ansible:测试registered_var.stderr不为空

Question

如何测试stderr是非空的::

- name: Check script
  shell: . {{ venv_name }}/bin/activate && myscritp.py
  args:
    chdir: "{{ home }}"
  sudo_user: "{{ user }}"
  register: test_myscript

- debug: msg='myscritp is Ok'
  when: not test_myscript.stderr

所以,如果没有错误,我可以阅读::

  TASK: [deploy | debug msg='critp is Ok] *******
  ok: [vagrant] => {
      "msg": "myscritp is Ok"
  }

并且cas stderr中的FATAL错误不为空.

Answer 1

你可以检查空字符串(当stderr为空时)

- name: Check script
  shell: . {{ venv_name }}/bin/activate && myscritp.py
  args:
    chdir: "{{ home }}"
  sudo_user: "{{ user }}"
  register: test_myscript

- debug: msg='myscritp is Ok'
  when: test_myscript.stderr == ""

如果要检查是否失败:

- debug: msg='myscritp has error: {{test_myscript.stderr}}'
  when: test_myscript.stderr != ""

另请参阅此stackoverflow问题

Answer 2

请参阅ansible-lint 默认规则。以下情况导致E602不与空字符串进行比较

when: test_myscript.stderr != ""

正确的语法以及“ Ansible Galaxy Warning-Free”选项是

when: test_myscript.stderr | length > 0

加法1. 当前规则说：

相反使用when: var而不是when: var != ""（或''' when: not var而不是``when：var ==“”

Answer 3

when: myvar | default('', true) | trim != ''

我| trim != ''用来检查变量是否具有空值。我也总是添加| default(..., true)检查以捕获何时myvar未定义。