分隔符可以是：“”； “_”
输入：
foo#foo_ehh113#The password of user 111 will expire within the next seven d
foo#foo_ehh204#The password of user 111 will expire within the next seven d
sadf#sadf this is a sample_text
foo#foo_ehh204#The password of user 222 will expire within the next seven da
foo - moreeee test
foo#foo_ehh113#The password of user 222 will expire within the next seven da
foo#foo_ehh113#The password of user 333 will expire within the next seven day
ldr#ldr_another sample text
foo#foo_ehh204#The password of user 333 will expire within the next seven day
输出：
foo#foo_ehh113#The password of user 111 will expire within the next seven d
foo#foo_ehh204#The password of user 111 will expire within the next seven d
foo#foo_ehh204#The password of user 222 will expire within the next seven da
foo#foo_ehh113#The password of user 222 will expire within the next seven da
foo#foo_ehh113#The password of user 333 will expire within the next seven day
foo#foo_ehh204#The password of user 333 will expire within the next seven day
因此给定行中的类似单词例如：
The
password
of
user
will
expire
within
the
我的问题：有没有什么方法可以只输出在一定程度上相似的行？前任。它们匹配 8 个单词
有没有任何 shell 脚本可以检测到这一点？
答案1
如果你想按顺序匹配文本，你可以尝试：
$ grep 'The.*password.*of.*user.*will.*expire.*within.*the' file
foo#foo_ehh113#The password of user 111 will expire within the next seven d
foo#foo_ehh204#The password of user 111 will expire within the next seven d
foo#foo_ehh204#The password of user 222 will expire within the next seven da
foo#foo_ehh113#The password of user 222 will expire within the next seven da
foo#foo_ehh113#The password of user 333 will expire within the next seven day
foo#foo_ehh204#The password of user 333 will expire within the next seven day
如果您有一个保存文本组的文件（名为 group.txt），您可以使用：
$ grep $(printf "%s.*" $(cat group.txt)) file
答案2
您可以在命令行中输入下一个，您可以根据需要提供任意多个关键字：
grep -E "one|two|three" file.txt
如果源是 dmesg 则执行以下操作：
dmesg | grep -E "one|two|three"