I know with the Gallery widget I was able to use getSelectedItemPosition(); to retrieve the current position, however it doesnt seem ViewPager has that.
I know I can setup a listener and retrieve the position when the page is switched. But I want the current view position.
You can use:
mViewPager.getCurrentItem()
Create a listener and set it on your viewpager:
/**
* Get the current view position from the ViewPager by
* extending SimpleOnPageChangeListener class and adding your method
*/
public class DetailOnPageChangeListener extends ViewPager.SimpleOnPageChangeListener {
private int currentPage;
#Override
public void onPageSelected(int position) {
currentPage = position;
}
public final int getCurrentPage() {
return currentPage;
}
}
Update 2019
Now you can set addOnPageChangeListener on View Pager to Observe change in Page position.
Since you wanted to setup a listener and retrieve the position when the page is switched
mViewPager.addOnPageChangeListener(object : OnPageChangeListener {
override fun onPageScrollStateChanged(state: Int) {}
override fun onPageScrolled(position: Int, positionOffset: Float, positionOffsetPixels: Int) {}
override fun onPageSelected(position: Int) {
pagePosition.setText("" + position + "/" + galleryAdapter!!.count)
}
})
My solution would only work if you have a TabLayout linked to your ViewPager.
This is how they're linked:
tabLayout.setupWithViewPager(viewPager);
And then to get current position, you can make use of:
tabLayout.getSelectedTabPosition()
I'm telling you now its a hack, so there is no reason to downvote for that reason. Meaning, it either will be helpful to you specifically or not. Either way, the description below will provide some insight and be helpful to the community. Also, this solution is good for the older APIs that don't have ViewPager.getCurrentItem().
First, a little information. If you iterate thru all children of a ViewPager with ViewPager.getChildAt(x); and print out with toString() (or getLeft()) each child View (a page) and then do this everytime you change pages, you will notice that the children will not be in the logical order that they are displayed in when you start going back pages (paging back to the beginning). Apparently, it will remove the unneeded child from the array then append the newest child to the array. So, for example, lets say you are looking at page 2 then changed to page 3, your list of children will be in this order page 2, page 3, page 4 meaning that ViewPager.getChildAt(1); will return the current page. But, if you then change back to page 2 (from page 3) your list of children will be in this order page 2, page 3, page 1 which means that ViewPager.getChildAt(1); does not return the current page. I have not yet been able to find simple logic to weed out the current page using this information. Because the order of the pages in the array behind getChildAt is in an arbitrary order based off of how the user has been paging around.
That being said, I developed a hack work-around. I have no clue if this function will work in all in environments, but it works for my current project. I would suspect if doesn't for you, then its an issue of different API level. But I dont actually suspect any issues for other environments.
Now, onto the meat. What I noticed was that the result of ViewPager.getChildAt(x).getLeft() will have some type of horizontal pixel coordinate relative to the parent. So, I used this information to weed out which view is the current one.
private int getCurrentPageIndex(ViewPager vp){
int first,second,id1,id2,left;
id1 = first = second = 99999999;
View v;
for ( int i = 0, k = vp.getChildCount() ; i < k ; ++i ) {
left = vp.getChildAt(i).getLeft();
if ( left < second ) {
if ( left < first ) {
second = first;
id2 = id1;
first = left;
id1 = i;
} else {
second = left;
id2 = i;
}
}
}
return id2;
}
This function is probably a questionable hack because it relies on the value of getLeft() to figure it all out. But, I grab the left coordinate of each child. I then compare this to the other values and store the first and second pages, returning the second page (current page) out of the function. It seems to work beautifully.
Why (might you ask) didn't I just use onClickListenter or whatever solution? Well, I was darned determined that there was a straight forward way to do this without having to include listeners, other classes, inconclusive focus, and other bloat. Unfortunately, this solution is not exactly straight forward. But, it does do away with the bloat, other classes and listeners. If I can figure out a more straight forward way, I will be rewriting this function. Or maybe, this will provide insight for someone else to have an epiphany.
Related
There are 2 drop-down lists. Each has a similar meaning, for example, "Jorge". Lists in different modules. When I need to fill in, for example, a list that is lower in the tree, then the first match is taken along the XPath path, on an undisclosed list.
