Using output piped from sed

Question

I have a sed command that is capturing a single line with sometext. The line in the file it is capturing ends with a linefeed. I am trying to utilize this variable in a pipeline, however, when I attempt to echo, or use it with other commands requiring an input, the result is a blank. Ex: sed '1,1!d' somefile.txt | echo "$1", I know the variable itself is not empty as I can replace echo "$1" with cat $1 and see the correct printout.

edit - I have tried piping to a tr -d and removing the newline. I have confirmed the newline character is gone, yet echos still show blank. Cats do not.

edit 2 - I piped the variable into an if statement ... | if [[ -z $1 ]]; then cat $1; fi it hits the if, is determined to be empty, so runs the cat, which prints a non-empty line to console. If the variable is empty why is cat still printing out information?

What is causing this inconsistency and how can I solve my problem? The ultimate goal is to run the output of one sed, through another to replace specific lines in a target file.

sed '1,1!d' somefile.txt | sed '2,1s/.*/'$1'/' targetfile.txt

Contents of somefile.txt:

these
are
words

Contents of targetfile.txt:

The next line should say these
This line should say these
The previous line should say these

Output of echo after sed:

<empty>

Output of cat after sed:

these

Output of 2nd sed, using input from 1st:

The next line should say these

the previous line should say these

Answer 1

You are confused about arguments and input data. Look at this:

$ echo "$1"

$ echo "foo" | if [[ -z $1 ]]; then cat $1; fi
foo

The first argument to my shell, $1 is empty so if [[ -z $1 ]] succeeds. The reason that cat $1 produces output is that you have a fundamental shell programming error in that statement - you aren't quoting your variable, $1. The correct syntax isn't cat $1, it's cat "$1". Look at the difference:

$ echo "foo" | if [[ -z $1 ]]; then cat "$1"; fi
cat: '': No such file or directory

We can simplify the code to make what's happening clearer:

$ echo "foo" | cat $1
foo

$ echo "foo" | cat "$1"
cat: '': No such file or directory

The reason that echo "foo" | cat $1 produces output is that the unquoted $1 is expanded by the shell to nothing before cat is called so that statement is equivalent to just echo "foo" | cat and so cat just copies the input coming in from the pipe to it's output.

On the other hand echo "foo" | cat "$1" generates an error because the shell expands "$1" to the null string before cat is called and so it's then asking cat to open a file named <null> and that of course does not exist, hence the error.

Always quote your shell variables unless you have a specific reason not to and fully understand all of the implications. Read a shell man page and/or google that if you're not sure what those implications are.

wrt another part of your code you have:

sed '1,1!d' somefile.txt | echo "$1"

but, unlike cat, echo neither reads it's input from a pipe nor from a file name passed as an argument. The input to echo is just the list of string arguments you provide it so while echo "foo" | cat will cause cat to read the input stream containing foo and output it, echo "foo" | echo will produce no output because echo isn't designed to read input from a pipe and so it'll just print a null string since you gave it no arguments.

It's not clear what you're really trying to accomplish but I think you might want to replace the 2nd line of targetfile.txt with the first line of somefile.txt. If so that's just:

awk '
    NR==FNR { if (NR==1) new=$0; next }
    FNR==2  { $0 = new }
    { print }
' somefile.txt targetfile.txt

Do not try to use sed to do it or you'll find yourself in escaping/quoting hell because, unlike awk, sed does not understand literal strings, see Is it possible to escape regex metacharacters reliably with sed.