使用sed，如何删除与任何一组模式匹配的所有初始行？

Question

I'm looking to use sed to delete all of the initial lines in a file that match a set of patterns. But, once none of the patterns are matched, the rest of the file should be passed through unmodified.

For example, if the patterns are:

^\s*#
^\s*$

Then for the file:

#!/bin/bash
# Some words

# The line above has spaces or tabs in it
    #
    # The above line has spaces before and after the #

main()
{ ... }

# Another comment

the sed script would generate:

main()
{ ... }

# Another comment

That is, in this case the set of patterns removes any initial Bash comments and, blank lines, and lines that just have whitespace in them.

While it might be interesting to hear how this might be done with other tools (such as awk ), I am mainly interested in solutions involving sed .

I'm looking for a specific solution to the example above as well as any guidance as to how to generalize this for a set of arbitrary patterns.

My environment is:

• CentOS 7.3 (3.10.0-514.6.1.el7.x86_64)
• bash-4.2.46-21.el7_3.x86_64
• sed-4.2.2-5.el7.x86_64

Answer 1

Delete lines that do match, but once you get past all the patterns, just print the rest using the loop :done;n;bdone

Here's the example:

test.c:

#!/bin/bash
# Some words

# The line above has spaces or tabs in it
    #
    # The above line has spaces before and after the #

DELETEME

main()
DELETEME - should stay
{ ... }



$ sed '/^\s*#/d; /DELETEME/d; /^\s*$/d; :done;n;bdone' test.c 
main()
DELETEME - should stay
{ ... }