我要按如下所示将键:值对中的特定元素作为目标,但我很困惑。我想求和并返回所有带有键的元素的值作为“项”。
[{ 'actcost': 5.56,
'bnfcode': '0101010G0AAABAB',
'bnfname': 'Co-MagaldroxSusp 195mg/220mg/5ml S/F',
'items': 2,
'nic': 5.98,
'practice': 'N81013',
'quantity': 1000},
{ 'actcost': 2.78,
'bnfcode': '0101010G0AAABAB',
'bnfname': 'Co-MagaldroxSusp 195mg/220mg/5ml S/F',
'items': 1,
'nic': 2.99,
'practice': 'N81029',
'quantity': 500},
{ 'actcost': 5.56,
'bnfcode': '0101010G0AAABAB',
'bnfname': 'Co-MagaldroxSusp 195mg/220mg/5ml S/F',
'items': 2,
'nic': 5.98,
'practice': 'N81088',
'quantity': 1000}]
答案 0 :(得分:1)
所以你先
DECLARE @NumberRecords INT
DECLARE @RowCount INT
DECLARE @Name NVARCHAR(MAX)
DECLARE @Command NVARCHAR(MAX)
DECLARE @Result int
DECLARE @Names TABLE (
[RowId] INT NOT NULL IDENTITY(1, 1),
[Name] NVARCHAR(MAX),
[Type] NVARCHAR(MAX)
)
INSERT INTO @Names
SELECT
QUOTENAME(SCHEMA_NAME([Objects].schema_id)) + '.' + QUOTENAME(OBJECT_NAME([Objects].object_id)) [Name],
type_desc [Type]
FROM sys.objects [Objects]
WHERE type_desc IN ('SQL_STORED_PROCEDURE',
'SQL_TRIGGER',
'SQL_SCALAR_FUNCTION',
'SQL_TABLE_VALUED_FUNCTION',
'SQL_INLINE_TABLE_VALUED_FUNCTION',
'VIEW')
ORDER BY [Name]
SET @RowCount = 1
SET @NumberRecords = (SELECT COUNT(*) FROM @Names)
WHILE (@RowCount <= @NumberRecords)
BEGIN
SELECT @Name = [Name]
FROM @Names
WHERE [RowId] = @RowCount
SET @Command = N'EXEC sp_refreshsqlmodule ''' + @Name + ''''
BEGIN TRY
EXEC @Result = sp_executesql @Command
IF @Result <> 0
BEGIN
RAISERROR('Failed', 16, 1)
END
ELSE
BEGIN
IF (NOT EXISTS (SELECT *
FROM sys.dm_sql_referenced_entities(@Name, 'OBJECT')
WHERE [is_incomplete] = 1))
BEGIN
DELETE
FROM @Names
WHERE [RowId] = @RowCount
END
END
END TRY
BEGIN CATCH
-- Nothing
END CATCH
SET @RowCount = @RowCount + 1
END
SELECT [Name],
[Type]
FROM @Names
该字符串,然后进行简单循环
obj = json.loads(txt)
可以缩写为
result = 0
for dct in obj:
result += dct.get('items', 0)
如果您想对其他字段求和,则
result = sum(dct.get('items', 0) for dct in obj)
可以缩写为
result = 0
for dct in obj:
if 'items' in dct:
result += dct['quantity']
答案 1 :(得分:0)
您可以使用ipython或闲置代码段:
vagrant@vagrant:~$ ipython
Python 3.6.7 (default, Oct 22 2018, 11:32:17)
Type 'copyright', 'credits' or 'license' for more information
IPython 7.4.0 -- An enhanced Interactive Python. Type '?' for help.
In [1]: d = [{'bnfcode': '0101010G0AAABAB', 'items': 2, 'practice': 'N81013', 'bnfname': 'Co-MagaldroxSusp 195mg/220mg/5m
...: l S/F', 'nic': 5.98, 'actcost': 5.56, 'quantity': 1000}, {'bnfcode': '0101010G0AAABAB', 'items': 1, 'practice': '
...: N81029', 'bnfname': 'Co-MagaldroxSusp 195mg/220mg/5ml S/F', 'nic': 2.99, 'actcost': 2.78, 'quantity': 500}, {'bnf
...: code': '0101010G0AAABAB', 'items': 2, 'practice': 'N81088', 'bnfname': 'Co-MagaldroxSusp 195mg/220mg/5ml S/F', 'n
...: ic': 5.98, 'actcost': 5.56, 'quantity': 1000}]
In [2]: sum([i.get('items', 0) for i in d])
Out[2]: 5