使用 AWK 查找并替换

Question

我有一个名为 sso_med.txt

的文件

        insert into brs.user_components (user_name, component_id, created_date, created_by, last_updated_date, last_updated_by) values ('502683504',16,current_date, '502264160', current_date, '502264160');
        insert into brs.user_components (user_name, component_id, created_date, created_by, last_updated_date, last_updated_by) values ('502689031',15,current_date, '502264160', current_date, '502264160');
        insert into brs.user_components (user_name, component_id, created_date, created_by, last_updated_date, last_updated_by) values ('502689031',16,current_date, '502264160', current_date, '502264160');
        insert into brs.user_components (user_name, component_id, created_date, created_by, last_updated_date, last_updated_by) values ('212462117', 15, current_date, '502264160', current_date, '502264160');
        insert into brs.user_components (user_name, component_id, created_date, created_by, last_updated_date, last_updated_by) values ('212462117', 16, current_date, '502264160', current_date, '502264160');

我想在'502689031',16,current_date,

之间介绍一个space

所以我尝试了下面的命令

awk '{gsub(/,[0-9a-zA-Z]/, ", "); print}' sso_med.txt

但我没有得到我期望的结果

insert into brs.user_components (user_name, component_id, created_date, created_by, last_updated_date, last_updated_by) values ('502683504', 6, urrent_date, '502264160', current_date, '502264160'); 
insert into brs.user_components (user_name, component_id, created_date, created_by, last_updated_date, last_updated_by) values ('502689031', 5, urrent_date, '502264160', current_date, '502264160'); 
insert into brs.user_components (user_name, component_id, created_date, created_by, last_updated_date, last_updated_by) values ('502689031', 6, urrent_date, '502264160', current_date, '502264160');

Answer 1

awk解决方案：

awk '{ gsub(/, */, ", ", [=10=]) }1' sso_med.txt

输出：

insert into brs.user_components (user_name, component_id, created_date, created_by, last_updated_date, last_updated_by) values ('502683504', 16, current_date, '502264160', current_date, '502264160');
insert into brs.user_components (user_name, component_id, created_date, created_by, last_updated_date, last_updated_by) values ('502689031', 15, current_date, '502264160', current_date, '502264160');
insert into brs.user_components (user_name, component_id, created_date, created_by, last_updated_date, last_updated_by) values ('502689031', 16, current_date, '502264160', current_date, '502264160');

gsub(/, */, ", ", [=12=]) - 将每个分隔符（即 ,）替换为/重新排列 =29=]

Answer 2

只给后面的参考:

awk '{gsub(/,[0-9a-zA-Z]/, " &"); print}' sso_med.txt

使用 AWK 查找并替换

Find with AWK and Replace

unix

awk

gawk

gsub