Question

我尝试使用whisper-merge合并2 wsp个文件。它们具有相同的保留策略，其中一个只有较旧的数据。

当我运行whisper-merge oldfile.wsp newfile.wsp时，我收到此错误

Traceback (most recent call last):
  File "/usr/local/src/whisper-0.9.12/bin/whisper-merge.py", line 32, in <module>
    whisper.merge(path_from, path_to)
  File "/usr/local/lib/python2.7/dist-packages/whisper.py", line 821, in merge
    (timeInfo, values) = fetch(path_from, fromTime, untilTime)
TypeError: 'NoneType' object is not iterable

有什么想法吗？

这里是2个文件的元数据输出：

＆＃34; from_file＆＃34; http://sprunge.us/dBHC
＆＃34; to_file＆＃34; http://sprunge.us/eIVG

Answer 1

对于包含多个档案的文件，whisper.py中的第812行被破坏。 https://github.com/graphite-project/whisper/blob/0.9.12/whisper.py#L812

fromTime = int(time.time()) - headerFrom['maxRetention']

要在第813行之后立即修复，请根据存档保留时间分配fromTime。 https://github.com/graphite-project/whisper/blob/0.9.12/whisper.py#L813

for archive in archives: # this line already exists
  fromTime = int(time.time()) - archive['retention'] # add this line

Answer 2

来自whisper.py

的代码段

def fetch(path,fromTime,untilTime=None):
    """fetch(path,fromTime,untilTime=None)

    path is a string
    fromTime is an epoch time
    untilTime is also an epoch time, but defaults to now.

    Returns a tuple of (timeInfo, valueList)
    where timeInfo is itself a tuple of (fromTime, untilTime, step)

    Returns None if no data can be returned
    """
    fh = open(path,'rb')
    return file_fetch(fh, fromTime, untilTime)

建议whisper.fetch()返回None，而后者（以及追溯中的最后一行）表明您的path_from文件存在问题。
展望得更深，whisper.file_fetch()似乎有两个地方可以返回None（至少明确地）：

def file_fetch(fh, fromTime, untilTime):
    header = __readHeader(fh)
    now = int( time.time() )
    if untilTime is None:
        untilTime = now
    fromTime = int(fromTime)
    untilTime = int(untilTime)

    # Here we try and be flexible and return as much data as we can.
    # If the range of data is from too far in the past or fully in the future, we
    # return nothing
    if (fromTime > untilTime):
        raise InvalidTimeInterval("Invalid time interval: from time '%s' is after until time '%s'" % (fromTime, untilTime))

    oldestTime = now - header['maxRetention']
    # Range is in the future
    if fromTime > now:
        return None               # <== Here
    # Range is beyond retention
    if untilTime < oldestTime:
        return None               # <== ...and here
    ...

得到错误＆＃34; TypeError：＆＃39; NoneType＆＃39;对象不可迭代＆＃34;当试图使用耳语合并

2 个答案: