Use sed instead of grep+cut to capture the passphrase

Current method would crop any passphrase containing an '=' sign.

For eg. wpa_passwphrase=Foo=Bar would return 'Foo' instead of 'Foo=Bar'.
This commit is contained in:
Nicolas Ramz 2024-10-22 17:27:31 +02:00
parent ba0bff9b9c
commit 0bb8a5400e

View File

@ -32,7 +32,7 @@ def ieee80211n():
return subprocess.run("cat /etc/hostapd/hostapd.conf | grep ieee80211n= | cut -d'=' -f2", shell=True, capture_output=True, text=True).stdout.strip()
def wpa_passphrase():
return subprocess.run("cat /etc/hostapd/hostapd.conf | grep wpa_passphrase= | cut -d'=' -f2", shell=True, capture_output=True, text=True).stdout.strip()
return subprocess.run("sed -En 's/wpa_passphrase=(.*)/\1/p' /etc/hostapd/hostapd.conf", shell=True, capture_output=True, text=True).stdout.strip()
def interface():
return subprocess.run("cat /etc/hostapd/hostapd.conf | grep interface= | cut -d'=' -f2 | head -1", shell=True, capture_output=True, text=True).stdout.strip()