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(C) XML Path Performance OptimizationsDiscusses some important things to know about using Chilkat paths in the Chilkat XML API.
#include <C_CkXml.h> void ChilkatSample(void) { HCkXml xml; BOOL success; int licCount; const char *s; int i; xml = CkXml_Create(); // Let's load XML containing the following: // <?xml version="1.0" encoding="utf-8"?> // <xyz> // <licenses> // <license> // <id>1234</id> // </license> // <license> // <id>1234</id> // </license> // ... // My sample XML contains 64,000 "license" nodes .. // ... // <license> // <id>1234</id> // </license> // <license> // <id>1234</id> // </license> // </licenses> // </xyz> // success = CkXml_LoadXmlFile(xml,"qa_output/large.xml"); if (success != TRUE) { printf("%s\n",CkXml_lastErrorText(xml)); CkXml_Dispose(xml); return; } // Iterating over the individual "license" nodes with this code snippet is // extremely slow: licCount = CkXml_NumChildrenHavingTag(xml,"licenses|license"); printf("license count = %d\n",licCount); i = 0; // If "10" is changed to licCount, then it becomes apparent that this loop gets slower with each iteration. while (i < 10) { CkXml_putI(xml,i); s = CkXml_getChildContent(xml,"licenses|license[i]|id"); printf("%d: %s\n",i,s); i = i + 1; } // The reason it is extremely slow is that the "license[i]" part of the path passed to GetChildContent // says: find the i'th child of "licenses" having the tag "license". Chilkat cannot assume that all // children of an XML node have the same tag. Therefore it's not possible to directly access the i'th child. // Internally, Chilkat must start at the 1st child and iterate until it reaches the i'th child having the // tag "license". // For example, imagine if the XML was like this: // <?xml version="1.0" encoding="utf-8"?> // <xyz> // <licenses> // <license> // <id>1234</id> // </license> // <somethingElse> // <a>abc</a> // </somethingElse> // <license> // <id>1234</id> // </license> // ... // In the above XML, the 1st "license" is the 1st child of "licenses", but the 2nd "license" // is the 3rd child of "licenses". // If you already know that all children have the same tag, there is a shortcut that allows // for direct access to that child. Just leave off the tag name, like this: i = 0; // If "10" is changed to licCount, then we can see the time for each loop is the same, and it's fast. while (i < 10) { CkXml_putI(xml,i); s = CkXml_getChildContent(xml,"licenses|[i]|id"); printf("%d: %s\n",i,s); i = i + 1; } // When we pass just the index "[i]", we're saying: Get the i'th child regardless of tag. // This is extremely fast because internally we can just access the i'th child directly. // Another performance improvement is to call NumChildrenAt rather than NumChildrenHavingTag. // For example: licCount = CkXml_NumChildrenAt(xml,"licenses"); printf("licCount = %d\n",licCount); // NumChildrenAt returns the total number of children at the tag path. If we already know // all children will have the same tag, we can just get the count CkXml_Dispose(xml); } |
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