[英]Ansible: Replace word in file
I'm making a Ansible playbook to setup CSF. 我正在制作Ansible手册来设置CSF。 I've got everything done except for the last part.
除了最后一部分,我已完成所有工作。
I'd like to disable port 22 in the /etc/csf/csf.conf
file. 我想在
/etc/csf/csf.conf
文件中禁用端口22。 So TCP_OUT = "20,21,22,25,53,80,110,113,443,587,993,995"
needs 22
removed. 因此,
TCP_OUT = "20,21,22,25,53,80,110,113,443,587,993,995"
需要删除22
。 I don't want to replace the entire line as some lines are different, some got port 2087
open, or 2222
for example. 我不想替换整行,因为某些行不同,例如某些端口打开了
2087
端口或2222
。 Is there any way I can only filter on 22
? 有什么办法只能过滤
22
吗?
Thank you in advance!! 先感谢您!!
You have several options: 您有几种选择:
This solution uses replace module, to look for a line beginning with TCP_OUT =
and replace ,22,
with ,
in the line. 此解决方案使用更换模块,以寻找与开头的行
TCP_OUT =
和替换,22,
与,
在该行。
tasks:
- name: Strip port 22
replace:
dest: /etc/csf/csf.conf
regexp: '^TCP_OUT\s*=\s*(.*),22,(.*)$'
replace: 'TCP_OUT = \1,\2'
Code working proof 代码工作证明
>>> TCP_OUT = '20,21,22,25,53,80,110,113,443,587,993,995,2087,2222,22'
>>> print(','.join([port for port in TCP_OUT.split(',') if port != '22']))
'20,21,25,53,80,110,113,443,587,993,995,2087,2222'
You could use template . 您可以使用template 。 Make a copy of your
/etc/csf/csf.conf
file and for the TCP_OUT line replace it with an ansible variable: 复制
/etc/csf/csf.conf
文件,并在TCP_OUT行中将其替换为ansible变量:
TCP_OUT = {{ port_list }}
Then set the list ahead of time in a variable with the ports you desire in the file. 然后,在变量中预先设置列表,并在文件中添加所需端口。
vars:
port_list = "20,21,25,53,80,110,113,443,587,993,995"
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