我有不同級別數（最多 4 級）的層次結構路徑：FACILITIES \ PARKROADS \ GRASS/TURF BLVD MAINTENANCEROADS \ SIDEWALKS \ REPLACEMENTFACILITIES \ PARKING - MAIN STRECREATION \ BANDSHELL \ PROPERTY \ BUILDING-GENERALFACILITIES \ FIREHALL \ PLUMBINGFACILITIES我想解析_\_分隔符處的級別并將值插入 Maximo 數據庫列中：WOEQ1WOEQ2WOEQ3WOEQ4（這些列的長度現在只有 10。稍后我將創建長度更長的適當的自定義列。）我嘗試過的：我已經弄清楚如何解析級別并將值插入到列中 - IF（這是一個很大的 IF）總是有 4 個級別：#Auto-script on WORKORDERh = mbo.getString("HIERARCHYPATH")mbo.setValue("WOEQ1", (h.split(' \\ '))[0][:10])mbo.setValue("WOEQ2", (h.split(' \\ '))[1][:10])mbo.setValue("WOEQ3", (h.split(' \\ '))[2][:10])mbo.setValue("WOEQ4", (h.split(' \\ '))[3][:10])但當然，我不會總是有 4 個級別。我可以有 1-4 之間任意數量的級別。如何解析反斜杠分隔的層次結構路徑（具有不同的級別數）？
2 回答
精慕HU
TA貢獻1845條經驗 獲得超8個贊
您可以讓“正?！辈鸱謥韯澐中?/p>
>>> s = """\
... FACILITIES \ PARK
... ROADS \ GRASS/TURF BLVD MAINTENANCE
... ROADS \ SIDEWALKS \ REPLACEMENT
... FACILITIES \ PARKING - MAIN ST
... RECREATION \ BANDSHELL \ PROPERTY \ BUILDING-GENERAL
... FACILITIES \ FIREHALL \ PLUMBING
... FACILITIES
... """
>>> for row in s.splitlines():
... print(row.split(" \\ "))
...
['FACILITIES', 'PARK']
['ROADS', 'GRASS/TURF BLVD MAINTENANCE']
['ROADS', 'SIDEWALKS', 'REPLACEMENT']
['FACILITIES', 'PARKING - MAIN ST']
['RECREATION', 'BANDSHELL', 'PROPERTY', 'BUILDING-GENERAL']
['FACILITIES', 'FIREHALL', 'PLUMBING']
['FACILITIES']
然后您可以迭代返回的列表，設置每個值
max_col_length = 10 # to be updated by author
for row in s.splitlines():
for index, atom in enumerate(row.split(" \\ "), 1): # count from 1
mbo = "create a new row" # not shown by author
mbo.setValue("WOEQ{}".format(index), atom[:max_col_length])
"INSERT row if processing rows individually"
"INSERT all rows if able to upload them all at the same time"
如果您需要始終在列表中提供至少 4 個（或 N 個）成員，您可以使用itertools.repeat填寫剩余值
>>> import itertools
>>> mylist = ['RECREATION', 'BANDSHELL', 'PROPERTY']
>>> mylist.extend(list(itertools.repeat(None, 4 - len(mylist))))
>>> print(mylist)
['RECREATION', 'BANDSHELL', 'PROPERTY', None]
守著一只汪
TA貢獻1872條經驗 獲得超4個贊
這是我最終使用的腳本（它是 @ti7 答案的變體）。
在新的層次結構路徑少于四個級別的情況下，腳本會將現有值清空。
parts = s.split(r' \ ')
for i, part in enumerate(parts, 1):
mbo.setValue(col_prefix + str(i), part)
for i in range(len(parts)+1, 5):
mbo.setValueNull(col_prefix + str(i))
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