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MIT App Inventor Help
1

how to save the coordinates resulting from IF, (X, Y) under a different name x1, y1
x2, y2
x3, y3
.....
xn, yn
where n = quantity from IF

then, from each X and Y, I will calculate new relationships
Google Photos

2

somehow I do not understand the question...
and as up to now you did not receive any answer, my guess is, others also do not understand...
so please elaborate... what is IF, (X,Y)... what are you calculating...
see also

Taifun

Trying to push the limits! Snippets, Tutorials and Extensions from Pura Vida Apps by icon24 Taifun.

3

I can see that the blocks are in Polish. Ask the question in Polish, maybe I understand the question at least.

4

This problem looks like a time series of (x,y) pairs, where the user wants to show the series in a List View.

To do this, I would use a list, where each item in the list is a dictionary with tags "x" and "y" for the x and y values respectively.

As each new pair (x,y) is calculated from the (x,y) values at length of list (the end),
use an Add item to list block to add a newly created dictionary with the new x and y tags/values.

Assigning the list to a ListView Elements hopefully will force each list item into a text vale. The text value of a dictionary should look like the JSON representation of that dictionary, {"x": 20, "y": 25}

5

Moje wyjaśnienia

6

Google Photos

Google Photos

Google Photos

Google Photos

Polish
w celu dokladnego wytłumaczenie wklejam wszystko, w celach testowych dla większej szybkości zostawiłem testowane elementy, reszta jest wyłączona bo będzie taka sama co do zasady działania.

  1. canvas1 360x480
    podzieliłem na 4 części
    po każdej części biega inna Ball
  2. Jeśli Ball znajdzie kolor czerwony RGB Wpisuje go globalCurentX CurentY,
  3. Linie w canvas1 sprawdzane są co 4 i co 4 px - aby dokładnie wyznaczyć środek kropki przeprowadzam dla niej następne próbkowanie, ale tym razem co każdą linie i px. Tu powinienem uzyskać wynik, ale mam błąd.
    Opisany przykład dotyczy tylko jedego punktu, co w przypadku kilku i to w 3 kolorach.
    POMOCY

for a thorough explanation, I paste everything, for testing purposes, for higher speed, I left the tested elements, the rest is turned off because it will be the same in principle.

  1. canvas1 360x480
    I divided it into 4 parts
    a different Ball runs on each part
  2. If Ball finds the color red RGB Enter it globalCurentX CurentY,
  3. The lines in canvas1 are checked every 4 and every 4 px - to accurately determine the center of the dot, I do another sampling for it, but this time every line and px. This is where I should get the result but I have a bug.
    The described example applies only to one point, which in the case of several, and in 3 colors.
    HELP
7
8

Sreen8. ......