[
{'_id': ObjectId('5abb4f9ca7e2c54c757b3e48'),
'amount': 8400,
'buyerEmail': 'otzYzwMh24edWk8NxSJOqCSZREe0',
'from': 'weixin',
'orderDate': datetime.datetime(2018, 3, 28, 8, 17, 28, 940000),
'orderid': '2018032816173212079',
'real': 8400,
'rechargeDate': datetime.datetime(2018, 3, 28, 8, 17, 41, 877000),
'status': 1,
'tradeNo': '4200000099201803287230332578',
'uid': ObjectId('5abb36051a62067bf7e30178')
},
{'_id': ObjectId('5b6699f6df03ec3294d7c0a4'),
'amount': 100,
'buyerEmail': 'otzYzwC3YwRdu7QrWLXqS3VRJybI',
'from': 'weixin',
'orderDate': datetime.datetime(2018, 8, 5, 6, 32, 21, 442000),
'orderid': '2018080514322245193',
'real': 100,
'rechargeDate': datetime.datetime(2018, 8, 5, 6, 33, 0, 381000),
'status': 1,
'tradeNo': '4200000148201808052403940202',
'uid': ObjectId('5a5738411a62061972e128cb')
},
.....
]
如何把同一个uid的整合一起,并只显示uid real from rechargeDate的元素?类似mongodb聚合的$group理想结果:
[
{'uid':xxxxxxxxxxxxxxxx:
[
{
'real':xxx,
'from':xxx,
'recharge':xxx,
},
{
'real':xxx,
'from':xxx,
'recharge':xxx,
},
{
'real':xxx,
'from':xxx,
'recharge':xxx,
},
]
},
{'uid':xxxxxxxxxxxxxxxx:
[
{
'real':xxx,
'from':xxx,
'recharge':xxx,
},
{
'real':xxx,
'from':xxx,
'recharge':xxx,
},
{
'real':xxx,
'from':xxx,
'recharge':xxx,
},
]
},
...
]
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