no root element found in xml что это
XML Parsing Error: no root element found Location in Console FF
I’m use ASP.NET MVC end I take this error only in FF. Why I take this error message? What is the cause of this? I don’t understand where the source of this error. Anyone have any thoughts?
9 Answers 9
Check this link for more information
in other words, this is a known Firefox issue
I changed the type of return value in Action method and it worked. More details can be found in my article on this topic.
For more details about status code check this page: List of HTTP status codes
I’ve seen the XML Parsing error: no root element error in a few different situations. It’s typically present in the Firefox developer console only.
Returning a 200 OK response code with an empty body. Your server should instead return a 204 No Content response code.
Returning a 204 No Content response code and then attaching content to the response. Your server should instead return a 200 OK response code.
If none of these fixes your issue, I would review all of your response headers for correctness. Other fields like Content-Length could presumable interfere as well.
Ошибка синтаксического анализа XML: корневой элемент не найден
Поэтому я делаю простое веб-приложение для входа / регистрации, но получаю следующую ошибку:
Я провел некоторое исследование об ошибках разбора XML, но мне все еще не удается заставить мой проект работать, я попробовал с Google Chrome и Firefox
Решение
АГА! Получил это сегодня по причине, которая заставит меня выглядеть довольно глупо, но который может быть однажды помогите кому-нибудь.
Настроив сервер Apache на моем компьютере, с PHP и так далее … Я получил эту ошибку … и затем понял, почему: я нажал на нужный HTML-файл (т. Е. Тот, который содержит Javascript / JQuery), поэтому в адресной строке браузера было показано «file: /// D: /apps/Apache24/htdocs/experiment/forms/index.html».
То, что вам нужно сделать, чтобы фактически использовать сервер Apache (при условии, что он работает и т. Д.), — это перейти «Http: //localhost/experiments/forms/index.html « в адресной строке браузера.
Для смягчения последствий я до сих пор использовал файл «index.php» и просто изменил на файл «index.html». Немного подвох, так как с первым вы должны получить к нему доступ «должным образом», используя localhost.
Другие решения
В Spring MVC Application возникла та же ситуация, что и в объявлении void, и изменив его на String, я решил проблему
Предполагая, что вы работаете с Javascript, вам нужно поместить заголовок перед отображением ваших данных:
Убедитесь, что ваш php-сервер работает и что php-код находится в соответствующей папке. Я столкнулся с этой же проблемой, если php не было там. Я также рекомендую поместить ваш html в ту же папку, чтобы предотвратить возникновение перекрестных ошибок при тестировании.
Если это не проблема, убедитесь, что каждый вызов SQL корректен в php, и что вы используете текущие стандарты php … Php меняется быстро, в отличие от html, css и Javascript, поэтому некоторые функции могут быть устаревшими.
Кроме того, я заметил, что вы, возможно, неправильно собираете свою переменную, что также может вызвать эту ошибку. Если вы отправляете переменные через форму, они должны быть в правильном формате и отправлены POST или GET, в зависимости от ваших предпочтений. Например, если бы у меня была страница входа в лабиринт:
HTML
JavaScript
Было бы полезно, если бы вы включили другие части кода, такие как HTML и JavaScript, так как мне не пришлось бы приводить свой собственный пример, подобный этому. Тем не менее, я надеюсь, что эти указатели помогут!
XML Parsing Error: no root element found Location
So I’m making a simple login/registration web application but I keep getting the following error:
here is my login.php
I’ve done some research about xml parsing errors but I still cant manage to make my project work, ive tried with Google Chrome and Firefox
4 Answers 4
AHA! Got this today for a reason which will make me look pretty silly but which might one day help someone.
Having set up an Apache server on my machine, with PHP and so on. I got this error. and then realised why: I had clicked on the HTML file in question (i.e. the one containing the Javascript/JQuery), so the address bar in the browser showed «file:///D:/apps/Apache24/htdocs/experiments/forms/index.html».
What you have to do to actually use the Apache server (assuming it’s running, etc.) is go «http://localhost/experiments/forms/index.html» in the browser’s address bar.
