pySerial捕获长响应

胜利者

嗨,大家好,我正在研究一个脚本,该脚本将使用数据通信标准由Vision Council的数据通信标准委员会镜头处理部门开发)通过串行端口从主机获取数据,并将数据传递给ModBus协议设备执行其操作。

由于我从财政上无权访问主机,因此我试图开发一个辅助脚本来模拟主机。我目前正处于需要从串行端口读取大量信息的阶段,而我仅获得部分数据。我希望在主机仿真器脚本send_job()函数上发送整个字符串

你们还可以告诉我这是否是个好方法吗?机器唯一要做的就是从主机响应中获取2个值,并将其分配给两个Modbus保持寄存器。

注意:初始化函数是硬编码的,因为它始终是相同的,并且除了状态之外,实际的响应数据都无关紧要。同样,作业请求是硬编码的,我只传递了我从Modbus保持寄存器中获得的作业#,关于主机如何解决此问题的确切逻辑应该无关紧要,我只需要以这种格式发送从设备扫描的作业编号。

主要脚本:

def request_job_modbus(job):
    data = F'[06][1c]req=33[0d][0a]job={job}[0d][0a][1e][1d]'.encode('ascii')
    writer(data)

def get_job_from_serial():
    response = serial_client.read_all()
    resp = response.decode()
    return resp




# TODO : SEND INIT SEQUENCE ONCE AND VERIFY IF REQUEST status=0
initiation_request()
init_response_status = get_init_status()
print('init method being active')
print(get_init_status())


while True:



  # TODO: get job request data
  job_serial = get_job_from_serial()
  print(job_serial)

主机仿真脚本:

def send_job():
        job_response = '''[06][1c]ans=33[0d]job=30925[0d]status=0;"ok"[0d]do=l[0d]add=;2.50[0d]ar=1[0d]
                        bcerin=;3.93[0d]bcerup=;-2.97[0d]crib=;64.00[0d]do=l[0d]ellh=;64.00[0d]engmask=;613l[0d]
                        erdrin=;0.00[0d]erdrup=;10.00[0d]ernrin=;2.00[0d]ernrup=;-8.00[0d]ersgin=;0.00[0d]
                        ersgup=;4.00[0d]gax=;0.00[0d]gbasex=;-5.30[0d]gcrosx=;-7.96[0d]kprva=;275[0d]kprvm=;0.55[0d]
                        ldpath=\\uscqx-tcpmain-at\lds\iot\do\800468.sdf[0d]lmatid=;151[0d]lmatname=;f50[0d]
                        lnam=;vsp_basic_fh15[0d]sgerin=;0.00[0d]sgerup=;0.00[0d]sval=;5.18[0d]text_11=;[0d]
                        text_12=;[0d]tind=;1.53[0d][1e][1d]'''.encode('ascii')

        writer(job_response)


def get_init_request():
        req = p.readline()
        print(req)

        request = req.decode()[4:11]
        # print(request)

        if request == 'req=ini':
            print('request ==  req=ini??? <<<<<<< cumple condicion y enviala respuesta')
            send_init_response()
            send_job()


while True:


        # print(get_init_request())
        get_init_request()

屏幕上显示的内容:主脚本

init method being active
     bce
     erd
condition was met init status=0
outside loop
     ers
condition was met init status=0
inside while loop
trigger reset <<<--------------------
5782
                    `:lmatid=;151[0d]lmatname=;f50[0d]
                        lnam=;vsp_basic_fh15[0d]sgerin=;0.00[0d]sgerup=;0.00[0d]sval=;5.18[0d]text_11=;[0d]
                        text_12=;[0d]tind=;1.53[0d][1e][1d]
outside loop

condition was met init status=0
outside loop

屏幕上显示的内容:主机仿真脚本

b'[1c]req=ini[0d][0a][1e][1d]'
request ==  req=ini??? <<<<<<< cumple condicion y enviala respuesta
b''
b'[06][1c]req=33[0d][0a]job=5782[0d][0a][1e][1d]'
b''
b''
b''
b''
b''
b''
亚伦

我怀疑您正在尝试一次写入太多的硬件缓冲区。特别是在处理低功耗硬件时,通常无法正确地假设您可以将整个消息填充到缓冲区中。即使是完整的现代PC有时也具有非常小的缓冲区,用于诸如串行端口之类的传统硬件。从开发切换到实际硬件时,您可能会发现需要使用RTS和DTR线来确定何时发送或接收数据。不幸的是,这取决于设计硬件的人,因为它们通常也被忽略。

我会尝试将您的数据传输分成较小的块,以进行测试以查看整个消息是否通过。这是一次快速且肮脏的初次尝试,可能会有错误,但是它应该使您走上正确的道路:

def get_job_from_serial():
    response = b'' #buffer for response
    while True:
        try:
            response += serial_client.read() #read any available data or wait for timeout
            #this technically could only be reading 1 char at a time, but any 
            #remotely modern pc should easily keep up with 9600 baud
        except serial.SerialTimeoutException: #timeout probably means end of data
            #you could also presumably check the length of the buffer if it's always 
            #a fixed length to determine if the entire message has been sent yet.
            break
    return response

def writer(command):
    written = 0 #how many bytes have we actually written
    chunksize = 128 #the smaller you go, the less likely to overflow
                    # a buffer, but the slower you go.
    while written < len(command):
        #you presumably might have to wait for p.dtr() == True or similar
        #though it's just as likely to not have been implemented.
        written += p.write(command[written:written+chunksize]) 
    p.flush() #probably don't actually need this

PS:我不得不去看源代码p.read_all(出于某种原因,我无法在线找到它),但是它并没有达到我认为您期望的那样。确切的代码是:

def read_all(self):
    """\
    Read all bytes currently available in the buffer of the OS.
    """
    return self.read(self.in_waiting)

没有等待完整消息的概念,它只是获取当前可用所有内容的捷径。

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