Thanks. This helps. Must I do anything special to move to the next value
in the list? Currently, I'm rendering the correct number of values, but
in each case only the first value. So, if my list consists of "a", "b",
"c", what is being rendered is "a", "a", "a" (which is the correct
number -- 3 -- but in each case only the first value). Here's the
relevant snippet:
List allsuggestions = ((Words)allwrongwords.get(i)).getSuggestions();
for (int k = 0; k < allsuggestions.size(); k++) {
final String onesuggestion = (String)allsuggestions.get(k);
listItem.add(new ListView("innerTable2", allsuggestions) {
protected void populateItem(ListItem listItem) {
MultiLineLabel listItemLabel = new
MultiLineLabel("one_suggestion", onesuggestion);
listItem.add(listItemLabel);
}
});
}
-- Geertjan
Johan Compagner wrote:
you have to use another listview inside a listitem
Because that is what you seem to do.
so something like
<tr wicket:id="outertable">
<td wicket:id="innertable"><span wicket:id="one_suggestion"></span></td>
</tr>
and then the current listItem must add another ListView instance
(called innertable)
johan
On 4/5/06, *Geertjan Wielenga* <[EMAIL PROTECTED]
<mailto:[EMAIL PROTECTED]>> wrote:
Hi all,
Part of my code goes like this:
for (int k = 0; k < allsuggestions.size();
k++) {
String onesuggestion =
(String)allsuggestions.get(k);
listItem.add(new
MultiLineLabel("one_suggestion", onesuggestion));
}
So, how do I render a new listitem for each iteration of this for
loop?
Currently only the first item in the iteration is rendered.
Geertjan
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