Grep字符串并删除第一个字符

Question

我想在grep一些字符串时删除文件中的第一个字符。但是需要在编辑之前将行更改为相同的位置。

示例文件：

#%PAM-1.0
auth            sufficient      pam_rootok.so
# Uncomment the following line to implicitly trust users in the "wheel" group.
#auth           sufficient      pam_wheel.so trust use_uid
auth            include         system-auth
account         sufficient      pam_succeed_if.so uid = 0 use_uid quiet

编辑后的预期视图：

#%PAM-1.0
auth            sufficient      pam_rootok.so
# Uncomment the following line to implicitly trust users in the "wheel" group.
auth           sufficient      pam_wheel.so trust use_uid
auth            include         system-auth
account         sufficient      pam_succeed_if.so uid = 0 use_uid quiet

在第四行的情况下，只需要删除“＃”，但我想在搜索字符串“足够的pam_wheel.so trust use_uid”时这样做，而不是在指向我要编辑的确切行时。

Answer 1

这是sed的作业：

$ sed -r 's/^#(.*sufficient\s+pam_wheel\.so trust use_uid.*)/\1/' file
#%PAM-1.0
auth            sufficient      pam_rootok.so
# Uncomment the following line to implicitly trust users in the "wheel" group.
auth           sufficient      pam_wheel.so trust use_uid
auth            include         system-auth
account         sufficient      pam_succeed_if.so uid = 0 use_uid quiet

Regexplanation：

s/                            # Substitute  
^#                            # A line starting with a #
(                             # Start capture group
.*                            # Followed by anything
sufficient                    # Followed by the word sufficient
\s+                           # Followed by whitespace
pam_wheel\.so trust use_uid   # Followed by the literal string (escaped .)
.*                            # Followed by anything
)                             # Stop capture group
/                             # Replace with 
\1                            # The first capture group 

因此，我们有效地匹配以包含字符串#的{​​{1}}开头并删除sufficient\s+pam_wheel.so trust use_uid

的行

注意：#标记用于扩展正则表达式，-r版本可能为-E，因此请检查sed。

如果您想将更改存储回man，请使用file选项：

-i

如果列对齐很重要，那么最多可以捕获$ sed -ri 's/^#(.*sufficient\s+pam_wheel\.so trust use_uid.*)/\1/' file ，然后获取2个捕获组并替换为sufficient。

\1 \2

修改

将字符串$ sed -r 's/^#(.*)(sufficient\s+pam_wheel\.so trust use_uid.*)/\1 \2/' file #%PAM-1.0 auth sufficient pam_rootok.so # Uncomment the following line to implicitly trust users in the "wheel" group. auth sufficient pam_wheel.so trust use_uid auth include system-auth account sufficient pam_succeed_if.so uid = 0 use_uid quiet 替换为AllowUsers support admin：

#AllowUsers support admin

Answer 2

perl -pi -e '$_=~s/^.//g if(/sufficient      pam_wheel.so trust use_uid/)' your_file

注意：这将进行就地替换。在运行命令后，您的文件将自动修改。

Answer 3

使用awk你可以这样做：

awk -F '#' '{ if ($0 ~ /^#.*sufficient +pam_wheel.so trust use_uid/) {$1=""; print;} else print}'  infile

OR（如果一开始只能有一个＃）：

awk -F '#' '{ if ($0 ~ /^#.*sufficient +pam_wheel.so trust use_uid/) print $2; else print}'  infile

Answer 4

这是使用sed的一种方式：

sed '/sufficient \+pam_wheel.so \+trust \+use_uid/s/^#//' file

结果：

#%PAM-1.0
auth            sufficient      pam_rootok.so
# Uncomment the following line to implicitly trust users in the "wheel" group.
auth           sufficient      pam_wheel.so trust use_uid
auth            include         system-auth
account         sufficient      pam_succeed_if.so uid = 0 use_uid quiet

Answer 5

sed命令只能执行RE匹配而不能进行字符串比较，因此要使用sed，您需要解析所需的字符串以逃避容易出错的所有RE元字符（例如，当前发布的解决方案都忽略了逃避“。”在pam_wheel.so）。

当您尝试匹配字符串时，最好只进行字符串比较，例如：

awk 'index($0,"sufficient pam_wheel.so trust use_uid"){sub/^#/,"")}1' file > tmp && mv tmp file