使用 when 条件匹配输出寄存器中的字符串(Ansible) [英] Using when conditional to match string in output register (Ansible)

问题描述

我无法在我的输出变量中搜索我用于 when 语句的指定字符串.下面的代码应该检查输出变量中的字符串distribute-list"，但是在运行剧本时它给出了错误.

Im unable to search my output variable for a specified string that im using for a when statement. The code below is supposed to check for the string "distribute-list" in the output variable but when run the playbook it gives the error.

fatal: [192.168.3.252]: FAILED! => {"failed": true, "msg": "The conditional check 'output | search(\"distribute-list\")' failed. The error was: Unexpected templating type error occurred on ({% if output | search(\"distribute-list\") %} True {% else %} False {% endif %}): expected string or buffer\n\nThe error appears to have been in '/home/khibiny/test4.yml': line 26, column 5, but may\nbe elsewhere in the file depending on the exact syntax problem.\n\nThe offending line appears to be:\n\n\n  - debug:\n    ^ here\n"}

这是导致问题的代码:

 - ios_command:
     commands: show run | sec ospf
     provider: "{{cli}}"
   register: output
 - debug:
     msg: "{{output.stdout_lines}}"
   when: output | search("distribute-list")                           

希望得到一些帮助.提前致谢.

Would appreciate some help. Thanks in advance.

推荐答案

search 需要字符串作为输入，但 output 是一个具有不同属性的字典.

search expects string as input, but output is a dict with different properties.

你应该擅长

when: output.stdout | join('') | search('distribute-list')

这里需要中间join，因为对于ios-family 模块stdout 是一个字符串列表，而stdout_lines 是一个列表列表(而对于通常的 command 模块，stdout 是一个字符串，stdout_lines 是一个字符串列表).

you need intermediate join here, because for ios-family modules stdout is a list of strings, and stdout_lines is a list of lists (whereas for usual command module stdout is a string and stdout_lines is a list of strings).

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