Is it possible to have two different tableviews scroll according to eachother? [solved]

[skammers]

I have two different tableviews containing QStrings. I need them to scroll according to eachother so they are always on the same height in the content. Is this possible? If not, what kind of an other approach could I use?
Im kind of new to this, so please help.

[JvdGlind]

In QML:

@import QtQuick 2.2
import QtQuick.Controls 1.2
import QtQuick.Window 2.1

Window {
visible: true
width: 360
height: 360

ListModel {
    id: someNumbersModel
}

function valueChanged(value) {
    firstTableView.__verticalScrollBar.value = value;
    secondTableView.__verticalScrollBar.value = value;
}

TableView {
    id: firstTableView

    anchors.top: parent.top;
    anchors.left: parent.left;
    anchors.right: parent.horizontalCenter;
    anchors.bottom: parent.bottom;

    TableViewColumn{ role: "location"  ; title: "Location" ; width: 100 }
    model: someNumbersModel

    flickableItem.onContentYChanged: secondTableView.flickableItem.contentY = flickableItem.contentY
    flickableItem.onContentXChanged: secondTableView.flickableItem.contentX = flickableItem.contentX
}
TableView {
    id: secondTableView

    anchors.top: parent.top;
    anchors.left: parent.horizontalCenter;
    anchors.right: parent.right;
    anchors.bottom: parent.bottom;

    TableViewColumn{ role: "location"  ; title: "Location" ; width: 100 }
    model: someNumbersModel

    flickableItem.onContentYChanged: firstTableView.flickableItem.contentY = flickableItem.contentY
    flickableItem.onContentXChanged: firstTableView.flickableItem.contentX = flickableItem.contentX
}

property int a;
Component.onCompleted: {
    for(a = 0 ; a < 100 ; a++)
        someNumbersModel.append({ "location": String(a) })
}

}
@

[skammers]

Thank you :) got it to work using:

flickableItem.onContentYChanged: secondTableView.flickableItem.contentY = flickableItem.contentY
flickableItem.onContentXChanged: secondTableView.flickableItem.contentX = flickableItem.contentX

:)

[JvdGlind]

No problem :). Please at [solved] to the title for the other readers.