我正在做一些使用libxml2库的小包装器。如果元素值混合,则存在更改元素值的问题。
在下面的示例中,我正在创建" child1"元素与"旧内容"价值和子孩子" child1.1"。然后我试图改变" child1"价值到"新内容"使用xmlNodeSetContent
函数(标有< ---的行)。问题是," child1.1"被xmlNodeSetContent
删除。我想只更改" OLD CONTENT"中的值。到"新内容"。
似乎xmlNodeSetContent
不是这样做的正确功能,或者libxml2中还有其他内容,我不知道。
问题:有没有办法在混合内容的情况下如何更改值?如果是这样,怎么做?
#include <stdio.h>
#include <string.h>
#include <libxml/parser.h>
#include <libxml/tree.h>
#define MY_ENCODING "ISO-8859-1"
xmlNodePtr createRoot(xmlDocPtr doc, char* name)
{
xmlNodePtr root;
root = xmlNewNode(NULL, BAD_CAST name);
xmlDocSetRootElement(doc, root);
return root;
}
xmlNodePtr addChild(xmlNodePtr parent, char* name, char* value)
{
return xmlNewChild(parent, NULL, BAD_CAST name, BAD_CAST value);
}
void setElementContent(xmlNodePtr element, char* value)
{
xmlNodeSetContent(element, BAD_CAST value);
}
void setElementAttribute(xmlNodePtr element, char* attribName, char* attribValue)
{
xmlSetProp(element, BAD_CAST attribName, BAD_CAST attribValue);
}
int main(int argc, char** argv)
{
xmlDocPtr doc = NULL;
xmlNodePtr root = NULL;
xmlNodePtr tmp1 = NULL;
xmlNodePtr tmp11 = NULL;
xmlNodePtr tmp2 = NULL;
xmlChar* xmlbuff;
int buffersize;
doc = xmlNewDoc(BAD_CAST "1.0");
root = createRoot(doc, "root");
//tmp1 = addChild(root, "child1", NULL);
tmp1 = addChild(root, "child1", "OLD CONTENT");
tmp11 = addChild(tmp1, "child1.1", NULL);
setElementContent(tmp11, "blablabla");
tmp2 = addChild(root, "child2", NULL);
setElementContent(tmp2, "22222");
setElementAttribute(tmp2, "id", "002");
setElementAttribute(tmp2, "lang", "en");
setElementContent(tmp1, "NEW CONTENT"); // <---
//xmlDocDumpFormatMemory(doc, &xmlbuff, &buffersize, 1);
xmlDocDumpFormatMemoryEnc(doc, &xmlbuff, &buffersize, MY_ENCODING, 1);
printf("%s", (char *) xmlbuff);
xmlCleanupParser();
xmlFree(xmlbuff);
xmlFreeDoc(doc);
return 0;
}
编辑:工作版本:
#include <stdio.h>
#include <string.h>
#include <libxml/parser.h>
#include <libxml/tree.h>
#define MY_ENCODING "ISO-8859-1"
xmlNodePtr createRoot(xmlDocPtr doc, char* name)
{
xmlNodePtr root;
root = xmlNewNode(NULL, BAD_CAST name);
xmlDocSetRootElement(doc, root);
return root;
}
xmlNodePtr addChild(xmlNodePtr parent, char* name, char* value)
{
return xmlNewChild(parent, NULL, BAD_CAST name, BAD_CAST value);
}
void setElementContent(xmlNodePtr element, char* value)
{
xmlNodePtr it = NULL;
int bFound = 0;
for (it = element->children; it != NULL; it = it->next)
{
if (it->type == XML_TEXT_NODE)
{
xmlNodeSetContent(it, BAD_CAST value);
bFound = 1;
}
}
if (!bFound)
{
xmlNodeSetContent(element, BAD_CAST value);
}
}
void setElementAttribute(xmlNodePtr element, char* attribName, char* attribValue)
{
xmlSetProp(element, BAD_CAST attribName, BAD_CAST attribValue);
}
void printXml(xmlDocPtr doc)
{
xmlChar* xmlbuff;
int buffersize;
//xmlDocDumpFormatMemory(doc, &xmlbuff, &buffersize, 1);
xmlDocDumpFormatMemoryEnc(doc, &xmlbuff, &buffersize, MY_ENCODING, 1);
printf("%s", (char *) xmlbuff);
xmlCleanupParser();
xmlFree(xmlbuff);
}
int main(int argc, char** argv)
{
xmlDocPtr doc = NULL;
xmlNodePtr root = NULL;
xmlNodePtr tmp1 = NULL;
xmlNodePtr tmp11 = NULL;
xmlNodePtr tmp2 = NULL;
xmlNodePtr tmp3 = NULL;
xmlNodePtr it = NULL;
doc = xmlNewDoc(BAD_CAST "1.0");
root = createRoot(doc, "root");
//tmp1 = addChild(root, "child1", NULL);
tmp1 = addChild(root, "child1", "OLD CONTENT");
tmp11 = addChild(tmp1, "child1.1", NULL);
setElementContent(tmp11, "blablabla");
tmp2 = addChild(root, "child2", NULL);
setElementContent(tmp2, "22222");
setElementAttribute(tmp2, "id", "002");
setElementAttribute(tmp2, "lang", "en");
tmp3 = addChild(root, "child3", NULL); /* create empty */
printf("\nBefore:\n");
printXml(doc);
setElementContent(tmp1, "NEW CONTENT");
setElementContent(tmp3, "333");
printf("\nAfter:\n");
printXml(doc);
xmlFreeDoc(doc);
return 0;
}
答案 0 :(得分:1)
混合内容由一系列文本节点和元素节点表示。您必须检索文本节点并仅替换它的内容:
xmlNodePtr textNode = &tmp1->children[0];
setElementContent(textNode, "NEW CONTENT");
//setElementContent(tmp1, "NEW CONTENT"); // <---