【发布时间】:2019-08-26 14:59:41
【问题描述】:
我正在尝试使用脚本连接到 KDB 并使用 qpython (https://github.com/exxeleron/qPython) 将行插入到 KDB 中的表中。我的表有以下列类型:"symbol","symbol","int","timestamp","string","string","symbol","symbol","string","string","string"
我尝试使用 '.u.upd' 但这没有返回任何内容并且不会更新表格:
time = [numpy.timedelta64((numpy.datetime64(datetime.now()) - today), 'ms') for x in range(1)]
row = [qlist(['test'], qtype=QSYMBOL_LIST), qlist(['test'], qtype=QSYMBOL_LIST), qlist([1], qtype=QINT_LIST), qlist(time, qtype=QTIME_LIST),
qlist([['t','e','s','t']], qtype=QSTRING_LIST), qlist([['t','e','s','t']], qtype=QSTRING_LIST), qlist(['test'], qtype=QSYMBOL_LIST), qlist(['test'], qtype=QSYMBOL_LIST),
qlist([['t','e','s','t']], qtype=QSTRING_LIST), qlist([['t','e','s','t']], qtype=QSTRING_LIST), qlist([['t','e','s','t']], qtype=QSTRING_LIST)]
result = self.q.sendSync('.u.upd', numpy.string_('tableName'), row)
当我尝试使用 insert 时,我收到错误 'type:
result = self.q('tableName insert (`test;`test;1i;2019.08.09D12:00:00.123123123;("t"; "e"; "s"; "t");("t"; "e"; "s"; "t");`test;`test;("t"; "e"; "s"; "t");("t"; "e"; "s"; "t");("t"; "e"; "s"; "t"))')
感谢任何帮助。
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