Accepted Solutions (1)
Hi,
Harish's suggestion may be a little easier to manage with the grouping/sorting aspects but I was able to do it in one map by mapping in this fashion and with three additional UDFs. For this example I ignored the 'header' node of the XML because it didn't appear to contain any data in your example. The field deliverynumber was used at the case context for all node mappings except 'idoc' and 'deliverynumber' See the screenshots and code below:
idoc:
case:
caseid:
deliveryid:
deliverynumber:
removeDuplicates UDF:
String tmp = "";
for(int i = 0; i < input.length; i++)
{
if(tmp.compareTo(input[i]) != 0)
{
result.addValue(input[i]);
tmp = input[i];
}
}
splitCaseByDelivery UDF:
String tmp = "";
ArrayList<String> al = new ArrayList<String>();
for(int i = 0; i < input.length; i++)
{
if(tmp.compareTo(input[i]) != 0)
{
result.addValue(input[i]);
tmp = input[i];
}
else
{
al.add(input[i]);
}
}
if(al.size() > 0)
{
splitCaseByDelivery(al.toArray(new String[0]), result, container);
}
splitDelivery UDF:
if(input.length == 0)
return;
String tmp = input[0];
ArrayList<String> al = new ArrayList<String>();
int i = 0;
while(i < input.length)
{
if(tmp.compareTo(input[i]) == 0)
{
result.addValue(input[i]);
for(int j = i+1; j < input.length; j++)
{
result.addValue(input[j]);
if(input[j].compareTo(ResultList.CC) == 0)
{
i = j;
break;
}
}
}
else
{
al.add(input[i]);
for(int j = i+1; j < input.length; j++)
{
al.add(input[j]);
if(input[j].compareTo(ResultList.CC) == 0)
{
i = j;
break;
}
}
}
i++;
}
if(al.size() > 0)
{
splitDelivery(al.toArray(new String[0]), result, container);
}
The UDFs for case id remove duplicate delivery so you'll get the right number of case ids and then it shifts additional occurrences of the same case id using recursion to the next idoc contexts when it must be created more than once. The UDF for deliverynumber does something similar where all the same numbers will be reorganized into the same idoc context but in the proper order.
Regards,
Ryan Crosby
Answers (2)
Hi,
Here is an adjusted map that should work no matter which deliveries you have matched to whatever cases in any order. First I'll note that for the previous UDFs you can delete splitDelivery & splitCaseByDelivery because I have setup a Function Library to manage the context rearrangement. See below for the adjusted maps for caseid & deliveryid:
caseid -
deliveryid -
Here are the function library details:
Attributes and Methods -
private HashMap<String, LinkedList<ArrayList<String>>> hm;
init -
hm = new HashMap<String, LinkedList<ArrayList<String>>>();
cleanup -
hm = null;
getQueueContexts -
// Get key and retrieve list of values from queue context
for(int i = 0; i < keys.length; i++)
{
LinkedList<ArrayList<String>> ll = hm.get(keys[i]);
ArrayList<String> al = ll.removeFirst();
// Add values to context
for(int j = 0; j < al.size(); j++)
result.addValue(al.get(j));
// Check size of queue
if(ll.size() == 0)
hm.remove(ll);
// Check hash map size - if there are still values then insert context change
if(hm.size() != 0)
result.addContextChange();
}
buildNodeMap -
// Insert or modify keys
int counter = 0;
ArrayList<String> al;
LinkedList<ArrayList<String>> ll;
// Iterate keys - if exists then modify otherwise insert new entry
for(int i = 0; i < keys.length; i++)
{
result.addValue(keys[i]);
al = new ArrayList<String>();
// Get or create new LinkedList (Queue)
if(hm.containsKey(keys[i]))
{
ll = hm.get(keys[i]);
}
else
{
ll = new LinkedList<ArrayList<String>>();
hm.put(keys[i], ll);
}
ll.add(al);
// Add list values for context length
while(counter < vals.length)
{
if(vals[counter].compareTo(ResultList.CC) != 0)
{
al.add(vals[counter]);
}
else
{
counter++;
break;
}
counter++;
}
}
Regards,
Ryan Crosby
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