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How to unfreeze a column that shouldn't be frozen
I have the following code that creates a grid:
var listGrid = new AW.Grid.Extended;
listGrid.setHeaderIndices([0,1]);
listGrid.setHeaderCount(2);
listGrid.setCellEditable = false;
listGrid.setCellText(listData);
listGrid.setHeaderText(headerRow,0);
listGrid.setHeaderTooltip(headerRow,0);
listGrid.setColumnCount(headerRow.length);
listGrid.setRowCount(listData.length);
listGrid.setId("listGrid");
listGrid.clearFixedModel();
document.write(listGrid);
I don't ever tell column 0 (leftmost column) to be fixed/frozen, but it is (in both ie and ff), and I tried listGrid.clearFixedModel(); but that does not unfix it.
I need this not to be fixed. I don't even know why it is being fixed in the first place.
John
Tuesday, February 23, 2010
Carlos
Tuesday, February 23, 2010
Thanks for the quick response!
I changed...
listGrid.clearFixedModel();
to
listGrid.setFixedLeft(-1);
and indeed it prevents anything from being fixed, but now there is what appears to be an extra column to the left of all the other columns.
(Also, now I see in the docs where it says that AW.Grid.Extended has a default-fixed column on the left, and so it makes sense why clearFixedModel() wouldn't make that go away (since that resets to the default, which is 1 fixed left)).
Any help with the extra-column would be appreciated. I can send screenshots of the problem (both on ff & ie) if needed.
John
Tuesday, February 23, 2010
Actually, just tried on a whim:
listGrid.setFixedLeft(0);
and that seems to do it, no fixed columns or extra blank ones.
I was under the misconception that setFixedLeft was referencing column-indices, when in fact it's just the number of columns to freeze. Sorry about that, thanks again for the response.
John
Tuesday, February 23, 2010
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