Question

我正在运行Bottle和python来从MongoDB中提取数据。我的输出是JSON对象格式我曾经有以下

{u'product': u'mortgage', u'Total_Product': 146533}
{u'product': u'debt collection', u'Total_Product': 65639}

但想要摆脱'u并获得以下格式：

'product':'mortgage', Total_Product: 146533
'product':'debt collection' 'Total_Product:65639

运行此代码后，我留下一个没有结果且没有错误消息的空白屏幕。有什么建议吗？

# find a single  with all aggregates 
choices = dict(
    aggr1 = db.complaints.aggregate([{"$group":{"_id":"$disputed", "Total_Disputed":{"$sum":1}}},{"$sort":{"Total_Disputed":-1}},{"$project":{"_id":0, "disputed":"$_id", "Total_Disputed":1}}]),
    aggr2 = db.complaints.aggregate([{"$group":{"_id":"$product", "Total_Product":{"$sum":1}}},{"$sort":{"Total_Product":-1}},{"$project":{"_id":0, "product":"$_id", "Total_Product":1}}]),
    aggr3 = db.complaints.aggregate([{"$group":{"_id":"$response", "Total_Response":{"$sum":1}}},{"$sort":{"Total_Response":-1}},{"$project":{"_id":0, "response":"$_id", "Total_Response":1}}]),
    aggr4 = db.complaints.aggregate([{"$group":{"_id":"$submitted", "Total_Submitted":{"$sum":1}}},{"$sort":{"Total_Submitted":-1}},{"$project":{"_id":0, "submitted":"$_id", "Total_Submitted":1}}]),
    aggr5 = db.complaints.aggregate([{"$group":{"_id":"$timely", "Total_Timely":{"$sum":1}}},{"$sort":{"Total_Timely":-1}},{"$project":{"_id":0, "Timely":"$_id", "Total_Timely":1}}])
)

def convert_keys_to_string(choices):
    """Recursively converts dictionary keys to strings."""
    if not isinstance(choices, dict):
        return choices
    return dict((str(k), convert_keys_to_string(v)) 
        for k, v in dictionary.items())
aggr1 = 'aggr1'
aggr2 = 'aggr2'
aggr3 = 'aggr3'
aggr4 = 'aggr4'
aggr5 = 'aggr5' 

    return bottle.template('analytics.tpl',  things1 = choices[aggr1], things2 = choices[aggr2], things3 = choices[aggr3], things4 = choices[aggr4], things5 = choices[aggr5])

bottle.debug(True)
bottle.run(host='localhost', port=8082)

tpl如下：

<!DOCTYPE html>
<html>
<head>
<title> @2015 </title>
</head>
<body>
<p>

</p>
<p>
<ul>
%for thing1 in things1:
<li>{{thing1}}</li>
%end
</ul></p>
<p>
<ul>
%for thing2 in things2:
<li>{{thing2}}</li>
%end
</ul></p>



</body>
</html>

最亲切的问候

Answer 1

您在模板{{thing1}}中使用转换为字典的字符串。

只需编写模板，根据需要格式化dict-entries：

%for thing1 in things1:
<li>'product':'{thing1.product}', Total_Product: {{thing1.Total_Product}}</li>
%end

或

%for thing1 in things1:
<li>'product':'{thing1.product}' 'Total_Product:{{thing1.Total_Product}}</li>
%end

您的问题不明确，您更喜欢哪一个。

Bottle，Python和MongoDB中的JSON输出

1 个答案: