meo*_*lic 5 gridview drag-and-drop qml
我遵循http://qt-project.org/wiki/Drag_and_Drop_within_a_GridView上给出的方法.
我有提出的计划.拖放工作正常,但在删除后,我的网格中出现空白空间.我怎样才能改善这个?
此外,我还注意到第46行中的函数indexAt()在拖动操作期间无法正常工作.因此我不得不添加gridArea.index的显式计算.
(编辑)在我看来,空的空间出现是因为在丢弃元素之后它返回到它的原始位置(它在可见元素下面).GridView未更新并以不规则状态完成.
import QtQuick 1.1
GridView {
id: mainGrid
cellWidth: 165; cellHeight: 95
width: 5*cellWidth; height: 4*cellHeight
model: myModel
delegate: myButton
ListModel {
id: myModel
function createModel() {
for (var i=1; i<=20; i++) {
myModel.append({"display": i, "uid": i})
}
}
Component.onCompleted: {createModel()}
}
Component {
id: myButton
Rectangle {
id: item
width: mainGrid.cellWidth-5; height: mainGrid.cellHeight-5;
border.width: 1
property int uid: (index >= 0) ? myModel.get(index).uid : -1
Text {
anchors.centerIn: parent
text: display
font.pixelSize: 48
}
states: [
State {
name: "active"; when: gridArea.activeId == item.uid
PropertyChanges {target: item; x: gridArea.mouseX-80; y: gridArea.mouseY-45; z: 10; smooth: false}
}
]
}
}
MouseArea {
id: gridArea
anchors.fill: parent
hoverEnabled: true
preventStealing : true
//property int index: mainGrid.indexAt(mouseX, mouseY) //WHY IS THIS NOT WORKING RELIABLE?
property int mX: (mouseX < 0) ? 0 : ((mouseX < mainGrid.width - mainGrid.cellWidth) ? mouseX : mainGrid.width - mainGrid.cellWidth)
property int mY: (mouseY < 0) ? 0 : ((mouseY < mainGrid.height - mainGrid.cellHeight) ? mouseY : mainGrid.height - mainGrid.cellHeight)
property int index: parseInt(mX/mainGrid.cellWidth) + 5*parseInt(mY/mainGrid.cellHeight) //item underneath cursor
property int activeId: -1 //uid of active item
property int activeIndex //current position of active item
onPressAndHold: {
activeId = mainGrid.model.get(activeIndex=index).uid
}
onReleased: {
activeId = -1
}
onPositionChanged: {
if (activeId != -1 && index != -1 && index != activeIndex) {
mainGrid.model.move(activeIndex, activeIndex = index, 1)
}
}
}
}
Run Code Online (Sandbox Code Playgroud)
在这里,我给出工作代码。我添加了额外的层次结构并将矩形放入项目中。此外,我必须重新调整矩形的父级。有许多类似的解决方案可用,例如此处。
在给定的解决方案中,函数indexAt()没有问题,因此不再需要gridArea.index的显式计算。
我将不胜感激对此解决方案的任何评论以及为什么这些更改如此重要。我仍然认为原来的解决方案很直观,我不明白为什么它不起作用。
import QtQuick 1.1
GridView {
id: mainGrid
cellWidth: 165; cellHeight: 95
width: 5*cellWidth; height: 4*cellHeight
model: myModel
delegate: myButton
ListModel {
id: myModel
function createModel() {
for (var i=1; i<=20; i++) {
myModel.append({"display": i, "uid": i})
}
}
Component.onCompleted: {createModel()}
}
Component {
id: myButton
Item {
id: item
width: mainGrid.cellWidth-5; height: mainGrid.cellHeight-5;
Rectangle {
id: box
parent: mainGrid
x: item.x; y: item.y;
width: item.width; height: item.height;
border.width: 1
property int uid: (index >= 0) ? myModel.get(index).uid : -1
Text {
anchors.centerIn: parent
text: display
font.pixelSize: 48
}
states: [
State {
name: "active"; when: gridArea.activeId == box.uid
PropertyChanges {target: box; x: gridArea.mouseX-80; y: gridArea.mouseY-45; z: 10}
}
]
}
}
}
MouseArea {
id: gridArea
anchors.fill: parent
hoverEnabled: true
preventStealing : true
property int index: mainGrid.indexAt(mouseX, mouseY) //item underneath cursor
property int activeId: -1 //uid of active item
property int activeIndex //current position of active item
onPressAndHold: {
activeId = mainGrid.model.get(activeIndex=index).uid
}
onReleased: {
activeId = -1
}
onPositionChanged: {
if (activeId != -1 && index != -1 && index != activeIndex) {
mainGrid.model.move(activeIndex, activeIndex = index, 1)
}
}
}
}
Run Code Online (Sandbox Code Playgroud)
| 归档时间: |
|
| 查看次数: |
4252 次 |
| 最近记录: |