libxml2的错误与命名空间和XPath [英] libxml2 error with namespaces and xpath

问题描述

我在这里粘贴一些code，与使用gcc file.c中-lxml2没有任何警告编译，假设libxml2的安装在您的系统。

I am pasting some code here that compiles with no warning using gcc file.c -lxml2, assuming that libxml2 is installed in your system.

#include <libxml/parser.h>
#include <libxml/xpath.h>
#include <assert.h>
#include <libxml/tree.h>
#include <libxml/xpathInternals.h>

xmlDocPtr
getdoc (char *docname) {
    xmlDocPtr doc;
    doc = xmlParseFile(docname);

    if (doc == NULL ) {
        fprintf(stderr,"Document not parsed successfully. \n");
        return NULL;
    }

    return doc;
}

xmlXPathObjectPtr
getnodeset (xmlDocPtr doc, xmlChar *xpath){

    xmlXPathContextPtr context;
    xmlXPathObjectPtr result;

    context = xmlXPathNewContext(doc);
    if (context == NULL) {
        printf("Error in xmlXPathNewContext\n");
        return NULL;
    }

    if(xmlXPathRegisterNs(context,  BAD_CAST "new", BAD_CAST "http://www.example.com/new") != 0) {
        fprintf(stderr,"Error: unable to register NS with prefix");
        return NULL;
    }

    result = xmlXPathEvalExpression(xpath, context);
    xmlXPathFreeContext(context);
    if (result == NULL) {
        printf("Error in xmlXPathEvalExpression\n");
        return NULL;
    }
    if(xmlXPathNodeSetIsEmpty(result->nodesetval)){
        xmlXPathFreeObject(result);
                printf("No result\n");
        return NULL;
    }
    return result;
}

int
main(int argc, char **argv) {

    char *docname;
    xmlDocPtr doc;
    xmlChar *xpath = (xmlChar*) "/new:book/section1";
    xmlNodeSetPtr nodeset;
    xmlXPathObjectPtr result;
    int i;
    xmlChar *keyword;

    if (argc <= 1) {
        printf("Usage: %s docname\n", argv[0]);
        return(0);
    }

    docname = argv[1];
    doc = getdoc(docname);
    result = getnodeset (doc, xpath);
    if (result) {
        nodeset = result->nodesetval;
        for (i=0; i < nodeset->nodeNr; i++) {
            keyword = xmlNodeListGetString(doc, nodeset->nodeTab[i]->xmlChildrenNode, 1);
        printf("keyword: %s\n", keyword);
        xmlFree(keyword);
        }
        xmlXPathFreeObject (result);
    }

    xmlFreeDoc(doc);
    xmlCleanupParser();
    return (1);
}

我的问题是，我想分析下面的XML

My problem is that I want to parse the following xml

<?xml version="1.0" encoding="UTF-8"?>
<book xmlns="http://www.example.com/new">
    <section1>Sec_1</section1>
    <section2>Sec_2</section2>
</book>

这本书元素定义元素内的命名空间。我想打印在XPath /电子书/ SECTION1值，并将其返回NULL。当我尝试了命名空间下返回元素我也得到错误，IE /新：书/ SECTION1

the book element defines a namespace inside that element. I want to print the value in the xpath /book/section1 and it returns NULL. When I am trying to return the element under a namespace I also get errors, ie /new:book/section1

我以为我的code，因为我没有使用正确的命名空间prefixes失败。我跑出来的时候。能否请你帮忙吗？

I assume that my code fails because I am not using correctly the namespace prefixes. I run out of time. Could you please help?

推荐答案

原来，当我从发现这里
它是不是真的的libxml的失败，这是一个问题，因为的libxml的正确的遵循XML / XPath规范的。

Turns out, as I found out from here, it is not really a failure of libXml, it's a problem because libXml correctly follows the XML/XPATH specifications.

由R Bourdeau提出的解决方案是正确的，但是，如果你有你解析XML文档的控制。

The solutions proposed by R Bourdeau are correct, however, if you have control of the xml document you are parsing.

有关XPath查询的上下文的独立的XML文档中的命名空间预选赛。默认的名称空间强制所有子标签进入一个命名空间;它们不需要在文件中的资格的 的，但必须在XPath查询合格。幸运的是，你注册了命名空间新与libxml的，所以cateof的解决方案应该工作。

The context for the XPATH query is independent of the namespace qualifiers in the xml document. The default namespace forces all child tags into a namespace; they don't require qualification in the document but must be qualified in the xpath query. Fortunately, you registered the namespace as new with libXml, so cateof's solution should work.

xmlXPathRegisterNs(context,  BAD_CAST "new", BAD_CAST "http://www.example.com/new"

xmlChar *xpath = (xmlChar*) "/new:book/new:section1";

我在这里内联XML能见度：

I'm inlining the xml here for visibility:

<?xml version="1.0" encoding="UTF-8"?>
<book xmlns="http://www.example.com/new">
    <section1>Sec_1</section1>
    <section2>Sec_2</section2>
</book>

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