How to Point to Linked Grid

[jennygong]

Hi I have a main grid, use linked definition to link a child grid. In main grid, what is the correct javascipt method to point to child linked gird? What i wanna do is find the syntax to point to child linked grid, and then xx.closedetailview(); Anyone knows how to do that? Thank you very much!

[GGandhi]

Re: How to Point to Linked Grid

getting id of the child grid is not that easy, as opposed to getting id of the parent.
take a look at video m53 version 11.
it might have some answers.

[jennygong]

Re: How to Point to Linked Grid

Hmmmm...Then do you know is there an easier way to accomplish what I want? I have a main grid, and linked a sub grid. When I select another record on main grid, the sub grid's grid part changes, but detail view of linked grid leaves the old one open... instead of displaying new record of selected... so that's why i want to close detail view of linked grid...

[GGandhi]

Re: How to Point to Linked Grid

if you have a div defined in the freeform area of the main parent grid and place a button to open the child grid by onClick event of the button into the div then it might close that and open the new child record.
test it out to see if it works.

[TheSmitchell]

Re: How to Point to Linked Grid

Hmmmm...Then do you know is there an easier way to accomplish what I want? I have a main grid, and linked a sub grid. When I select another record on main grid, the sub grid's grid part changes, but detail view of linked grid leaves the old one open... instead of displaying new record of selected... so that's why i want to close detail view of linked grid...

You could add a javascript function to the parent grid that would call the code you wanted to execute in one of the child grid's client-side events, such as onGridInitializeComplete.

var po = {grid.object}.getParentObject();
if (po) {
    po.__refreshChildGrid = function (grid) {
        return function () {
            grid.refresh();
        };
    } ({grid.object});
}

You *might* be able to get away without the external wrapper function (grid) { return [...] } ({grid.object}). I'd play around with it to see if it was necessary. Even if it's not necessary, there is no harm in defining the function this way --- the exterior function returns a function where grid is set to {grid.object}. The ({grid.object}) part is necessary - it executes the function immediately, setting __refreshChildGrid to the return value, function() { grid.refresh(); }, where grid = {grid.object}.

In the parent grid, you would call the function like this:

if (typeof ({grid.object}.__refreshChildGrid) != 'undefined') {
    {grid.object}.__refreshChildGrid();
}

The typeof ({grid.object}.__refreshChildGrid) != 'undefined' is a check to make sure the member exists. If it does exist, then you execute the function.

Good luck!
---
Sarah

PS: I realize that you're trying to close the detail view, so you'll decidedly need to modify this code to do that, but the concept is the same.