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(Objective-C) XML Path Performance OptimizationsDiscusses some important things to know about using Chilkat paths in the Chilkat XML API.
#import <CkoXml.h> #import <NSString.h> CkoXml *xml = [[CkoXml alloc] init]; // 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> // BOOL success = [xml LoadXmlFile: @"qa_output/large.xml"]; if (success != YES) { NSLog(@"%@",xml.LastErrorText); return; } // Iterating over the individual "license" nodes with this code snippet is // extremely slow: int licCount = [[xml NumChildrenHavingTag: @"licenses|license"] intValue]; NSLog(@"%@%d",@"license count = ",licCount); NSString *s = 0; int i = 0; // If "10" is changed to licCount, then it becomes apparent that this loop gets slower with each iteration. while (i < 10) { xml.I = [NSNumber numberWithInt: i]; s = [xml GetChildContent: @"licenses|license[i]|id"]; NSLog(@"%d%@%@",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) { xml.I = [NSNumber numberWithInt: i]; s = [xml GetChildContent: @"licenses|[i]|id"]; NSLog(@"%d%@%@",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 = [[xml NumChildrenAt: @"licenses"] intValue]; NSLog(@"%@%d",@"licCount = ",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 |
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