Question

我有一个json数据，如下所示，我在每个字段中都有一个uri要匹配。如果找到匹配项，则不要进行[fields]的下一次迭代。概念是用户在搜索后单击了URL。如果用户单击第三个URL，则收集第1，第2，第3个数据值。不要追求第4个值。

如果用户单击了第5个网址，则收集第1至第5个字段uri并退出该json对象。采取一个新的json对象并执行相同的过程。

[document]-> [fields1] -> [uri]
[document]-> [fields2] -> [uri]
[document]-> [fields3] -> [uri]
..... 
.. till 20-30 times. 

    I have written below code, but the above logic is not working. Kindly help on this. 



 uri='http://abcd.com/123.html'
    print(uri)
    for index_srch_log,row_srch_log in df_search_log_mongo.iterrows():
        RESPONSE = row_srch_log['RESPONSE']
        json_response = json.loads(RESPONSE)
        if 'documents' in json_response:
            field_data=json_response['documents']

            for row_resp_list in field_data:
                print('uri:',row_resp_list['fields']['uri'])
                match_found=False
                for i in row_resp_list['fields']['uri']:
                    print('i',i)

                    if uri == i:
                        print('yes matched')
                        match_found=True
                        break
                        print('found')
                    else:
                        print('not matched')
                        match_found=False
                    if match_found==True:
                        break

输出：

uri: ['http://abcddsc779072.html']
i value: http://abcddsc779072.html
not matched
uri: ['http://abcddsc932618.html']
i value: http://abcddsc932618.html
yes matched

-应该在此处停止并从DF获取下一个响应对象。 -但它再次继续下一个[fields]数据。

  uri: ['http://abcddsc988555.html']
    i value:  http://abcddsc988555.html
    not matched
    uri: ['http://abcddsc1094909.html']
    i value: http://abcddsc1094909.html
    not matched

Answer 1

您没有打破外部循环。考虑以下更改：

    if 'documents' in json_response:
        field_data=json_response['documents']

        for row_resp_list in field_data:
            print('uri:',row_resp_list['fields']['uri'])
            match_found=False
            for i in row_resp_list['fields']['uri']:
                print('i',i)

                if uri == i:
                    print('yes matched')
                    match_found=True
                    break
                else:
                    print('not matched')

            if match_found:
                break

在python

1 个答案: