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(Lianja) XML Path Performance OptimizationsDiscusses some important things to know about using Chilkat paths in the Chilkat XML API.
loXml = createobject("CkXml") // 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> // llSuccess = loXml.LoadXmlFile("qa_output/large.xml") if (llSuccess <> .T.) then ? loXml.LastErrorText release loXml return endif // Iterating over the individual "license" nodes with this code snippet is // extremely slow: lnLicCount = loXml.NumChildrenHavingTag("licenses|license") ? "license count = " + str(lnLicCount) i = 0 // If "10" is changed to licCount, then it becomes apparent that this loop gets slower with each iteration. do while i < 10 loXml.I = i s = loXml.GetChildContent("licenses|license[i]|id") ? str(i) + ": " + s i = i + 1 enddo // 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. do while i < 10 loXml.I = i s = loXml.GetChildContent("licenses|[i]|id") ? str(i) + ": " + s i = i + 1 enddo // 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: lnLicCount = loXml.NumChildrenAt("licenses") ? "licCount = " + str(lnLicCount) // 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 release loXml |
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