xml-parsing

$.parseXML not working with valid xml

女生的网名这么多〃 提交于 2019-12-05 03:34:17
XML: <?xml version="1.0"?> <choices> <choice> <start>39</start> <duration>6</duration> <path> <name></name> <complete></complete> </path> <path> <name></name> <complete></complete> </path> </choice> </choices> $.ajax({ url: 'choices.xml', context: this, async: false, success: function(response) { var xmlDoc = $.parseXML(response); console.log(xmlDoc); // null } }); The XML is reported as valid, and no error is thrown. I know I can use $(response), but I don't need that. jQuery 1.7.2 dataType Default: Intelligent Guess (xml, json, script, or html) "xml": Returns a XML document that can be

StAX XML all content between two required tags

无人久伴 提交于 2019-12-05 03:29:37
Starting learning the StAX, using XMLStreamReader, I faced with some problem. How can I get ALL content between tags as Text? I mean, I know name of needed tag, and when I find it, I must go to the close tag, and everything I found between them I must append at some string. For example, we have something like <rootTag> ... <someTag> Some text content and other tags here… </someTag > <tagINeed> <someinternalTag1> <someinternalTag11> Some text content.. </someinternalTag11> ... </someinternalTag1> <someinternalTag2> Something here </someinternalTag2> </tagINeed> ... <somethingAnother> ... <

Uncaught exception 'DOMException' with message 'Hierarchy Request Error'

自古美人都是妖i 提交于 2019-12-05 01:30:05
I'm getting error while replacing or adding a child into a node. Required is : I want to change this to.. <?xml version="1.0"?> <contacts> <person>Adam</person> <person>Eva</person> <person>John</person> <person>Thomas</person> </contacts> like this <?xml version="1.0"?> <contacts> <person>Adam</person> <p> <person>Eva</person> </p> <person>John</person> <person>Thomas</person> </contacts> error is Fatal error: Uncaught exception 'DOMException' with message 'Hierarchy Request Error' my code is function changeTagName($changeble) { for ($index = 0; $index < count($changeble); $index++) { $new =

Java create InputStream from ZipInputStream entry

风格不统一 提交于 2019-12-05 01:28:27
问题 I would like to write a method that read several XML files inside a ZIP, from a single InputStream. The method would open a ZipInputStream, and on each xml file, get the corresponding InputStream, and give it to my XML parser. Here is the skeleton of the method : private void readZip(InputStream is) throws IOException { ZipInputStream zis = new ZipInputStream(is); ZipEntry entry = zis.getNextEntry(); while (entry != null) { if (entry.getName().endsWith(".xml")) { // READ THE STREAM } entry =

How to access a specific element and attribute in xml with vba in excel?

生来就可爱ヽ(ⅴ<●) 提交于 2019-12-05 00:50:28
问题 I'm struggling with getting excel to parse through an xml file. I've found a plethora of examples, but none seem to be quite what I am looking for and I can't seem to get past the error "Object variable or With block variable not set" The xml is well formed and looks like the following: <xml tag> <PLMXML> <WorkflowTemplate name=""> <argument name=""> </argument> </WorkflowTemplate > <WorkflowTemplate name=""> etc. I'm trying to use VBA to get to the value of all the Children's names

How can I get Nokogiri to parse and return an XML document?

[亡魂溺海] 提交于 2019-12-05 00:46:01
问题 Here's a sample of some oddness: #!/usr/bin/ruby require 'rubygems' require 'open-uri' require 'nokogiri' print "without read: ", Nokogiri(open('http://weblog.rubyonrails.org/')).class, "\n" print "with read: ", Nokogiri(open('http://weblog.rubyonrails.org/').read).class, "\n" Running this returns: without read: Nokogiri::XML::Document with read: Nokogiri::HTML::Document Without the read returns XML, and with it is HTML? The web page is defined as "XHTML transitional", so at first I thought

Access nested children in xml file parsed with ElementTree

我怕爱的太早我们不能终老 提交于 2019-12-05 00:11:32
I am new to xml parsing. This xml file has the following tree: FHRSEstablishment |--> Header | |--> ... |--> EstablishmentCollection | |--> EstablishmentDetail | | |-->... | |--> Scores | | |-->... |--> EstablishmentCollection | |--> EstablishmentDetail | | |-->... | |--> Scores | | |-->... but when I access it with ElementTree and look for the child tags and attributes, import xml.etree.ElementTree as ET import urllib2 tree = ET.parse( file=urllib2.urlopen('http://ratings.food.gov.uk/OpenDataFiles/FHRS408en-GB.xml' % i)) root = tree.getroot() for child in root: print child.tag, child.attrib I

Free Real-Time Currency conversion Exchange Rate xml feed url [closed]

独自空忆成欢 提交于 2019-12-04 23:58:13
问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 5 years ago . Any link offering the real-time updating xml data feed to integrate in iphone app 回答1: You can try these xml files http://www.ecb.int/stats/eurofxref/eurofxref-daily.xml http://www.ecb.int/stats/exchange/eurofxref/html/index.en.html http://themoneyconverter.com/rss-feed/USD/rss.xml 回答2: you can try yahoo , its

How to select multiple nodes in different levels?

那年仲夏 提交于 2019-12-04 22:56:18
Having this (simplified) XML: <?xml version="1.0" encoding="UTF-8"?> <kml> <Document> <Placemark> <name>Poly 1</name> <Polygon> <coordinates> -58.40844625779582,-34.60295278618136,0 </coordinates> </Polygon> </Placemark> <Placemark> <name>Poly 2</name> <Polygon> <coordinates> -58.40414334150432,-34.59992445476809,0 </coordinates> </Polygon> </Placemark> </Document> </kml> How can I select the name and coordinates of each Placemark? Right now I can select their name with the following XPath expression: //Document//Placemark//name How can I select both without any other data? You can use a union

Android REST XML result to Listview

不羁岁月 提交于 2019-12-04 22:17:39
I have a REST web service that returns an xml result like this: - <MyCategories xmlns="http://schemas.datacontract.org/2004/07/ceva" xmlns:i="http://www.w3.org/2001/XMLSchema-instance"> - <Category> <CategoryName>First category</CategoryName> <Id>1</Id> </Category> - <Category> <CategoryName>Second category</CategoryName> <Id>2</Id> </Category> - <Category> <CategoryName>Third category</CategoryName> <Id>3</Id> </Category> </MyCategories> I acces the web service like this: HttpClient httpclient = new DefaultHttpClient(); HttpGet request = new HttpGet(WebServiceURL); request.addHeader("deviceId