xml-parsing

How to parse xml string with xmlpullparser android

无人久伴 提交于 2019-12-05 09:39:30
问题 I want to parse an XML string and display it in a EditText but I can't, I not understand what could be the problem, please a bit help, my code : private String xmlc = "<game><cel>5</cel><val>2</val></game>"; private CharSequence readXML(String xmlc2) throws XmlPullParserException { XmlPullParserFactory factory = XmlPullParserFactory.newInstance(); XmlPullParser xpp = factory.newPullParser(); String results = ""; String celda = ""; String valor = ""; xpp.setInput(new StringReader (xmlc2)); int

csproj: how to get all resources?

别说谁变了你拦得住时间么 提交于 2019-12-05 09:26:16
问题 How to get all recources,wich copies to created binary? I think that it is all elements like this (has CopyToOutputDirectory tag): <ItemGroup> <None Include="Configs\Config.config"> <CopyToOutputDirectory>Always</CopyToOutputDirectory> </None> </ItemGroup> and like this: <ItemGroup> <Resource Include="Resources\Icons\icon4.png" /> </ItemGroup> <ItemGroup> <Resource Include="Resources\Icons\icon5.png" /> </ItemGroup> <ItemGroup> <Resource Include="Resources\Icons\icon6.png" /> </ItemGroup>

Address Out of bounds error when reading xml

寵の児 提交于 2019-12-05 08:40:33
I am getting a weird segfault when using libxml to parse a file. This code worked previously when I compiled it as a 32bit application. I changed it to a 64 bit application and it stops working. The seg fault comes in at "if (xmlStrcmp(cur->name, (const xmlChar *) "servers"))" cur->name is a const xmlChar * and it points to an address that says its out out bounds. But when I debug and go to that memory location, that data is correct. int XmlGetServers() { xmlDocPtr doc; xmlNodePtr cur; doc = xmlParseFile("Pin.xml"); if (doc == NULL) { std::cout << "\n Pin.xml not parsed successfully." << std:

Is there something like BeautifulSoup for Android? [closed]

十年热恋 提交于 2019-12-05 08:07:00
It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center . Closed 7 years ago . I want to find an xml parser that is extremely forgiving. Like Python's BeautifulSoup. Is there anything out there? Okay, I found Jsoup. A Java version of a permissive XML parser like BeautifulSoup. http://jsoup.org/ I will try that and see how it goes. thx 来源: https://stackoverflow.com/questions/13165883/is

What does LIBXML_NOENT do (and why isn't it called LIBXML_ENT)?

吃可爱长大的小学妹 提交于 2019-12-05 07:52:31
In PHP, one can pass optional arguments to various XML parsers, one of them being LIBXML_NOENT . The documentation has this to say about it: LIBXML_NOENT (integer) Substitute entities Substitute entities isn't very informative (what entities? when are they substituted?). But I think it's fair to assume that NOENT is short for NO_ENTITIES or NO_EXTERNAL_ENTITIES , so to me it seems to be a fair assumption that this flag disables the parsing of (external) entities. But that is indeed not the case: $xml = '<!DOCTYPE root [<!ENTITY c PUBLIC "bar" "/etc/passwd">]> <test>&c;</test>'; $dom = new

How to Parse XML File using xmlparsing on iphone?

倖福魔咒の 提交于 2019-12-05 07:30:46
问题 This question was migrated from Skeptics Stack Exchange because it can be answered on Stack Overflow. Migrated 8 years ago . How do I access following XML file using xmlparsing on an iphone? <?xml version="1.0" encoding="iso-8859-1"?> <chapter> <TITLE>Title &plainEntity;</TITLE> <para> <informaltable> <tgroup cols="3"> <tbody> <row><entry>a1</entry><entry morerows="1">b1</entry><entry>c1</entry></row> <row><entry>a2<<?xml version="1.0" encoding="iso-8859-1"?> <ADDRESS> <CONTACT> <NAME FIRST=

XML parsing : encoding utf-8 & UTF-8

亡梦爱人 提交于 2019-12-05 06:04:45
问题 I am trying to parse the values from this LINK, whose xml encoding is like this <?xml version="1.0" encoding="utf-8"?> when I tried to get response throws message in logcat as shown 11-19 17:25:13.350: W/System.err(3360): This is not valid URL 11-19 17:25:13.350: W/System.err(3360): java.lang.NullPointerException When I tried with some other LINK ,whose encoding is like this <?xml version="1.0" encoding="UTF-8"?> It works fine, I can parse the values. is xml parsing failing due encoding not

Which XML parser to use here?

只谈情不闲聊 提交于 2019-12-05 05:32:53
I am receving an XML file as an input, whose size can vary from a few KBs to a lot more. I am getting this file over a network. I need to extract a small number of nodes as per my use, so most of the document is pretty useless for me. I have no memory preferences, I just need speed. Considering all this, I concluded : Not using DOM here (due to possible huge size of doc , no CRUD requirement, and source being network) No SAX as I only need to get a small subset of data. StaX can be a way to go, but I am not sure if it is the fastest way. JAXB came up as another option - but what sort of parser

C# Adding a root to an XDocument

谁都会走 提交于 2019-12-05 04:52:19
I have a string that contains an XML, lets say like this: <Novels> <Book> <Title>Cat in hat</Title> <Price>12</Price> </Book> </Novels> I want to make an XDocument that looks like this: <Booklist> <Novels> <Book> <Title>Cat in hat</Title> <Price>12</Price> </Book> </Novels> </Booklist> I can load the xml string into an XDocument using XDocument doc = XDocument.Parse(xmlString); How would I load the document under a new root. I can think of something like creating a new XDocument with the root I want and then using a for loop to add the nodes as children, but is there an easier way of doing

Extract elements from XML file using Python

南笙酒味 提交于 2019-12-05 04:24:17
问题 The link below gives us the list of ingredients in recipelist. I would like to extract the names of the ingredient and save it to another file using python. http://stream.massey.ac.nz/file.php/6087/Eva_Material/Tutorials/recipebook.xml So far I have tried using the following code, but it gives me the complete recipe not the names of the ingredients: from xml.sax.handler import ContentHandler import xml.sax import sys def recipeBook(): path = "C:\Users\user\Desktop" basename = "recipebook.xml"