使用sed删除空白行之前的所有内容

Question

Lets say I have a file which is something like this:

"Testing is important"

Nothing is impossible

The output should be:

Nothing is impossible

This means the sed removed everything before new line. Also, I need to make sure it works on bash on windows.

Please help.

Answer 1

You can try this

sed '1,/^\s*$/d' file  

\\s is whitespace, it's same with

sed '1,/^[[:blank:]]*$/d' file  

Answer 2

Sed supports addressing lines both as numbers and as matching regex. In your case, you can delete all lines starting from 1, and ending with an empty line:

sed -e '1,/^$/d'

On Windows your files may contain contain carriage returns, in which case you can use:

sed -e '1,/^\r*$/d'

(assuming GNU sed)

Answer 3

To allow for more than one blank line and multiple lines after the final blank line, you could use something like this:

awk 'BEGIN{RS=ORS=""}{a=$0}END{print a}' file

This unsets the Record Separator RS , so that each block/paragraph is treated as a separate record. It assigns each record to the variable a , then prints the last value of a once the file has been processed. The Output Record Separator ORS is also unset, so that no newline is appended to the final block.