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Can I use sed to check the first line of some command's output (to stdout) and delete this very first line if it matches a certain pattern?
Say, the command's output is something like this:
"AB"
"CD"
"E"
"F"
I want it to become:
"CD"
"E"
"F"
But when the first line is "GH", I don't want to delete the line.
I tried this, but it doesn't work:
<some_command> |sed '1/<pattern>/d'
The shell told me:
sed: 0602-403 1/<pattern>/d is not a recognized function.
I only want to use sed to process the first line, leaving the other lines untouched.
What is the correct syntax here?
923
In sed, p prints the addressed line(s), while P prints only the first part (up to a newline character \n ) of the addressed line. If you have only one line in the buffer, p and P are the same thing, but logically p should be used.
Substitution command In some versions of sed, the expression must be preceded by -e to indicate that an expression follows. The s stands for substitute, while the g stands for global, which means that all matching occurrences in the line would be replaced.
You need to use regular expression to remove digits or number from input. For example, 's/[0-9]*//g' will remove all numbers from the input.
This might work for you:
sed -e '1!b' -e '/GH/!d' file
You want to reference the 1st line, then say delete:
$ sed '1 d' file
No need for any pattern if you know which line you want to delete.
With a pattern, use this syntax:
$ sed '0,/pattern/ d' file
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