YAML to JSON with Python, Kubernetes api call

Question

I want to convert this file into json format, does anyone know how to do it?

This is the yaml file :

apiVersion: v1
kind: Namespace
metadata:
  name: theiaide
---
apiVersion: networking.k8s.io/v1beta1
kind: Ingress
metadata:
  #name: rewrite
  name: theiaide
  namespace: theiaide
  annotations:
    #nginx.ingress.kubernetes.io/rewrite-target: /$2
    kubernetes.io/ingress.class: nginx
spec:
  rules:
  - host: ide.quantum.com
    http:
      paths:
      - path: /
        backend:
          serviceName: theiaide
          servicePort: 80
---
apiVersion: v1
kind: Service
metadata:
 name: theiaide
 namespace: theiaide
spec:
 ports:
 - port: 80
   targetPort: 3000
 selector:
   app: theiaide
---
apiVersion: apps/v1
kind: Deployment
metadata:
  labels:
    app: theiaide
  name: theiaide
  namespace: theiaide
spec:
  selector:
    matchLabels:
      app: theiaide
  replicas: 1
  template:
    metadata:
      labels:
        app: theiaide
    spec:
      containers:
      - image: theiaide/theia-python
        imagePullPolicy: IfNotPresent
        name: theiaide
        ports:
        - containerPort: 3000

code.py

import json,yaml
txt=""
with open(r"C:\Users\77922\PycharmProjects\ide-ingress.yaml",'r') as f:
    for a in f.readlines():
        txt=txt+a
print(yaml.dump(yaml.load_all(txt),default_flow_style=False))
print(json.dumps(yaml.load_all(txt),indent=2,sort_keys=True))

when I run python code.py ,and I got the error:

TypeError: can't pickle generator objects

I don’t know if it is the reason for this --- delimiter, because there are multiple --- delimiters in my yaml file

Then I tried the following function:

def main():

    # config.load_kube_config()

    f = open(r"C:\Users\77922\PycharmProjects\ide-ingress.yaml","r")
    generate_dict  = yaml.load_all(f,Loader=yaml.FullLoader)
    generate_json = json.dumps(generate_dict)
    print(generate_json)
    # dep = yaml.load_all(f)
    # k8s_apps_v1 = client.AppsV1Api()
    # resp = k8s_apps_v1.create_namespaced_deployment(
    #     body=dep, namespace="default")
    # print("Deployment created. status='%s'" % resp.metadata.name)
if __name__ == '__main__':
    main()

 raise TypeError(f'Object of type {o.__class__.__name__} '
TypeError: Object of type generator is not JSON serializable

I just want to use this yaml file to call kubernetes api to generate namespace

Why do you need to dump it to JSON? If you already have that YAML, using subprocess.call(['kubectl', 'apply', '-f', 'ide-ingress.yaml']) could load it, or as you show in the last example, the Kubernetes Python client has native data structures and APIs for Kubernetes objects. — David Maze
– David Maze, Commented Jul 23, 2020 at 10:24
@Layne Wu is the solution posted by Ewerton solved your issue? — Dawid Kruk
– Dawid Kruk, Commented Jul 23, 2020 at 15:37
@DavidMaze In fact, I just want to use the API to call, but currently create deployment fails. The creation of igress namespace and service is successful. — Layne Wu
– Layne Wu, Commented Jul 24, 2020 at 2:57
@DawidKruk Yes, Ewerton helped me solve this file conversion problem, but I seem to be discussing something else with David Maze. Do I need to open a new question to discuss it? — Layne Wu
– Layne Wu, Commented Jul 24, 2020 at 3:00

ewertoncodes · Accepted Answer · 2020-07-23 03:34:11Z

3

Your file contains more than one document. You should use the safe_load_all function rather than yaml.load and list rather than json.dumps

import json,yaml
txt=""
with open(r"C:\Users\77922\PycharmProjects\ide-ingress.yaml",'r') as f:
    for a in f.readlines():
        txt=txt+a
print(yaml.dump_all(yaml.safe_load_all(txt),default_flow_style=False))
print(list(yaml.safe_load_all(txt)))

edited Jul 23, 2020 at 3:34

answered Jul 23, 2020 at 2:30

ewertoncodes

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1 Comment

Layne Wu Over a year ago

I tried it and it seems that there were other errors. I updated it.

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