Question

我有一个代码，它将excel表中的用户名作为输入发送，该输入存储在变量xy中的instagram API中，后者给出了结果。我将URL加载到JSON以在json模式中获取它。

例如，当我的变量xy包含＆＃34; shawn_123＆＃34; API的结果是：

{
"meta":  {
"code": 200
},
"data":  [
 {
  "username": "shawn_123",
  "profile_picture": "https://scontent.cdninstagram.com/t51.2885-  19/s150x150/11417456_1610194859266611_592197892_a.jpg",
  "id": "641567093",
  "full_name": "shawn ritz"
},
 {
  "username": "shawn_12345",
  "profile_picture": "https://scontent.cdninstagram.com/t51.2885-19/s150x150/11324946_808347519273018_2073555780_a.jpg",
  "id": "2074312361",
  "full_name": "shawney"
}
]
}

我的代码是

for r in range(1,10):
var=r,sheet.cell(row=r,column=1).value
xy=var[1]
myopener=Myopener()
url=myopener.open('https://api.instagram.com/v1/users/search?q='+xy+'&count=1&access_token=641567093.1fb234f.a0ffbe574e844e1c818145097050cf33')
beta=json.load(url)
for item in beta['data']:
    print(item['id'])

因为它从同一用户名检索两个输出。

注意：我想要一个正则表达式，它从json中搜索确切的用户名并保存该记录的user_id。

Answer 1

您可以在打印ID之前检查username是否匹配：

for item in beta['data']:
    if item['username'] == xy: # here check the username from your input
        print(item['id'])

或者，使用next运算符：

user_id = next((item['id'] for item in beta['data'] if item['username'] == xy), None)

Answer 2

d = {"data":  [ {  "username": "shawn_123",  "profile_picture": "https://scontent.cdninstagram.com/t51.2885-  19/s150x150/11417456_1610194859266611_592197892_a.jpg",  "id": "641567093",  "full_name": "shawn ritz"}, {  "username": "shawn_12345",  "profile_picture": "https://scontent.cdninstagram.com/t51.2885-19/s150x150/11324946_808347519273018_2073555780_a.jpg",  "id": "2074312361",  "full_name": "shawney" }]}

#### To get list
user_name = 'something'
print [i['username'] for i in d['data'] if i['username'] == user_name]

#### To get the username

print [i['username'] for i in d['data'] if i['username'] == user_name][0]

Answer 3

试试这个

"shawn_123".*?"id":\s"(\d+)"

Regex demo

<强>解释
.：除了换行符sample之外的任何字符 *：零次或多次sample
?：一次或无sample
\s：“空格字符”：空格，制表符，换行符，回车符，垂直制表符sample
( … )：捕获小组sample
\：逃脱一个特殊字符sample
+：一个或多个sample

的Python：

import re
p = re.compile(ur'"shawn_123".*?"id":\s"(\d+)"', re.DOTALL)
test_str = u"{\n\"meta\":  {\n\"code\": 200\n},\n\"data\":  [\n {\n  \"username\": \"shawn_123\",\n  \"profile_picture\": \"https://scontent.cdninstagram.com/t51.2885-  19/s150x150/11417456_1610194859266611_592197892_a.jpg\",\n  \"id\": \"641567093\",\n  \"full_name\": \"shawn ritz\"\n},\n {\n  \"username\": \"shawn_12345\",\n  \"profile_picture\": \"https://scontent.cdninstagram.com/t51.2885-19/s150x150/11324946_808347519273018_2073555780_a.jpg\",\n  \"id\": \"2074312361\",\n  \"full_name\": \"shawney\"\n}"
m = re.findall(p, test_str)
print m
#[u'641567093']

在循环时查找存储在变量中的许多用户名的完全匹配

3 个答案: