requests.get(url) 从 kubernetes api 返回错误代码 404，而响应可以通过 curl/GET 获取

Question

我在 kubernetes api 上使用 requests.get() 遇到了一个问题

url = 'http://10.69.117.136:8080/api/v1/namespaces/"default"/pods/tas-core/'
json = requests.get(url)
print json.content

错误代码 404 将返回为： {"kind": "Status","apiVersion": "v1","metadata": {},"status": "Failure","message": "pods \"tas-core\" not found","原因”：“未找到”，“详细信息”：{“名称”：“tas-core”，“种类”：“pods”}，“代码”：404}

但如果我使用 GET/curl，则可以成功返回响应：

curl http://10.69.117.136:8080/api/v1/namespaces/"default"/pods/tas-core/

{"kind": "Pod","apiVersion": "v1","metadata": {"name": "tas-core","namespace":"default","selfLink": "/api /v1/namespaces/default/pods/tas-core","uid":"a264ce8e-a956-11e5-8293-0050569761f2","re​​sourceVersion":"158546","creationTimestamp":"2015-12-23T09:22 :06Z","labels": {"app": "tas-core"},"annotations": {"ctrl": "dynamic","oam": "dynamic"}},"spec": {"volumes "：[ ...

更短的网址也可以正常工作

url = 'http://10.69.117.136:8080/api/v1/namespaces/'
json = requests.get(url)
print json.content

{"kind":"NamespaceList","apiVersion":"v1","metadata":{"selfLink":"/api/v1/namespaces/","re​​sourceVersion":"220452"},"items ":[{"元数据":{"name":"default","selfLink":"/api/v1/namespaces/default","uid":"74f89440-a94a-11e5-9afd-0050569761f2","re​​sourceVersion ":"6","creationTimestamp":"2015-12-23T07:54:55Z"},"spec":{"finalizers":["kubernetes"]},"status":{"phase":"Active "}}]}

我哪里做错了？

Answer 1

从requests 发出请求并从命令行将其发送到不同的网址。

来自 Python 代码的 requests 请求确实尝试使用包含引号的 url。

curl 来自命令行确实会去掉引号（在其他情况下它会转义引号）。

我无法测试您的真实网址以获取真实请求，但我想，以下可能可行：

url = 'http://10.69.117.136:8080/api/v1/namespaces/default/pods/tas-core/'
json = requests.get(url)
print json.content