data = {
'ids': [12, 3, 4, 5, 6 , ...]
}
urllib2.urlopen("http://abc.com/api/posts/create",urllib.urlencode(data))
I want to send a POST request, but one of the fields should be a list of numbers. How can I do that ? (JSON?)
我想发送一个POST请求,但是其中一个字段应该是数字列表。我怎么做呢?(JSON ?)
5 个解决方案
#1
125
If your server is expecting the POST request to be json, then you would need to add a header, and also serialize the data for your request...
如果您的服务器期望POST请求是json,那么您需要添加一个header,并为您的请求序列化数据……
Python 2.x
Python 2. x
import json
import urllib2
data = {
'ids': [12, 3, 4, 5, 6]
}
req = urllib2.Request('http://example.com/api/posts/create')
req.add_header('Content-Type', 'application/json')
response = urllib2.urlopen(req, json.dumps(data))
Python 3.x
Python 3. x
https://*.com/a/26876308/496445
https://*.com/a/26876308/496445
If you don't specify the header, it will be the default application/x-www-form-urlencoded
type.
如果没有指定标题,它将是默认的应用程序/x-www-form-urlencode类型。
#2
88
I recommend using the incredible requests
module.
我建议使用不可思议请求模块。
http://docs.python-requests.org/en/v0.10.7/user/quickstart/#custom-headers
http://docs.python-requests.org/en/v0.10.7/user/quickstart/定义header
url = 'https://api.github.com/some/endpoint'
payload = {'some': 'data'}
headers = {'content-type': 'application/json'}
response = requests.post(url, data=json.dumps(payload), headers=headers)
#3
32
for python 3.4.2 I found the following will work:
对于python 3.4.2,我发现以下方法可以工作:
import urllib.request
import json
body = {'ids': [12, 14, 50]}
myurl = "http://www.testmycode.com"
req = urllib.request.Request(myurl)
req.add_header('Content-Type', 'application/json; charset=utf-8')
jsondata = json.dumps(body)
jsondataasbytes = jsondata.encode('utf-8') # needs to be bytes
req.add_header('Content-Length', len(jsondataasbytes))
print (jsondataasbytes)
response = urllib.request.urlopen(req, jsondataasbytes)
#4
4
You have to add header,or you will get http 400 error. The code works well on python2.6,centos5.4
您必须添加标题,否则您将得到http 400错误。代码在python2.6,centos5.4上运行得很好
code:
代码:
import urllib2,json
url = 'http://www.google.com/someservice'
postdata = {'key':'value'}
req = urllib2.Request(url)
req.add_header('Content-Type','application/json')
data = json.dumps(postdata)
response = urllib2.urlopen(req,data)
#5
3
This works perfect for Python 3.5
, if the URL contains Query String / Parameter value,
这对于Python 3.5来说是完美的,如果URL包含查询字符串/参数值,
Request URL = https://bah2.com/ws/rest/v1/concept/
Parameter value = 21f6bb43-98a1-419d-8f0c-8133669e40ca
请求URL = https://bah2.com/ws/rest/v1/concept/参数值= 21f6bb43-98a1-419d-8f0c-8133669e40ca
import requests
url = 'https://bahbah2.com/ws/rest/v1/concept/21f6bb43-98a1-419d-8f0c-8133669e40ca'
data = {"name": "Value"}
r = requests.post(url, auth=('username', 'password'), verify=False, json=data)
print(r.status_code)
#1
125
If your server is expecting the POST request to be json, then you would need to add a header, and also serialize the data for your request...
如果您的服务器期望POST请求是json,那么您需要添加一个header,并为您的请求序列化数据……
Python 2.x
Python 2. x
import json
import urllib2
data = {
'ids': [12, 3, 4, 5, 6]
}
req = urllib2.Request('http://example.com/api/posts/create')
req.add_header('Content-Type', 'application/json')
response = urllib2.urlopen(req, json.dumps(data))
Python 3.x
Python 3. x
https://*.com/a/26876308/496445
https://*.com/a/26876308/496445
If you don't specify the header, it will be the default application/x-www-form-urlencoded
type.
如果没有指定标题,它将是默认的应用程序/x-www-form-urlencode类型。
#2
88
I recommend using the incredible requests
module.
我建议使用不可思议请求模块。
http://docs.python-requests.org/en/v0.10.7/user/quickstart/#custom-headers
http://docs.python-requests.org/en/v0.10.7/user/quickstart/定义header
url = 'https://api.github.com/some/endpoint'
payload = {'some': 'data'}
headers = {'content-type': 'application/json'}
response = requests.post(url, data=json.dumps(payload), headers=headers)
#3
32
for python 3.4.2 I found the following will work:
对于python 3.4.2,我发现以下方法可以工作:
import urllib.request
import json
body = {'ids': [12, 14, 50]}
myurl = "http://www.testmycode.com"
req = urllib.request.Request(myurl)
req.add_header('Content-Type', 'application/json; charset=utf-8')
jsondata = json.dumps(body)
jsondataasbytes = jsondata.encode('utf-8') # needs to be bytes
req.add_header('Content-Length', len(jsondataasbytes))
print (jsondataasbytes)
response = urllib.request.urlopen(req, jsondataasbytes)
#4
4
You have to add header,or you will get http 400 error. The code works well on python2.6,centos5.4
您必须添加标题,否则您将得到http 400错误。代码在python2.6,centos5.4上运行得很好
code:
代码:
import urllib2,json
url = 'http://www.google.com/someservice'
postdata = {'key':'value'}
req = urllib2.Request(url)
req.add_header('Content-Type','application/json')
data = json.dumps(postdata)
response = urllib2.urlopen(req,data)
#5
3
This works perfect for Python 3.5
, if the URL contains Query String / Parameter value,
这对于Python 3.5来说是完美的,如果URL包含查询字符串/参数值,
Request URL = https://bah2.com/ws/rest/v1/concept/
Parameter value = 21f6bb43-98a1-419d-8f0c-8133669e40ca
请求URL = https://bah2.com/ws/rest/v1/concept/参数值= 21f6bb43-98a1-419d-8f0c-8133669e40ca
import requests
url = 'https://bahbah2.com/ws/rest/v1/concept/21f6bb43-98a1-419d-8f0c-8133669e40ca'
data = {"name": "Value"}
r = requests.post(url, auth=('username', 'password'), verify=False, json=data)
print(r.status_code)