
This is taken from a Micro$oft example:
HOW TO: Serialize and Deserialize XML in Visual Basic .NET
http://support.microsoft.com/kb/316730/EN-US/#3
<?xml version="1.0" encoding="utf-8"?>
<clsProduct xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<Name>Widget</Name>
<Description>Faster, better, cheaper</Description>
<Quantity>5</Quantity>
</clsProduct>
Bob
What would be interesting to know is how important is this information to the deserialization process:
xmlns:xsd="http://www.w3.org/2001/XMLSchema"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
This is taken from a Micro$oft example:
HOW TO: Serialize and Deserialize XML in Visual Basic .NET
http://support.microsoft.com/kb/316730/EN-US/#3
<?xml version="1.0" encoding="utf-8"?>
<clsProduct xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<Name>Widget</Name>
<Description>Faster, better, cheaper</Description>
<Quantity>5</Quantity>
</clsProduct>
Bob
ASKER
Hey LearnedOne! Happy Holidays!
Line 58 is
If ser.CanDeserialize(reader)
On your suggestion I tried this
<?xml version='1.0'?><MyType xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"><InnerClass><
Still get the same error.
Also, may not necessarily be xml that I have previously serialized. I am trying to deserialize it and catching the error in order to validate the structure. It no errors happen, I am deserializing it into an object to do further validation on the values. If all passes, the original xml document will be sent on to another program to process.
Line 58 is
If ser.CanDeserialize(reader)
On your suggestion I tried this
<?xml version='1.0'?><MyType xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"><InnerClass><
Still get the same error.
Also, may not necessarily be xml that I have previously serialized. I am trying to deserialize it and catching the error in order to validate the structure. It no errors happen, I am deserializing it into an object to do further validation on the values. If all passes, the original xml document will be sent on to another program to process.
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ASKER
LearnedOne:
I noticed your example used a hard file rather than a memory stream. This got me thinking that maybe there was an issue with the reader using a memory stream to deserialize. Perhaps it was not recognizing the stream as xml?
So I modified the code to the following:
Dim myobj As New MyType
Dim ser As New Xml.Serialization.XmlSeria
Dim st As New System.IO.MemoryStream
Dim writer As New Xml.XmlTextWriter(st, New System.Text.UTF8Encoding)
writer.WriteString(xmltext
st.Position = 0
'begin change
Dim context As Xml.XmlParserContext
context = New Xml.XmlParserContext(Nothi
Dim reader As New Xml.XmlTextReader(xmltext,
'end change
reader.Normalization = True
st.Position = 0
If ser.CanDeserialize(reader)
myobj = ser.Deserialize(reader)
Else
MY_Deserialize = "Can't deserialize"
End If
This gives me a way to tell the reader that this is an xml document without having to specify any details about the document.
I seems to work fine this way!
Thanks for your example! It got me thinking on the right track.
-TN
I noticed your example used a hard file rather than a memory stream. This got me thinking that maybe there was an issue with the reader using a memory stream to deserialize. Perhaps it was not recognizing the stream as xml?
So I modified the code to the following:
Dim myobj As New MyType
Dim ser As New Xml.Serialization.XmlSeria
Dim st As New System.IO.MemoryStream
Dim writer As New Xml.XmlTextWriter(st, New System.Text.UTF8Encoding)
writer.WriteString(xmltext
st.Position = 0
'begin change
Dim context As Xml.XmlParserContext
context = New Xml.XmlParserContext(Nothi
Dim reader As New Xml.XmlTextReader(xmltext,
'end change
reader.Normalization = True
st.Position = 0
If ser.CanDeserialize(reader)
myobj = ser.Deserialize(reader)
Else
MY_Deserialize = "Can't deserialize"
End If
This gives me a way to tell the reader that this is an xml document without having to specify any details about the document.
I seems to work fine this way!
Thanks for your example! It got me thinking on the right track.
-TN
ASKER
LearnedOne:
Please post in these threads to pick up the rest of the points.
https://www.experts-exchange.com/questions/21223988/Easy-points-for-a-true-guru.html
https://www.experts-exchange.com/questions/21219538/125-point-question-on-xmlexception.html
Please post in these threads to pick up the rest of the points.
https://www.experts-exchange.com/questions/21223988/Easy-points-for-a-true-guru.html
https://www.experts-exchange.com/questions/21219538/125-point-question-on-xmlexception.html
<?xml version="1.0" encoding="ISO-8859-1"?>
in any case you may use this to validate your xml -
http://www.ltg.ed.ac.uk/~richard/xml-check.html
hope this helps.