Not lists, but values in drop-down lists!
There are 2 drop-down lists. Each has a similar meaning, for example, "Jorge". Lists in different modules. When I need to fill in, for example, a list that is lower in the tree, then the first match is taken along the XPath path, on an undisclosed list.
Not lists, but values in drop-down lists!
I wanted to implement it in Java this way:
Example:
if (findElement(By.xpath("(//example//example)")).isDisplayed()) {
findElement(By.xpath("(//example//example)")).click();
}
But in this case, the element is not displayed.
How to implement a search of all values similar to the XPath path in order to get the one that is displayed?
I tried to do something like this: (//example//example)1 (//example//example)[2] (//example//example)[3]
In my case, we have that 1 - the element does not exist [2] - exists, but is not displayed (isDisplayed = false) [3] - exists, is displayed (isDisplayed = true)
iterating through the values in the loop for [n] cannot be implemented, because, for example, the value 1 is not.
Described as difficult as possible :D. Excuse me.
If someone understands my nonsense, please help me. How to implement my requirement?
enter image description here
UPD:
The problem was solved (for me) by substituting the first value into the expression ()"{1}" immediately.
Now I'm interested in why I get an exception after the first iteration:
Method threw 'org.openqa.selenium.ElementNotInteractableException' exception.
Code:
int number = 1;
String option = "(//ul[contains(#style, 'display: block')]//li//span[contains(text(),'" + valueField + "') or strong[contains(text(),'" + valueField.toUpperCase() + "')]])";
findElement(By.xpath(option+"["+number+"]"));
String[] words = valueField.split(" ");
StringBuilder builder = new StringBuilder();
for (int i = 0; i < words.length; i++) {
builder.append(words[i]);
setFieldByLabel(nameModule, nameLabel, builder.toString());
fastWaitLoading();
for (int y = 0; y < 10; y++) {
if (findElement(By.xpath(option+"["+number+"]")).isDisplayed()) {
new Actions(browser.getWebDriver())
.moveToElement(findElement(option))
.click()
.build()
.perform();
break;
}
number++;
}
}
So I am trying to fully understand your question, and I don't. What I would recommend for a situation like this is, iterate through all elements by creating a list with: findElements(By.xpath ... )
This way you will get a list of webelements and you can iterate through them. Then apply a foreach, assert if element is displayed (it exists as it has been found with findElements) and you should be able to interact with it.
Yeah, everything is in a prominent place)
Missed it
new actions(browser.getWebDriver()) .moveToElement(findElement(**option**)) .click() .build() .perform(); break;
Here
new actions(browser.getWebDriver())
.moveToElement(findElement(**option + "[" + number+"]"**))
.click()
.build()
.perform();
break;
In my app, I want a counter from 0 to 8 to decide the number of players in a game.
Below there are 8 possible fields to write a name inside, which are all set to invisible. If the players-counter is set to 3 players, there should be the first 3 fields visible. Depending on the actual number of the counter, the visibility of the fields changes (1player = first field, 5 players = first 5 fields).
When the +1 (player) button is clicked, a certain method is activated. I tried to run a for-loop everytime the button is clicked. In this for-loop from 0 to "whatever amount" (max. 8 players) the actual fields should be found with "findById" and set to visible.
I tried it with a string resource (.xml) and I can get the text of the resource but with my thought process, I have to update the string resource to every number of the field (if 3 players: "field_" + "1", "field_" + "2", "field_" + "3").
How can I get and (most importantly) set/update a string resource for this specific purpose?
(Switch is too inefficient and I can't use a string with the findViewBy Id()-method by updating the String (not string resource) like mentioned before.
Please help, and accept the fact that I'm new to Android Studio for one week!)
You can use "getIdentifier" which takes a String parameter. So you can set the type as "id" in the second parameter of this method. This method returns the id of the view you want, but beware, it will throw a "FATAL EXCEPTION" if the id of the View doesn't exist. With this id, you can use findViewById to fetch the TextView and change its visibility. The "getIdentifier" method can be called from the "getResources()" method.
