使用libxml2创建和解析xml文件

毕业设计需要用到xml文件来组织和存放数据，

对于Linux环境下，有libxml2可供使用。

经过一段时间查询文档和网站，

基本掌握创建xml文档和解析xml的操作，

简单做一下记录。

创建xml

例子如下：

#include <stdio.h>
#include <libxml/parser.h>
#include <libxml/tree.h>

int main(int argc, char **argv)
{
        xmlDocPtr doc = NULL;
        xmlNodePtr root_node = NULL, node = NULL, node1 = NULL;

        doc = xmlNewDoc(BAD_CAST "1.0"); // create a new xml document.
        root_node = xmlNewNode(NULL, BAD_CAST "root"); // create a root node.
        xmlDocSetRootElement(doc, root_node);

        xmlNewChild(root_node, NULL, BAD_CAST "node1", BAD_CAST "content of node1");
        //xmlNewChild(root_node, NULL, BAD_CAST "node2", NULL);

        node = xmlNewChild(root_node, NULL, BAD_CAST "node3", BAD_CAST "node3 has attributes");
        xmlNewProp(node, BAD_CAST "attribute", BAD_CAST "yes");

        node = xmlNewNode(NULL, BAD_CAST "node4");
        node1 = xmlNewText(BAD_CAST
                   "other way to create content (which is also a node)");
        xmlAddChild(node, node1);
        xmlAddChild(root_node, node);

        xmlSaveFormatFileEnc(argc > 1 ? argv[1] : "-", doc, "UTF-8", 1);

        xmlFreeDoc(doc);

        xmlCleanupParser();

        xmlMemoryDump();
        return(0);
}

 libxml的api使用 const unsigned char* 。

而string literal 只能隐式转换到 const char*。

所以libxml提供一个BAD_CAST用来作显示转换。

代码应该不难看懂，生成的xml文件如下：

<?xml version="1.0" encoding="UTF-8"?>
<root>
  <node1>content of node1</node1>
  <node3 attribute="yes">node3 has attributes</node3>
  <node4>other way to create content (which is also a node)</node4>
</root>

xml文件和创建xml的代码对照着看就很容易看懂如何生成节点以及属性了。

解析xml

代码如下：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <libxml/parser.h>
#include <libxml/xmlmemory.h>

int main(int argc, char **argv)
{
        xmlDocPtr doc;
        xmlNodePtr curNode;

        xmlKeepBlanksDefault(0);
        doc = xmlReadFile("mine.xml", "UTF-8", XML_PARSE_RECOVER); // open a xml doc.
        curNode = xmlDocGetRootElement(doc); // get root element.

        if (curNode == NULL)
        {
                fprintf(stderr, "open file failed. 
");
                xmlFreeDoc (doc);
                return -1;
        }

        if (xmlStrcmp(curNode->name, "root")) // if the same,xmlStrcmp return 0, else return 1
        {
                fprintf(stderr, "check rootElement failed. 
");
                xmlFreeDoc (doc);
                return -1;
        }

        curNode = curNode->children; // move to root element's children.
        char *nodeName;
        char *content;

        if (curNode != NULL)
        {
                nodeName = (char *) curNode->name; 
                content = (char *) xmlNodeGetContent(curNode);
                printf ("Current node name:%s,	 the content is:%s.

", nodeName, content);
        }

        curNode = curNode->next;
        char *attr;
        if (curNode != NULL)
        {
                nodeName = (char *) curNode->name;
                content = (char *) xmlNodeGetContent(curNode);
                attr = (char *) xmlGetProp(curNode, (const xmlChar *)"attribute"); // get node attribute
                printf ("Current node name:%s,	 the content is:%s,	 AND THE ATTR IS:%s.

", nodeName, content,attr); 
        }

        curNode = curNode->next;
        if (curNode != NULL)
        {
                nodeName = (char *) curNode->name;
                content = (char *) xmlNodeGetContent(curNode);
                printf ("Current node name:%s,	 the content is:%s.

", nodeName, content);   
        }

        xmlFree(curNode);
        xmlFreeDoc(doc);
        return 1;
}

上面的代码是简单的按生成的xml结构来解析，

正确的用法应该是写成一个函数来调用，

可以解析任何的已知根节点的xml文件。

解析的结果输入如下：

[nigelzeng@ubuntu xml-learning]$ ./xml-mine-parse 
Current node name:node1,         the content is:content of node1.

Current node name:node3,         the content is:node3 has attributes,    AND THE ATTR IS:yes.

Current node name:node4,         the content is:other way to create content (which is also a node).

参考：

http://xmlsoft.org/index.html

http://www.cppblog.com/lymons/archive/2009/03/30/37553.html

http://www.4ucode.com/Study/Topic/1622022