CODEWITHSUNDEEP
xpagesxpages-ssjs
jjtbsomhorst
jjtbsomhorst

Reputation: 1667

pager control. Hide previous/next programmatically

On a custom control I have a repeat control which iterates over a vector. This repeat control does have a pager bound to it. I want to hide the previous and next links when the pager is on the first or on the last page of the control.

To hide the previous is ofcourse easy >> Add a rendered property to getpageCount() > 0. The next link is a bit of a problem. The pager class does not have a method getCurrentPage(). therefore I can't find out which page I'm currently at on the pager.

Is there someone who does have a fix / idea on how to hide/show the next / previous links on a pager using SSJS?

Upvotes: 0

Views: 549

Answers (2)

Ramkumar
Ramkumar

Reputation: 842

document.getElementById("#id:pager2}").firstChild.nextSibling.lastChild.lastChild.textContent)

Try the above, Actually above is for getting the value of the Group's value of last element.

I guess that using the client side javascript has more easier than SSJS for manipulating the DOM tree,

Here, We have <previous><Group><Next>,

So we need to get the second element .And it is a span tag. I am getting the lastChild of lasChild.

Note: Do not hide it by setRendered(true). Because after that DOM tree will loose its children. Use display:none property.

Upvotes: 0

Mervin Thomas
Mervin Thomas

Reputation: 46

You can get the number of items in Vector (A) You also know the number of items you are going to display in 1 page of your repeat control(B) So total number of pages will be mod(A/B)+1 . If you are in that page, you can hide the next button.

Mod(modulo) means to skip the decimal part or integer division.

Upvotes: 3

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