Shell 分割行并获取第一部分的脚本

Question

我这里的任务是从一个大约有 1000 行的日志文件中提取文件名，在日志中，每一行都以文件名开头，后跟其他详细信息，我现在想提取每个文件名(绝对路径，从每一行的 './') 开始并将其放入文件中。示例日志文件包含以下数据。

./plugins-src/rabbitmq-management/src/rabbit_mgmt_wm_overview.erl:1:%%   The contents of this file are subject to the Mozilla Public License
./plugins-src/rabbitmq-management/src/rabbit_mgmt_old_db.erl:1:%%   The contents of this file are subject to the Mozilla Public License
./plugins-src/rabbitmq-management/src/rabbit_mgmt_wm_exchange.erl:1:%%   The contents of this file are subject to the Mozilla Public License
./plugins-src/rabbitmq-management/src/rabbit_mgmt_wm_channel.erl:1:%%   The contents of this file are subject to the Mozilla Public License
./plugins-src/rabbitmq-management/src/rabbit_mgmt_wm_vhosts.erl:1:%%   The contents of this file are subject to the Mozilla Public License
./plugins-src/rabbitmq-management/src/rabbit_mgmt_wm_permission.erl:1:%%   The contents of this file are subject to the Mozilla Public License
./plugins-src/rabbitmq-management/src/rabbit_mgmt_util.erl:1:%%   The contents of this file are subject to the Mozilla Public License
./plugins-src/rabbitmq-management/src/rabbit_mgmt_wm_queue_purge.erl:1:%%   The contents of this file are subject to the Mozilla Public License
./plugins-src/rabbitmq-management/src/rabbit_mgmt_format.erl:1:%%   The contents of this file are subject to the Mozilla Public License
./plugins-src/rabbitmq-management/src/rabbit_mgmt_wm_exchanges.erl:1:%%   The contents of this file are subject to the Mozilla Public License
./plugins-src/rabbitmq-management/src/rabbit_mgmt_wm_bindings.erl:1:%%   The contents of this file are subject to the Mozilla Public License
./plugins-src/rabbitmq-management/src/rabbit_mgmt_wm_definitions.erl:1:%%   The contents of this file are subject to the Mozilla Public License
./plugins-src/rabbitmq-management/src/rabbit_mgmt_wm_queue_get.erl:1:%%   The contents of this file are subject to the Mozilla Public License
./plugins-src/rabbitmq-federation-management/src/rabbit_federation_mgmt.erl:1:%%  The contents of this file are subject to the Mozilla Public License
./plugins-src/rabbitmq-mqtt/src/rabbit_mqtt_processor.erl:1:%% The contents of this file are subject to the Mozilla Public License
./plugins-src/rabbitmq-mqtt/src/rabbit_mqtt_util.erl:1:%% The contents of this file are subject to the Mozilla Public License
./plugins-src/rabbitmq-mqtt/src/rabbit_mqtt_collector.erl:1:%% The contents of this file are subject to the Mozilla Public License
./plugins-src/rabbitmq-mqtt/src/rabbit_mqtt_frame.erl:1:%% The contents of this file are subject to the Mozilla Public License
./plugins-src/rabbitmq-mqtt/src/rabbit_mqtt_sup.erl:1:%% The contents of this file are subject to the Mozilla Public License
./plugins-src/rabbitmq-mqtt/src/rabbit_mqtt.erl:1:%% The contents of this file are subject to the Mozilla Public License

有一个冒号 (:) 可以用作分隔符，在每一行中准确地结束文件名，但我没有 shell 脚本的专业知识来分割它并提取文件名.

Answer 1

 awk -F':' '{print }' filename.log

 # OR 

 cut -d':' -f1 filename.log

Answer 2

另一个使用 bash 的途径是：

while read -r line; do echo "${line%%:*}"; done <filename

它使用带子字符串删除的参数扩展，这是一组内置的字符处理例程。基本上：

var="123:456:789"
echo "${var#*:}"   # 456:789  remove from left to 1st occurrence of ':'
echo "${var%:*}"   # 123:456  remove from right to 1st occurrence of ':'
echo "${var##*:}"  # 789  remove from left to last occurrence of ':'
echo "${var%%:*}"  # 123  remove from right to last occurrence of ':'

(注意:扩展中通配符的位置)

它们甚至可以嵌套。

Shell 分割行并获取第一部分的脚本

Shell script to slice lines and get the first part

shell

delimiter

slice