bash - 从多行文件中获取子字符串

Question

我有一个包含以下内容的文件：

# Disabling this option will break firewall functionality that relies on
# stateful packet inspection (e.g. DNAT, PACKET_FILTER) and makes the firewall
# less secure
#
# This option should be set to "1" in all other circumstances
LF_SPI = "1"

# Allow incoming TCP ports
TCP_IN = "20,21,22,25,53,80,110,143,443,465,587,"

# Allow outgoing TCP ports
TCP_OUT = "20,21,22,25,53,80,110,113,143,443,465,587,"

# Allow incoming UDP ports
UDP_IN = "20,21,53,111"

# Allow outgoing UDP ports
# To allow outgoing traceroute add 33434:33523 to this list 
UDP_OUT = "20,21,53,113,123,111"

我需要在该行之间TCP_IN = "和,"末尾获取文本。

我试过ALLOWEDPORTS=$(sed -e 's/.*TCP_IN = \"\(.*\)TCP_OUT.*/\1/' csf.conf)了，但我得到了整个文件。

我也试过ALLOWEDPORTS=$(sed -e 's/TCP_IN = \"\(.*\)TCP_OUT/\1/' csf.conf)，但我也得到了整个文件。

有什么帮助吗？我需要20,21,22,25,53,80,110,143,443,465,587,分配给变量。

标签： bash

Answer 1

你可以使用这个sed：

out=$(sed -n 's/.*TCP_IN = "\([^"]*\)".*/\1/p' file)

echo "$out"

20,21,22,25,53,80,110,143,443,465,587,