libxml2 无法从节点获取内容

libxml2是一个开源的XML解析库，可用于解析和处理XML文档。然而，有时候我们可能会遇到一个问题，即无法从节点中获取到所需的内容。本文将介绍这个问题，并提供一个案例代码来解决这个问题。

c
#include 
#include 
#include 
void printNodeContent(xmlNodePtr node) {
    xmlNodePtr current = node->xmlChildrenNode;
    while (current != NULL) {
        if (current->type == XML_TEXT_NODE) {
            xmlChar* content = xmlNodeGetContent(current);
            printf("%s\n", content);
            xmlFree(content);
        }
        current = current->next;
    }
}
int main() {
    xmlDocPtr doc;
    xmlNodePtr root, node;
    xmlXPathContextPtr xpathCtx;
    xmlXPathObjectPtr xpathObj;
    // 解析XML文档
    doc = xmlReadFile("example.xml", NULL, 0);
    if (doc == NULL) {
        fprintf(stderr, "Failed to parse XML document\n");
        return 1;
    }
    // 获取根节点
    root = xmlDocGetRootElement(doc);
    // 创建XPath上下文
    xpathCtx = xmlXPathNewContext(doc);
    if (xpathCtx == NULL) {
        fprintf(stderr, "Failed to create XPath context\n");
        xmlFreeDoc(doc);
        return 1;
    }
    // 使用XPath表达式选择节点
    xpathObj = xmlXPathEvalExpression((xmlChar*)"//title", xpathCtx);
    if (xpathObj == NULL) {
        fprintf(stderr, "Failed to evaluate XPath expression\n");
        xmlXPathFreeContext(xpathCtx);
        xmlFreeDoc(doc);
        return 1;
    }
    // 遍历选择的节点并打印内容
    int i;
    for (i = 0; i < xpathObj->nodesetval->nodeNr; i++) {
        node = xpathObj->nodesetval->nodeTab[i];
        printf("");
        printNodeContent(node);
        printf("");
    }
    // 释放资源
    xmlXPathFreeObject(xpathObj);
    xmlXPathFreeContext(xpathCtx);
    xmlFreeDoc(doc);
    return 0;
}