Python中的十六进制转换错误
我遇到了一个问题。
我有一个页面,可以向传感器网络发送命令。
当我点击这段代码时:
<a href='javascript:void(send_command_to_network("{{net.id}}", "restartnwk"));'>Restart Network <i class="icon-repeat"></i> </a>
我会调用一个JavaScript函数,具体是:
function send_command_to_network(net, command) {
$.ajax({url: "/networks/" + net + "/send?command=" + command,
type: "GET",
async: true,
dataType: "json",
success: function(json_response) {
var err = json_response['error'];
if (err) {
show_alert('error', err);
return;
}
var success = json_response['success'];
if (success) {
show_alert('success', success);
return;
}
show_alert('alert', "This should not happen!");
}
});
}
这个函数会生成一个网址,用来调用一个在Tornado这个用Python写的网页服务器里的处理程序。这个处理程序是:
class NetworkSendHandler(BaseHandler):
# Requires authentication
@tornado.web.authenticated
def get(self, nid):
# Get the command
command = self.get_argument('command').upper();
# The dictionary to return
ret = {}
#Check if the command is available
if command not in ['RESTARTNWK']:
raise tornado.web.HTTPError(404, "Unknown command: " + str(command))
#Command ZDP-RestartNwk.Request
if command == 'RESTARTNWK':
op_group = "A3"
op_code = "E0"
packet_meta = "*%s;%s;%s;#"
pkt_len = hextransform(16, 2)
packet = packet_meta % (op_group, op_code, pkt_len)
packet = packet.upper()
op_group_hex=0xA3
op_code_hex=0xE0
cmdjson = packet2json(op_group_hex,op_code_hex, packet)
self.finish()
在Tornado的调试控制台里,我收到了这个错误:
Traceback (most recent call last):
File "/usr/lib/python2.6/site-packages/tornado/web.py", line 988, in _execute
getattr(self, self.request.method.lower())(*args, **kwargs)
File "/usr/lib/python2.6/site-packages/tornado/web.py", line 1739, in wrapper
return method(self, *args, **kwargs)
File "./wsn.py", line 859, in get
cmdjson = packet2json(op_group_hex,op_code_hex, packet)
File "./wsn.py", line 188, in packet2json
fcs = fcs ^ int('0x' + payload[(i - 1):(i + 1)], 16)
ValueError: invalid literal for int() with base 16: '0x*A'
我觉得错误出现在调用packet2json之后。我认为这个错误是在函数内部的转换过程中,因为错误信息说'0x*A'不是一个有效的整数。我觉得这个'*'就是问题所在……我该怎么解决这个问题呢?
补充说明
抱歉,我忘记了函数的内容:
# Transform an integer into a string with HEX symbols in the format that is
# understandable by Quantaservd
def hextransform(data, length):
data = hex(data).lstrip('0x').rstrip('L')
assert(len(data) <= length)
# zero-padding
data = ('0' * (length - len(data))) + data
print data
# Swap 'bytes' in the network ID
data = list(data)
for i in range(0, length, 2):
tmp = data[i]
data[i] = data[i + 1]
data[i + 1] = tmp
# Reverse the whole string (TODO: CHECK)
data.reverse()
data = "".join(data)
return data
def packet2json(op_group, op_code, payload):
#op_group =
#op_code =
#payload="" # stringa ascii 2 char per byte, senza checksum
lun = len(payload) / 2
fcs = op_group ^ op_code ^ lun #fcs = XOR of op_groip, op_code, lenght and payload
for i in range(len(payload)):
if ((i % 2) == 1):
fcs = fcs ^ int('0x' + payload[(i - 1):(i + 1)], 16)
print int('0x' + payload[(i - 1):(i + 1)], 16)
s=cmd_payload(op_group,op_code, lun, payload, fcs)
#p = r '{"data": "' + s + r '"}'
p=r'{"data": "'+s+r'"}'
return p
1 个回答
0
问题出在 packet2json()
这个函数上:
for i in range(len(payload)):
if ((i % 2) == 1):
fcs = fcs ^ int('0x' + payload[(i - 1):(i + 1)], 16)
在你的程序中,你是这样调用这个函数的:
packet2json(0xa3, 0xe0, '*A3;E0;10;#')
现在上面的循环会从第一个字符开始(索引为0),因为它的索引不是奇数,所以会被忽略。接着会检查第二个字符(索引为1),它的索引是奇数(1%2 == 1)。在 if
语句的内容中,使用了 int('0x' + payload[(i-1):(i+1)], 16)
,这实际上是 int('0x' + payload[0:2], 16)
,也就是 int('0x*A',16)
,这会导致程序崩溃...
你需要重新写这个 packet2json()
函数,使用正确的字符串索引,或者在调用时确保负载中没有前面的 *
。