private String getRepetitionText(String type, FormIndex index, boolean newrep) {
if (element instanceof GroupDef
&& ((GroupDef) element).getRepeat()
&& index.getElementMultiplicity() >= 0) {
GroupDef g = (GroupDef) element;
String title = getLongText();
int ix = index.getElementMultiplicity() + 1;
int count = getNumRepetitions();
String caption = null;
if ("header".equals(type)) {
caption = g.entryHeader;
} else if ("choose".equals(type)) {
caption = g.chooseCaption;
if (caption == null) {
caption = g.entryHeader;
}
}
if (caption == null) {
return title + " " + ix + "/" + count;
}
HashMap<String, Object> vars = new HashMap<String, Object>();
vars.put("name", title);
vars.put("i", Integer.valueOf(ix));
vars.put("n", Integer.valueOf(count));
vars.put("new", new Boolean(newrep));
return form.fillTemplateString(caption, index.getReference(), vars);
} else {
return null;
}
}
public boolean canCreateRepeat(TreeReference repeatRef, FormIndex repeatIndex) {
GroupDef repeat = (GroupDef) this.getChild(repeatIndex);
// Check to see if this repeat can have children added by the user
if (repeat.noAddRemove) {
// Check to see if there's a count to use to determine how many children this repeat
// should have
if (repeat.getCountReference() != null) {
int currentMultiplicity = repeatIndex.getElementMultiplicity();
// get the total multiplicity possible
long fullcount =
((Integer) this.getInstance().getDataValue(repeat.getCountReference()).getValue())
.intValue();
if (fullcount <= currentMultiplicity) {
return false;
}
} else {
// Otherwise the user can never add repeat instances
return false;
}
}
// TODO: If we think the node is still relevant, we also need to figure out a way to test that
// assumption against
// the repeat's constraints.
return true;
}
public int getMultiplicity() {
return index.getElementMultiplicity();
}
