小编Xan*_*xir的帖子

Ansible:将评估变量传递给包含

我有一个Ansible Playbook,可以调用另一个传递参数的playbook.这很好,但是当我尝试传入变量时,Ansible会爆炸.这是我的剧本:

---
- name: This is the toplevel play for orchestration
  hosts: nogroup
  vars:
    customer: myCustomer
  sudo: yes
- include: amazon.yml application=FooServer instance_type=t2.micro instance_count=1 customer={{ customer }}
- include: amazon.yml application=BarServer instance_type=t2.micro instance_count=1 customer={{ customer }}
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我试着跑:

ansible-playbook orchestration.yml -e "customer=otherCustomer"
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我收到此错误:

Traceback (most recent call last):
  File "/usr/local/Cellar/ansible/1.7.2/libexec/bin/ansible-playbook", line 5, in <module>
    pkg_resources.run_script('ansible==1.7.2', 'ansible-playbook')
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/Extras/lib/python/pkg_resources.py", line 492, in run_script
    self.require(requires)[0].run_script(script_name, ns)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/Extras/lib/python/pkg_resources.py", line 1350, in run_script
    execfile(script_filename, namespace, namespace)
  File "/usr/local/Cellar/ansible/1.7.2/lib/python2.7/site-packages/ansible-1.7.2-py2.7.egg/EGG-INFO/scripts/ansible-playbook", line 309, in <module>
    sys.exit(main(sys.argv[1:])) …
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ansible

8
推荐指数
2
解决办法
2万
查看次数

只从Alexa Skills Kit获取单个单词参数

我正在写一个Alexa技能,我只能在我的代码中获得单字参数.

这是intent模式:

    {
  "intents": [
    {
      "intent": "HeroQuizIntent",
      "slots": [
        {
          "name": "SearchTerm",
          "type": "SEARCH_TERMS"
        }
      ]
    },
    {
      "intent": "HeroAnswerIntent",
      "slots": [
        {
          "name": "SearchTerm",
          "type": "SEARCH_TERMS"
        }
      ]
    },
    {
      "intent": "AMAZON.HelpIntent"
    }
  ]
}
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我的样本话语是:

HeroQuizIntent quiz me
HeroAnswerIntent is it {SearchTerm}
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对于HeroAnswerIntent,我正在检查SearchTerm插槽,我只会在那里找到单个单词.

因此,"彼得帕克"给出"帕克","史蒂夫罗杰斯"给出"罗杰斯","托尼斯塔克"给出"斯塔克".

如何在插槽中接受多个单词?

alexa alexa-skill alexa-skills-kit

7
推荐指数
1
解决办法
3616
查看次数

只有我可以从自定义斜杠命令中看到我的输出

我在Slack中创建了一个自定义斜杠命令.后端代码,而不是它的重要性,是AWS中的Python中的Lambda函数.

我的问题是,当我输入斜杠命令时,我是唯一可以看到该消息的人.否则,它完美地运作.有没有办法让别人看到我的自定义斜杠命令的输出?

slack-api slack

5
推荐指数
1
解决办法
2172
查看次数

在CloudFormation yaml中创建BucketPolicy时出错

我正在尝试使用以下yaml在CloudFormation中创建S3 Bucket策略:

  cloudTrailBucketPolicy:
    Type:  "AWS::S3::BucketPolicy"
    Properties:
      Bucket:
        Ref: cloudtrailBucket
      PolicyDocument:
        -
          Action:
            - "s3:GetBucketAcl"
          Effect:  Allow
          Resource:
            Fn::Join:
              - ""
              -
                - "arn:aws:s3:::"
                -
                  Ref: cloudtrailBucket
                - "/*"
          Principal:  "*"
        -
          Action:
            - "s3:PutObject"
          Effect:  Allow
          Resource:
            Fn::Join:
              - ""
              -
                - "arn:aws:s3:::"
                -
                  Ref: cloudtrailBucket
                - "/*"
          Principal:
            Service:  cloudtrail.amazonaws.com
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当我尝试这样做时,我收到一条消息"属性PolicyDocument的值必须是一个对象"

有人有主意吗?

yaml amazon-s3 aws-cloudformation

4
推荐指数
2
解决办法
3588
查看次数