Android 如何使imageView离开屏幕[Kotlin]
所以我是Kotlin的新手,我正在尝试制作一个超级简单的应用程序。它所做的就是当我点击右键时,它会向右移动,和左键一样。问题是,当我单击任一按钮(例如右键)时,我可以单击它,直到图像完全脱离屏幕。那么,我如何实现一个代码,一旦它击中屏幕边缘,它就停止移动 我的代码Android 如何使imageView离开屏幕[Kotlin],android,kotlin,Android,Kotlin,所以我是Kotlin的新手,我正在尝试制作一个超级简单的应用程序。它所做的就是当我点击右键时,它会向右移动,和左键一样。问题是,当我单击任一按钮(例如右键)时,我可以单击它,直到图像完全脱离屏幕。那么,我如何实现一个代码,一旦它击中屏幕边缘,它就停止移动 我的代码 package com.example.change_position_circle import android.animation.ObjectAnimator import androidx.appcompat.app.App
package com.example.change_position_circle
import android.animation.ObjectAnimator
import androidx.appcompat.app.AppCompatActivity
import android.os.Bundle
import android.widget.Button
import kotlinx.android.synthetic.main.activity_main.*
class MainActivity : AppCompatActivity() {
override fun onCreate(savedInstanceState: Bundle?) {
super.onCreate(savedInstanceState)
setContentView(R.layout.activity_main)
//val picture = findViewById<ImageView>(R.id.SpongeBob)
val right_button = findViewById<Button>(R.id.right)
val left_button = findViewById<Button>(R.id.left)
right_button.setOnClickListener()
{
//ObjectAnimator.ofFloat(SpongeBob, "x", 100)
SpongeBob.animate().setDuration(90).translationXBy(100f)
}
left_button.setOnClickListener()
{
//ObjectAnimator.ofFloat(SpongeBob, "translationXBy", 100f).apply {
//duration = 200
// start()
SpongeBob.animate().setDuration(90).translationXBy(-100f)
//}
}
}
}
感谢您的帮助欢迎来到科特林
所以是的,你已经有了你的海绵宝宝动画,现在你需要一些逻辑来控制动画。这里的问题是你并不总是希望完整的动画发生,对吧?如果他离屏幕边缘太近,你希望按钮只能将他移动到看不见的墙,如果他正好靠在墙上,那就意味着根本不移动
动画和绘图系统不会对可以放置视图的位置设置任何限制,因此由您自己处理。您基本上需要在单击按钮时执行此操作:
获取海绵宝宝的位置坐标,这是你现在真正关心的X
计算出你关心的边的位置,视图的坐标描述了左上角的位置,因此如果你在看右边,你需要X坐标+视图的宽度
计算出屏幕或父布局边缘的X坐标,无论您希望海绵宝宝包含在其中
如果海绵宝宝边和屏幕边之间的距离小于正常动画移动,则需要将其更改为剩余距离
您还需要计算出适当的持续时间,如果他移动的距离为正常距离的一半,则动画所需的时间应为正常距离的一半
这需要做很多工作,有几种方法可以做到这一点,但这里有一种方法只是使用屏幕边缘作为边界
import android.os.Bundle
import android.view.View
import android.widget.Button
import androidx.appcompat.app.AppCompatActivity
import kotlin.math.abs
import kotlin.math.min
private const val MOVE_DISTANCE = 100
private const val MOVE_TIME = 90
class MainActivity : AppCompatActivity() {
private var screenWidth = 0
private lateinit var spongeBob : View
override fun onCreate(savedInstanceState: Bundle?) {
super.onCreate(savedInstanceState)
setContentView(R.layout.spongebob)
// store this when the Activity is created, if the device is rotated the Activity
// will be recreated and this method gets run again
screenWidth = applicationContext.resources.displayMetrics.widthPixels
//val picture = findViewById<ImageView>(R.id.SpongeBob)
val right_button = findViewById<Button>(R.id.right)
val left_button = findViewById<Button>(R.id.left)
spongeBob = findViewById(R.id.spongeBob)
right_button.setOnClickListener()
{
// two possible values - the distance to the edge, and the normal amount we move
// we want the smaller of the two (i.e. always move the normal amount, unless
// the edge is closer than that
val distance = min(distanceToEdge(left = false), MOVE_DISTANCE)
moveSpongeBob(distance)
}
left_button.setOnClickListener()
{
val distance = min(distanceToEdge(left = true), MOVE_DISTANCE)
// we're moving left so we need to use a negative distance
moveSpongeBob (-distance)
}
}
private fun distanceToEdge(left: Boolean): Int {
// Get the Spongebob's top-left position - the call is a method on the View class,
// I'm assuming SpongeBob is a View, and you need to pass an array in because
// that's just how it works for whatever reason...
