Working out the API for Xsp Classes in XPages
Tuesday, November 10, 2009 at 1:30AM One of the biggest problems I have is interacting with the various components of an XPage programatically. It's not that I can't get hold of them, but it is that I don't know what classes they are from a Java point of view, and therefore don't know what methods and properties I have available to me.
So my solution is to first create an error. Say, for example, I want to programatically control a Pager control. I'll add a button to my XPage with an onClick event something like this:
getComponent("pager1").setStart(1);
I know this will fail, but if I have the default error page enabled then I'll get an error message like this:
And from the error message I can now tell that the object I want to talk to is of the class "com.ibm.xsp.component.xp.XspPager" and when I add that to the variable definition I'll get full type ahead of the API for that class in Domino Designer. So now I know that I can write something like this:
var pager:com.ibm.xsp.component.xp.XspPager = getComponent("pager1");
pager.gotoFirst();
Controls, api, documentation in Tips and Tricks 
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Reader Comments (5)
cool tip, that's what you call Try and Error
For what it's worth, the following syntax will also tell you the class of any control:
typeof getComponent("pager1")
So if you want to extract the class name without triggering an error, you can create a computed field and set its value to return the class name of another control via typeof.
Nice technique Matt! Thanks for sharing it....
-John
Other technique:
var file = getComponent("fileUpload1");
print(file);
you get in console the class
10/11/2009 13:36:08 HTTP JVM: com.ibm.xsp.component.xp.XspFileUpload@78c778c7
There is even a proper command for that :)
var obj = getComponent("XSPID");
_dump(obj);