使用 python 向 RESTful API 发出请求

Question

我有一个 RESTful API，我使用 EC2 实例上的 Elasticsearch 实现公开了它来索引内容语料库。我可以通过在我的终端 (MacOSX) 上运行以下命令来查询搜索：

curl -XGET 'http://ES_search_demo.com/document/record/_search?pretty=true' -d '{
  "query": {
    "bool": {
      "must": [
        {
          "text": {
            "record.document": "SOME_JOURNAL"
          }
        },
        {
          "text": {
            "record.articleTitle": "farmers"
          }
        }
      ],
      "must_not": [],
      "should": []
    }
  },
  "from": 0,
  "size": 50,
  "sort": [],
  "facets": {}
}'

如何使用python/requests 或python/urllib2 将上述请求转换为API 请求（不确定该选择哪一个- 一直在使用urllib2，但听说请求更好......）？我是作为标题传递还是以其他方式传递？

Answer 1

使用requests：

import requests
url = 'http://ES_search_demo.com/document/record/_search?pretty=true'
data = '''{
  "query": {
    "bool": {
      "must": [
        {
          "text": {
            "record.document": "SOME_JOURNAL"
          }
        },
        {
          "text": {
            "record.articleTitle": "farmers"
          }
        }
      ],
      "must_not": [],
      "should": []
    }
  },
  "from": 0,
  "size": 50,
  "sort": [],
  "facets": {}
}'''
response = requests.post(url, data=data)

根据您的 API 返回的响应类型，您可能需要查看response.text 或response.json()（或者可能首先检查response.status_code）。请参阅快速入门文档here，尤其是this section。

Answer 2

使用requests 和json 让事情变得简单。

1. 调用 API
2. 假设 API 返回一个 JSON，将 JSON 对象解析为 Python dict 使用json.loads 函数
3. 遍历字典以提取信息。

Requests 模块为您提供有用的函数来循环成功和失败。

if(Response.ok)：将帮助您确定您的 API 调用是否成功（响应码 - 200）

Response.raise_for_status() 将帮助您获取从 API 返回的 http 代码。

以下是进行此类 API 调用的示例代码。也可以在github 中找到。该代码假定 API 使用摘要式身份验证。您可以跳过此步骤或使用其他适当的身份验证模块来对调用 API 的客户端进行身份验证。

#Python 2.7.6
#RestfulClient.py

import requests
from requests.auth import HTTPDigestAuth
import json

# Replace with the correct URL
url = "http://api_url"

# It is a good practice not to hardcode the credentials. So ask the user to enter credentials at runtime
myResponse = requests.get(url,auth=HTTPDigestAuth(raw_input("username: "), raw_input("Password: ")), verify=True)
#print (myResponse.status_code)

# For successful API call, response code will be 200 (OK)
if(myResponse.ok):

    # Loading the response data into a dict variable
    # json.loads takes in only binary or string variables so using content to fetch binary content
    # Loads (Load String) takes a Json file and converts into python data structure (dict or list, depending on JSON)
    jData = json.loads(myResponse.content)

    print("The response contains {0} properties".format(len(jData)))
    print("\n")
    for key in jData:
        print key + " : " + jData[key]
else:
  # If response code is not ok (200), print the resulting http error code with description
    myResponse.raise_for_status()

Answer 3

因此，您想在 GET 请求的正文中传递数据，最好在 POST 调用中进行。您可以通过使用这两个请求来实现这一点。

原始请求

GET http://ES_search_demo.com/document/record/_search?pretty=true HTTP/1.1
Host: ES_search_demo.com
Content-Length: 183
User-Agent: python-requests/2.9.0
Connection: keep-alive
Accept: */*
Accept-Encoding: gzip, deflate

{
  "query": {
    "bool": {
      "must": [
        {
          "text": {
            "record.document": "SOME_JOURNAL"
          }
        },
        {
          "text": {
            "record.articleTitle": "farmers"
          }
        }
      ],
      "must_not": [],
      "should": []
    }
  },
  "from": 0,
  "size": 50,
  "sort": [],
  "facets": {}
}

请求调用示例

import requests

def consumeGETRequestSync():
data = '{
  "query": {
    "bool": {
      "must": [
        {
          "text": {
            "record.document": "SOME_JOURNAL"
          }
        },
        {
          "text": {
            "record.articleTitle": "farmers"
          }
        }
      ],
      "must_not": [],
      "should": []
    }
  },
  "from": 0,
  "size": 50,
  "sort": [],
  "facets": {}
}'
url = 'http://ES_search_demo.com/document/record/_search?pretty=true'
headers = {"Accept": "application/json"}
# call get service with headers and params
response = requests.get(url,data = data)
print "code:"+ str(response.status_code)
print "******************"
print "headers:"+ str(response.headers)
print "******************"
print "content:"+ str(response.text)

consumeGETRequestSync()

Answer 4

下面是python中执行rest api的程序-

import requests
url = 'https://url'
data = '{  "platform": {    "login": {      "userName": "name",      "password": "pwd"    }  } }'
response = requests.post(url, data=data,headers={"Content-Type": "application/json"})
print(response)
sid=response.json()['platform']['login']['sessionId']   //to extract the detail from response
print(response.text)
print(sid)