Below you can see what it would be like to make visible a TextView that has the id "textView1":
int id = getResources().getIdentifier("textView1", "id", getPackageName());
TextView textView = findViewById(id);
textView.setVisibility(View.VISIBLE);
Below you can see how you would make 8 TextView with id 1 to 8 visible:
TextView textView;
for (int i = 1; i <= 8; i++) {
int id = getResources().getIdentifier("textView" + i, "id", getPackageName());
textView = findViewById(id);
textView.setVisibility(View.VISIBLE);
}
So, just put the limit at i <= x , with x being the limit of players who will play:
TextView textView;
for (int i = 1; i <= totalPlayers; i++) {
int id = getResources().getIdentifier("textView" + i, "id", getPackageName());
textView = findViewById(id);
textView.setVisibility(View.VISIBLE);
}
Do you just want to make some EditTexts visible and others not? Personally I'd keep it simple, do the lookups once (in onCreate or wherever) and store the references in a list. Then when you need to display n fields, you can just iterate over the list and set the first n to VISIBLE and the rest to INVISIBLE.
I feel like it's fine to just list all the EditText IDs (R.id.field_1 etc) and generate your list of actual Views from that, but if that repetition bothers you, there's a few things you could do. Like:
set a tag attribute on each field in the XML, and use findViewWithTag to look them up, generating the lookup strings programmatically, like "field_" + i
do a similar thing with the resource ID, like in #Moises's answer
lookup their containing layout, use getChildCount and [getChildAt] to iterate over the views in that layout, and use isInstance to collect all the EditTexts in order3
create and add the EditTexts in code - you probably don't want to do this, but you could!
I'm not really sure what you mean about the string resource or what you're trying to do - I'd honestly just make a list of R.id.field_1 etc, iterate over that to do findViewById on each and store those in a new list, and you're done. Also my Java's a bit rusty so sorry no example code!
I am creating a simple game of 15 puzzle (Please see here: https://en.wikipedia.org/wiki/15_puzzle). I have an array of 4 by 4 buttons and want to know the best way to handle a user clicking on a button and moving around the numbers.
I have started off by creating a JavaFX gridpane to display the buttons as well as a HashMap (suggested by multiple people) to store a mapping between the displayed number (1-15) and the button at that location.
I populate the map with class objects "Btn"
public class Btn {
private int ID;
private int x;
private int y;
public Button button;
for (int i = 1; i <= 4; i++) {
for (int j = 1; j <= 4; j++) {
Btn btn = new Btn(count, j, i, new Button("" + count));
map.put(count, btn);
btn.button.setMinSize(100, 100);
btn.button.setOnAction(e -> {
if (isMoveLegal(btn) == true) {
move(btn);
}
});
count++;
grid.add(btn.button, j, i);
}
}
So far it has worked quite well. In the isMoveLegal() method I check if the empty field is above, left, right, or below the clicked button. Which gets quite complicated because of the HashMaps nature of not preserving any order. But my real problem lies in the move() method. Since there is no order, I can't just switch 2 map nodes or can I?
if (MoveDir.equals("Right")) {
map.get((btn.getYs()+1)*4 - (4-btn.getXs())).button.setText("" + map.get(btn.getID()).getID());
map.get(btn.getID()).button.setText("" + map.get((btn.getYs()+1)*4 - (4-btn.getXs())).getID());
int tempID = map.get(btn.getID()).getID();
map.get(btn.getID()).setID(16);
map.get((btn.getYs()+1)*4 - (4-btn.getXs())).setID(tempID);
// System.out.println(map.get(btn.getID()).getID());
// System.out.println(map.get((btn.getYs()+1)* 4 - (4-btn.getXs())).getID());
map.get(btn.getID()).setYs(map.get(btn.getID()).getYs()+1);
map.get((btn.getYs()+1)* 4 - (4-btn.getXs())).setYs(map.get((btn.getYs()+1)* 4 - (4-btn.getXs())).getYs()-1);
This is what I have come up with so far, which extremely complicated and tedious. To quickly sum it up, I first switch the labels of the buttons, then I change the ID variables of the 2 buttons, and then I change the cord variables (x & y) to indicate their new location in map/grid.
I'm pretty sure I have gone way too far with this and there is a simpler solution to all of this but I am just unsure what that would be. Should I ditch the HashMap entirely and just use an object array of Btns?
