在pyqt中显示接收的图像

Question

我正在尝试显示我从pyqt窗口内的测试本地服务器收到的图像。我试过这个链接，但我无法让它工作。你可以帮忙吗？提前致谢。

Displaying Image (in bytes) in PyQt

客户代码

import socket
from PIL import Image
import sys
from PyQt5.QtWidgets import QApplication, QWidget, QLabel
from PyQt5.QtGui import QIcon, QPixmap

app = QApplication(sys.argv)
widget = QWidget()
label = QLabel(widget)

HOST = socket.gethostbyname(socket.gethostname())
PORT = 5000

s = socket.socket()
s.connect((HOST, PORT))

while True:
    image_binaries = s.recv(100000000)
    if not image_binaries:
        break
    #img = Image.open(BytesIO(image_binaries))


pixmap = QPixmap()
pixmap.loadFromData(image_binaries)
label.setPixmap(pixmap)
widget.resize(pixmap.width(), pixmap.height())

widget.show()
sys.exit(app.exec_())

服务器代码

import socket

HOST = "0.0.0.0"
PORT = 5000

s = socket.socket()
s.bind((HOST, PORT))

data = open(r'Screenshot_1.jpg', 'rb').read()
print("Waiting  for connection...")

s.listen(5)
while True:
    client, client_address = s.accept()
    print(client_address, "is connected")
    client.send(data)

Answer 1

此代码中存在不同的问题。

首先是读取循环是错误的。 TCP是一种流协议。这意味着数据可能包含您必须连接的可变数量的数据包。所以它应该是：

image_binaries = b""
while True:
    chunk = s.recv(100000000)
    if not chunk:
        break
    image_binaries += chunk
s.close()

然后你有一个可能的竞争条件服务器端。一旦所有数据排队等待传输，client.send(data)将立即返回。如果立即发生另一个连接，将重用client，这可能导致先前的连接在发送所有内容之前关闭。你应该在这里使用graceful shutdown：

while True:
    client, client_address = s.accept()
    print(client_address, "is connected")
    client.send(data)
    client.shutdown(socket.SHUT_WR)   # notify that nothing more is to be sent
    _ = client.recv(16)               # wait for the client to close when everything has been received
    client.close()                    # Ok, we can explicitely close