In mitigation I have up to now been using an «index.php» file and just changed to an «index.html» file. Bit of a gotcha, since with the former you are obliged to access it «properly» using localhost.
I had same situation in Spring MVC Application as it was declared as void, changing it to return String solved the issue
Assuming you are working with javascript, you need to put a header in front of echoing your data:
Make sure you’re php server is running and that the php code is in the appropriate folder. I ran into this same issue if the php was not there. I also recommend putting your html in that same folder to prevent cross-origin errors when testing.
If that is not the issue, ensure that every SQL call is correct in the php, and that you are using current php standards. Php changes quickly, unlike html, css and Javascript, so some functions may be deprecated.
Also, I noticed that you may not be collecting your variable correctly, which can also cause this error. If you are sending variables via form, they need to be in proper format and sent either by POST or GET, based on your preference. For example, if I had a login page for a maze game:
HTML
JavaScript
It would help if you included the other parts of the code such as the html and JavaScript as I wouldn’t have to give my own example like this. However, I hope these pointers help!
Root element is missing
I am reading xml from xxx URl but i am getting error as Root element is missing.
My code to read xml response is as follows:
and the xml response is as follows:
Can you please tell where i am going wrong.
6 Answers 6
Just in case anybody else lands here from Google, I was bitten by this error message when using XDocument.Load(Stream) method.
Make sure the stream position is set to 0 (zero) before you try and load the Stream, its an easy mistake I always overlook!
Make sure you XML looks like this:
Also, a blank XML file will return the same Root elements is missing exception. Each XML file must have a root element / node which encloses all the other elements.
Hi this is odd way but try it once
I try with your code and same XML without adding any XML declaration it works for me
As stated by Phil in below answer please set the xmlStream position to zero if it is not zero.
If you are loading the XML file from a remote location, I would check to see if the file is actually being downloaded correctly using a sniffer like Fiddler.
I wrote a quick console app to run your code and parse the file and it works fine for me.
check the available space on the server. sometimes this is the problem.
I had the same problem when i have trying to read xml that was extracted from archive to memory stream.
Problem was in these lines:
And solution is (Thanks to @Phil):
Not the answer you’re looking for? Browse other questions tagged c# asp.net xml or ask your own question.
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Cannot find root element in XML file
(1) The XML for which querying root returns null:
The thing is that I would very much like to be able to read in XML as in
(1), to allow clients to validate their XML before publishing. For instance
in XmlSpy I can verify the XML content to be valid before publishing it. Of
course verification is no longer necessary once my program tries to read it,
it is assumed that published XML should be valid.
Kind regards,
—
Tom Tempelaere.
TT (Tom Tempelaere) wrote:
Well if you simply want to access the root element then
xmlDoc.DocumentElement
will do whatever the name of the element is and whatever namespace that
element has (where the namespace in your XML sample is the reason that
your SelectSingleNode does not find the element).
Thus if you need to root element simply use
XmlElement rootElement = xmlDoc.DocumentElement;
I don’t think it makes sense to use XPath and SelectSingleNode to look
for something the object model gives as a property.
As for SelectSingleNode to find an element with XPath in a certain
namespace you would need e.g.
XmlNamespaceManager namespaceManager = new
XmlNamespaceManager(xmlDoc.NameTable);
namespaceManager.AddNamespace(«pf», «http://tempuri.org/XMLSchema.xsd»);
XmlElement rootElement =
xmlDoc.SelectSingleNode(«pf:IncidentDefinitions», namespaceManager) as
XmlElement;
Thanks Martin, that helped.
Kind regards,
—
Tom Tempelaere.
I may have cried Victory a little too early. None of your alternative
suggestions works (I should have tested first). See inline for some comment.
«Martin Honnen» wrote:
Well if you simply want to access the root element then
xmlDoc.DocumentElement
will do whatever the name of the element is and whatever namespace that
element has (where the namespace in your XML sample is the reason that
your SelectSingleNode does not find the element).
Thus if you need to root element simply use
XmlElement rootElement = xmlDoc.DocumentElement;
I don’t think it makes sense to use XPath and SelectSingleNode to look
for something the object model gives as a property.