Thanks for the help.
The problem you're facing is that your UI and your data structure are too tightly coupled. You need to separate them, give them their own responsibilities, and use communication (method calls) between them to make things happen.
For example, the UI should only be responsible for displaying the numbers and allowing the user to click a number to perform a move. Using a gridpane of buttons seems reasonable for this. Each button only needs to know it's x,y coordinate.
The data model should be a 2-dimensional array that contains the number at that location. For example:
+-+-+-+
|4|1|8|
+-+-+-+
|2|3|7|
+-+-+-+
| |5|6|
+-+-+-+
(You can figure out how to represent the empty space. Maybe a null, or the number '0' or '-1')
The data model can be wrapped in a data object that handles questions about and manipulations to the data model. For example, one method it might contain is getTheNumberAtLocation(x,y) which the buttons will call, and will return the number at that location. The buttons will use this to determine what number they should show.
Another method it might contain is isThereASpaceNextToLocation(x,y). This would return whether or not a space exists next to the location given by the x,y. The button will call this method to determine whether or not it can be clicked.
Finally, you could have a moveLocationToEmptySpace(x,y) which will manipulate the data to put the number at the current location into the empty space. After that, all the buttons should call getTheNumberAtLocation(x,y) to update the number they are showing.
By arranging your code like this you've nicely separated the concern of the UI (show text, handle button clicks) from the concern of maintaining the game state (checking if a move is valid, performing a move, determining the state of the board). Now when you write code it will be easier to write because each method will be responsible for a small, manageable piece of logic.
To summarise:
Gridpane (4x4)
-> Button
-> label [calls GameData.getTheNumberAtLocation]
-> click [calls GameData.isThereASpaceNextToLocation, moveLocationToEmptySpace]
GameData
-> Array (4x4) (private)
-> Integer (1-15)
-> getTheNumberAtLocation(x,y)
-> isThereASpaceNextToLocation(x,y)
-> moveLocationToEmptySpace(x,y)
I am trying to create "AI" for Nine Men's Morris but I got hardstuck on minMax algorithm. Summing up, I was trying to find the issue for over 10h but didn't manage to. (debugging this recursion is nasty or I am doing it badly or both)
Since I started doubting everything I wrote I decided to post my issue so someone can find anything wrong in my version of minMax. I realise it is really hard task without the whole application so any suggestions where I should triple check my code are also very welcome.
Here is link to the video, explaining minMax, on which I based my implementation: https://www.youtube.com/watch?v=l-hh51ncgDI (First video that pops up on yt after searching for minmax - just in case you want to watch the video and don't want to click the link)
My minMax without alpha-beta pruning:
//turn - tells which player is going to move
//gameStage - what action can be done in this move, where possible actions are: put pawn, move pawn, take opponent's pawn
//depth - tells how far down the game tree should minMax go
//spots - game board
private int minMax(int depth, Turn turn, GameStage gameStage, Spot[] spots){
if(depth==0){
return evaluateBoard(spots);
}
//in my scenario I am playing as WHITE and "AI" is playing as BLACK
//since heuristic (evaluateBoard) returns number equal to black pawns - white pawns
//I have decided that in my minMax algorithm every white turn will try to minimize and black turn will try to maximize
//I dont know if this is correct approach but It seems logical to me so let me know if this is wrong
boolean isMaximizing = turn.equals(Turn.BLACK);
//get all possible (legal) actions based on circumstances
ArrayList<Action> children = gameManager.getAllPossibleActions(spots,turn,gameStage);
//this object will hold information about game circumstances after applying child move
//and this information will be passed in recursive call
ActionResult result;
//placeholder for value returned by minMax()
int eval;
//scenario for maximizing player
if(isMaximizing){
int maxEval = NEGATIVE_INF;
for (Action child : children){
//aplying possible action (child) and passing its result to recursive call
result = gameManager.applyMove(child,turn,spots);
//evaluate child move
eval = minMax(depth-1,result.getTurn(),result.getGameStage(),result.getSpots());
//resets board (which is array of Spots) so that board is not changed after minMax algorithm
