How to implement collision in pygame?

Question

pygame

I need to collide the green square so that the player's center cannot go through its walls.
You can't just do it with ifs, because the object will then be moved or even deleted.

import sys, pygame

pygame.init()

FPS = 60
SCREEN = (640, 480)
screen = pygame.display.set_mode(SCREEN)
rect = pygame.Rect(100, 100, 200, 200) # объект для хранения прямоугольных координат
player = pygame.Rect(20, 20, 40, 40)

x_player = 50
y_player = 50
speed = 1

while True:
    for event in pygame.event.get():
        if event.type == pygame.QUIT: # выход
            pygame.quit()
            sys.exit()
    # передвижение персонажа ↓
    keys = pygame.key.get_pressed()
    if keys[pygame.K_d] and x_player < SCREEN[0]:
        x_player += speed
    if keys[pygame.K_a] and x_player > 0:
        x_player -= speed
    if keys[pygame.K_s] and y_player < SCREEN[1]:
        y_player += speed
    if keys[pygame.K_w] and y_player > 0:
        y_player -= speed
    if keys[pygame.K_LSHIFT] and speed < 2:
        speed += 1
    if not keys[pygame.K_LSHIFT] and speed > 1:
        speed -= 1
    # передвижение персонажа ↑

    # правильная коллизия ↓
    if x_player <= -1:
        x_player = 0
    if y_player <= -1:
        y_player = 0
    if x_player >= SCREEN[0]:
        x_player = SCREEN[0]
    if y_player >= SCREEN[1]:
        y_player = SCREEN[1]


    # коллизия на объекты просто так не работает, нужно сделать отдельные 8 префабов для коллизии
    if rect.collidepoint(x_player, y_player):
        y_player -= speed
        print("collision")

    # правильная коллизия ↑


    # print(x_player, y_player)

    screen.fill((0,0,0))
    pygame.draw.circle(screen, "YELLOW", (x_player, y_player), 10)
    pygame.draw.rect(screen, "GREEN", (100, 100, 200, 200))
    pygame.display.update()
    pygame.time.Clock().tick(FPS)

Answer 1

You need to use the search on the site with a request

pygame collision

and work through the results. This is a common question and there are many answers. If what has already been answered is not enough for you, I doubt the success in the case of your question.