As for SelectSingleNode to find an element with XPath in a certain
namespace you would need e.g.
XmlNamespaceManager namespaceManager = new
XmlNamespaceManager(xmlDoc.NameTable);
namespaceManager.AddNamespace(«pf», «http://tempuri.org/XMLSchema.xsd»);
XmlElement rootElement =
xmlDoc.SelectSingleNode(«pf:IncidentDefinitions», namespaceManager) as
XmlElement;
Doesn’t work either. It does return an element, but querying for any of the
documents nodes fails.
In the meanwhile I’ll have to ask my clients to not publish any XML files
with attributes for the root element.
Kind regards,
—
Tom Tempelaere.
Maybe it isn’t very obvious, but in the xsi:schemaLocation attribute there
is a space between the two strings in it «http://tempuri.org/XMLSchema.xsd»
and «C:/IncidentDefinitions.xsd». I thought that was not valid for XML to do,
but XmlSpy generates it that way. Could that be the problem?
Kind regards,
Tom Tempelaere.
TT (Tom Tempelaere) wrote:
Well to be honest I thought this would solve the problem, but it actually
doesn’t. If instead of querying the ‘IncidentDefinitions’ root element I use
xmlDocument.DocumentElement I still have a problem: I can’t find any other
nodes that are in the document.
Doesn’t work either. It does return an element, but querying for any of the
documents nodes fails.
If those child or descendant elements are in the namespace
http://tempuri.org/XMLSchema.xsd then you again need to use the
namespace manager and you need to use the prefix in your XPath
expressions, that is how XPath works. I don’t understand why you claim
the suggested ways don’t work if you say they give you the element. If
you then write additional code that does not do what you want then that
additional code needs to be fixed obviously. Use that namespace manager,
use that prefix in your XPath expressions to qualify element names (e.g.
pf:elementname), pass the namespace manager to the SelectSingleNode or
SelectNodes method calls and if you have the proper path then you will
find nodes.
—
Well to be honest I thought this would solve the problem, but it actually
doesn’t. If instead of querying the ‘IncidentDefinitions’ root element I use
xmlDocument.DocumentElement I still have a problem: I can’t find any other
nodes that are in the document.
>As for SelectSingleNode to find an element with XPath in a certain
namespace you would need e.g.
XmlNamespaceManager namespaceManager = new
XmlNamespaceManager(xmlDoc.NameTable);
namespaceManager.AddNamespace(«pf», «http://tempuri.org/XMLSchema.xsd»);
XmlElement rootElement =
xmlDoc.SelectSingleNode(«pf:IncidentDefinitions «, namespaceManager) as
XmlElement;
Doesn’t work either. It does return an element, but querying for any of the
documents nodes fails.
If those child or descendant elements are in the namespace
http://tempuri.org/XMLSchema.xsd then you again need to use the
namespace manager and you need to use the prefix in your XPath
expressions, that is how XPath works. I don’t understand why you claim
the suggested ways don’t work if you say they give you the element.
They don’t work because from what I understood I only needed the namespace
thingy for querying the root node. The nodes in the XML file aren’t qualified
so why would I have to qualify them to look them up?
Makes me wonder how I support both types of XML files, those with attributed
root and those without.
Anyway I don’t really understand what it means: «If those child or
descendant elements are in the namespace
http://tempuri.org/XMLSchema.xsd. «. I’m far from being an expert in XML
(hence my questions) but I would say that the nodes that I defined in my
schema aren’t «child» or «descendant» elements in the namespace
«http://tempuri.org/XMLSchema.xsd». Or am I just not seeing things correctly?
If
you then write additional code that does not do what you want then that
additional code needs to be fixed obviously. Use that namespace manager,
use that prefix in your XPath expressions to qualify element names (e.g.
pf:elementname), pass the namespace manager to the SelectSingleNode or
SelectNodes method calls and if you have the proper path then you will
find nodes.
Well my code that reads in the XML works fine if the root node doesn’t have
the attributes, so my code isn’t wrong per se.
Anyway, since I’m too noob to understand your answers, I’ll go the easy way.
Away with those stupid attributes.