在Python中使用GSM调制解调器拨打电话时收到"NO CARRIER"错误

Question

我想用我的GSM调制解调器拨打电话.所以我写了下面的程序:

import time
import serial

recipient = "+98xxxxxxxxxx"

phone = serial.Serial("COM10",  115200, timeout=5)
try:
    time.sleep(0.5)
    phone.write(b'ATZ\r')
    time.sleep(1)
    phone.write(b'ATD"'+recipient.encode() +b'"\r')
    while(1):
        print(phone.readline())
    time.sleep(0.5)
finally:
    phone.close()

但是当我运行它时,我收到了这个输出:

>>> ================================ RESTART ================================
>>> 
b'ATZ\r\r\n'
b'OK\r\n'
b'ATDxxxxxxxxxx\r\r\n'
b'NO CARRIER\r\n'

这个"NO CARRIER"错误意味着什么？

请注意,我可以成功发送短信.

---

这是我用来发送短信的程序:

import time
import serial

recipient = "+98xxxxxxxxxx"
message = "Test"

phone = serial.Serial("COM10",  115200, timeout=5)


try:
    time.sleep(0.5)
    phone.write(b'ATZ\r')
    time.sleep(0.5)
    phone.write(b'AT+CMGF=1\r')
    time.sleep(0.5)
    phone.write(b'AT+CMGS="' + recipient.encode() + b'"\r')
    time.sleep(0.5)
    phone.write(message.encode() + b"\r")
    time.sleep(0.5)
    phone.write(bytes([26]))
    time.sleep(0.5)
finally:
    phone.close()

Answer 1

我找到了错误的起源:

语法ATD+98xxxxxxxxxx;之后是终止字符串.我忘记在数字后面加分号.

所以我更换

phone.write(b'ATD"'+recipient.encode() +b'"\r')

同

phone.write(b'ATD"'+recipient.encode() +b';"\r')

现在它工作正常.

根据本文档中的括号,我认为使用";" 是可选的.但似乎我错了.