val location = IntArray(2)
spongeBob.getLocationOnScreen(location)
val x = location[0]
// I'm just using the View.getWidth() call here (Kotlin style) but I don't know
// what the SpongeBob class is, so you'll need to handle this
// You could set this once, like when we get the screen width, but width will be 0 until
// the View is laid out - so you can't do it in #onCreate, #onViewCreated should work
val spongeBobWidth = spongeBob.width
// the left edge is just the x position, however far that is from zero
return if (left) x
// the right edge is the x position plus the width of the bob
else screenWidth - (x + spongeBobWidth)
}
// Actually move the view, by the given distance (negative values to move left)
private fun moveSpongeBob(distance: Int) {
// work out how much this distance relates to our standard move amount, so we can
// adjust the time by the same proportion - converting to float so we don't get
// integer division (where it's rounded to a whole number)
val fraction = distance.toFloat() / MOVE_DISTANCE
// distance can be negative (i.e. moving left) so we need to use the abs function
// to make the duration a postitive number
val duration = abs(MOVE_TIME * fraction).toLong()
spongeBob.animate().setDuration(duration).translationXBy(distance.toFloat())
}
}
有更好的东西你可以做,海绵宝宝应该被称为海绵宝宝,并成为一个视图,但这是最基本的。在坐标系上也可能对您有所帮助。欢迎来到Kotlin
所以是的,你已经有了你的海绵宝宝动画,现在你需要一些逻辑来控制动画。这里的问题是你并不总是希望完整的动画发生,对吧?如果他离屏幕边缘太近,你希望按钮只能将他移动到看不见的墙,如果他正好靠在墙上,那就意味着根本不移动
动画和绘图系统不会对可以放置视图的位置设置任何限制,因此由您自己处理。您基本上需要在单击按钮时执行此操作:
获取海绵宝宝的位置坐标,这是你现在真正关心的X
计算出你关心的边的位置,视图的坐标描述了左上角的位置,因此如果你在看右边,你需要X坐标+视图的宽度
计算出屏幕或父布局边缘的X坐标,无论您希望海绵宝宝包含在其中
如果海绵宝宝边和屏幕边之间的距离小于正常动画移动,则需要将其更改为剩余距离
您还需要计算出适当的持续时间,如果他移动的距离为正常距离的一半,则动画所需的时间应为正常距离的一半
这需要做很多工作,有几种方法可以做到这一点,但这里有一种方法只是使用屏幕边缘作为边界
import android.os.Bundle
import android.view.View
import android.widget.Button
import androidx.appcompat.app.AppCompatActivity
import kotlin.math.abs
import kotlin.math.min
private const val MOVE_DISTANCE = 100
private const val MOVE_TIME = 90
class MainActivity : AppCompatActivity() {
private var screenWidth = 0
private lateinit var spongeBob : View
override fun onCreate(savedInstanceState: Bundle?) {
super.onCreate(savedInstanceState)
setContentView(R.layout.spongebob)
// store this when the Activity is created, if the device is rotated the Activity
// will be recreated and this method gets run again
screenWidth = applicationContext.resources.displayMetrics.widthPixels
//val picture = findViewById<ImageView>(R.id.SpongeBob)
val right_button = findViewById<Button>(R.id.right)
val left_button = findViewById<Button>(R.id.left)
spongeBob = findViewById(R.id.spongeBob)
right_button.setOnClickListener()
{
// two possible values - the distance to the edge, and the normal amount we move
// we want the smaller of the two (i.e. always move the normal amount, unless
// the edge is closer than that
val distance = min(distanceToEdge(left = false), MOVE_DISTANCE)
moveSpongeBob(distance)
}
left_button.setOnClickListener()
{
val distance = min(distanceToEdge(left = true), MOVE_DISTANCE)
// we're moving left so we need to use a negative distance
moveSpongeBob (-distance)
}
}
private fun distanceToEdge(left: Boolean): Int {
// Get the Spongebob's top-left position - the call is a method on the View class,
// I'm assuming SpongeBob is a View, and you need to pass an array in because
// that's just how it works for whatever reason...
val location = IntArray(2)
spongeBob.getLocationOnScreen(location)
val x = location[0]
// I'm just using the View.getWidth() call here (Kotlin style) but I don't know
// what the SpongeBob class is, so you'll need to handle this
// You could set this once, like when we get the screen width, but width will be 0 until
// the View is laid out - so you can't do it in #onCreate, #onViewCreated should work
val spongeBobWidth = spongeBob.width
// the left edge is just the x position, however far that is from zero
return if (left) x
// the right edge is the x position plus the width of the bob
else screenWidth - (x + spongeBobWidth)
}
// Actually move the view, by the given distance (negative values to move left)
private fun moveSpongeBob(distance: Int) {
// work out how much this distance relates to our standard move amount, so we can
// adjust the time by the same proportion - converting to float so we don't get
// integer division (where it's rounded to a whole number)
val fraction = distance.toFloat() / MOVE_DISTANCE
// distance can be negative (i.e. moving left) so we need to use the abs function
// to make the duration a postitive number
val duration = abs(MOVE_TIME * fraction).toLong()
spongeBob.animate().setDuration(duration).translationXBy(distance.toFloat())
}
}
有更好的东西你可以做,海绵宝宝应该被称为海绵宝宝,并成为一个视图,但这是最基本的。在坐标系上也可能对您有所帮助。谢谢!它按预期工作,并感谢添加评论,以帮助我更好地理解它。之所以没有将海绵宝宝初始化为视图,是因为我使用的是kotlinx.android.synthetic.main.activity_main。*我在Kotlin中学到,只要使用该软件包,就不必像在Java中那样初始化视图。哦,好吧,我没有弄乱这些扩展!我对视图的意思是,我使用的是视图方法和属性,比如.width,所以代码假设无论Spongebob是什么类,它都是从视图派生的-所以如果它是ImageView或其他什么,那就好了!非常感谢。它按预期工作,并感谢添加评论,以帮助我更好地理解它。之所以没有将海绵宝宝初始化为视图,是因为我使用的是kotlinx.android.synthetic.main.activity_main。*我在Kotlin中学到,只要使用该软件包,就不必像在Java中那样初始化视图。哦,好吧,我没有弄乱这些扩展!我对视图的意思是,我使用的是视图方法和属性,比如.width,所以代码假设无论Spongebob是什么类,它都是从视图派生的-所以如果它是ImageView或其他什么,那就好了!