//because I am working on the original board to avoid time consuming copies
gameManager.unapplyMove(child,turn,spots,result);
if(maxEval<eval){
maxEval = eval;
//assign child with the biggest value to global static reference
Instances.theBestAction = child;
}
}
return maxEval;
}
//scenario for minimizing player - the same logic as for maximizing player but for minimizing
else{
int minEval = POSITIVE_INF;
for (Action child : children){
result = engine.getGameManager().applyMove(child,turn,spots);
eval = minMax(depth-1,result.getTurn(),result.getGameStage(),result.getSpots());
engine.getGameManager().unapplyMove(child,turn,spots,result);
if(minEval>eval){
minEval=eval;
Instances.theBestAction = child;
}
}
return minEval;
}
}
Simple heuristic for evaluation:
//calculates the difference between black pawns on board
//and white pawns on board
public int evaluateBoard(Spot[] spots) {
int value = 0;
for (Spot spot : spots) {
if (spot.getTurn().equals(Turn.BLACK)) {
value++;
}else if(spot.getTurn().equals(Turn.WHITE)){
value--;
}
}
return value;
}
My issue:
//the same parameters as in minMax() function
public void checkMove(GameStage gameStage, Turn turn, Spot[] spots) {
//one of these must be returned by minMax() function
//because these are the only legal actions that can be done in this turn
ArrayList<Action> possibleActions = gameManager.getAllPossibleActions(spots,turn,gameStage);
//I ignore int returned by minMax() because,
//after execution of this function, action choosed by minMax() is assigned
//to global static reference
minMax(1,turn,gameStage,spots);
//getting action choosed by minMax() from global
//static reference
Action aiAction = Instances.theBestAction;
//flag to check if aiAction is in possibleActions
boolean wasFound = false;
//find the same action returned by minMax() in possibleActions
//change the flag upon finding one
for(Action possibleAction : possibleActions){
if(possibleAction.getStartSpotId() == aiAction.getStartSpotId() &&
possibleAction.getEndSpotId() == aiAction.getEndSpotId() &&
possibleAction.getActionType().equals(aiAction.getActionType())){
wasFound = true;
break;
}
}
//when depth is equal to 1 it always is true
//because there is no other choice, but
//when depth>1 it really soon is false
//so direct child of root is not chosen
System.out.println("wasFound?: "+wasFound);
}
Is the idea behind my implementation of minMax algorithm correct?
I think the error might exist in that you are updating Instances.theBestAction even while evaluating child moves.
For example, lets say 'Move 4' is the true best move that will eventually be returned, but while evaluating 'Move 5', theBestAction is set to the best child action of 'Move 5'. From this point on, you won't update the original theBestAction back to 'Move 4'.
Perhaps just a simple condition that only sets theBestAction when depth == originalDepth?
Rather than using a global, you could also consider returning a struct/object that contains both the best score AND the move that earned the score.
I am using a CellTable in GWT and trying to add styling to a row whenever certain events happen. The code to add styling is as follows:
Range range = playlistTable.getVisibleRange();
int start = range.getStart();
for (int i = start; i < start + playlistTable.getPageSize(); i++) {
if (playlistData.get(i) == currentSong) {
playlistTable.getRowElement(i).addClassName("highlightRow");
} else {
playlistTable.getRowElement(i).removeClassName("highlightRow");
}
}
When the app loads and the first page of the cell table is displayed, everything works perfectly. However when I scroll to the next page and make a call to the above code, GWT throws an IndexOutOfBoundsException. It does not like my index when I make the call to getRowElement, which only occurs if the index is "not on the current page." It's as though getRowElement always thinks the current page is the first page. If I scroll back to the first page, everything behaves fine. Has anybody encountered this? Am i missing something or is this a bug in GWT?
I suspect the problem is that start is 0 on the page that works. What happens if you do playlistTable.getRowElement(i - start) ? You may have to throw a + 1 or a - 1 in somewhere.
Not a direct answer to your question, but your code would be much simpler with RowStyles
// do it only once!
playlistTable.setRowStyles(new RowStyles<Song>() {
#Override
public String getStyleNames(Song row, int rowIndex) {
return (row == currentSong) ? "highlightRow": "";
}
});
You might also want to use equals() and/or some ProvidesKey object to compare the row with the currentSong.