嗨,我是python编程的新手。在这里,我使用转储从我的mongodb获取输出。但是我在PostMan中收到格式错误的JSON字符串错误。
我的代码是:
cursor = db.Details.find()
for document in (yield cursor.to_list(length=100)):
self.write(bson.json_util.dumps(document))
我的输出是:
{"Name": "Will","_id": {"$oid": "55a8f075a382c32392b75bad"}}
{"Name": "Alex", "_id": {"$oid": "55acc2205d8882ef8a667d34"}}
{"data": null, "status": "success"}
我希望输出如何:
{"data": [
{"Name": "Will","_id": {"$oid": "55a8f075a382c32392b75bad"}},
{"Name": "Alex", "_id": {"$oid": "55acc2205d8882ef8a667d34"}}
], "status": "success"}
请帮帮我。
提前致谢
PostMan的My Output屏幕截图
答案 0 :(得分:2)
如何首先将所有内容保存在列表中,然后将该列表转储到json
?
data = []
cursor = db.Details.find()
for document in (yield cursor.to_list(length=100)):
data.append(document)
self.write(bson.json_util.dumps({"data": data}))
修改强>:
要获得所需输出的success
变量,可以尝试
data = []
status = ""
cursor = db.Details.find()
for document in (yield cursor.to_list(length=100)):
if 'status' in document: # check if key 'status' in document
status = document[status]
else:
data.append(document)
self.write(bson.json_util.dumps({"data": data, "status": status}))