Read each line, replace string and save into a new file

[igormonteiro]

I want to replace a string based on each line of list.txt and create a new file with the name of the string collected.

*Content of list.txt*

Output:
new_server new_server2 new_server3

*Content of zabbix_agentd.conf*

Output:
Hostname=server

fin = open("zabbix_agentd.conf", "rt") fout = open(new_server_string_collected, "wt") for line in fin:#	fout.write(line.replace('Hostname=server', 'Hostname=new_server')) fin.close() fout.close() fh = open('list.txt') while True: line = fh.readline() print(line) if not line: break fh.close()

*Results expected*

1. python sample.py
2. file created: new_server
3. cat new_server

Output:
Hostname=new_server

[Axel_Erfurt]

you want replace one word only?

fin = open("zabbix_agentd.conf", "r") fout = open(new_server_string_collected, "w") str_in = fin.read() fin.close() str_out = str_in.replace('Hostname=server', 'Hostname=new_server') fout.write(str_out) fout.close()

[buran]

something within these lines

zabix_config = 'zabbix_agentd.conf' server_list = 'list.txt' with open(zabix_config) as zbx: zbx_config = zbx.read() # read full config file content with open(server_list) as f: for line in f: line = line.strip() # remove the new line ending with open(f'{line}') as out_f: new_zbx = zbx_config.replace('Hostname=server', f'Hostname={line}') out_f.write(new_zbx)