I'd like to insert multiple lines above a target line in a file by using sed.
The file.txt
below contains one line "target line
". My initial version is using the single quote:
sed '/target line/ i\
inserted line1;\
inserted line2;\
inserted line3;' file.txt
The result is:
inserted line1;
inserted line2;
inserted line3;
target line
This version works as expected that the newline at the end of each line is escaped by \
to a literal newline instead of a command terminator. Refer to here.
Then I'd like to use shell variable in the replacement string, so I tried to use double quotes to enable the variable expansion:
sed "/target line/ i\
inserted line1;\
inserted line2;\
inserted line3;" file.txt
But this time the newline and the first four spaces disappeared:
inserted line1; inserted line2; inserted line3;
target line
How do I correctly insert a newline in double quotes here?
With single-quotes:
The backslash followed by the newline are transmitted as-is to sed. Then sed actually uses the backslash to escape the raw newline character into the string rather than terminating the command. See:
$ printf %s 'hello\
world' | hexdump -C
Which clearly shows the backslash 5c
followed by the newline 0a
contained in the string.
00000000 68 65 6c 6c 6f 5c 0a 77 6f 72 6c 64 |hello\.world|
0000000c
With double-quotes:
The backslash has special meaning in double-quotes. It causes the following newline character to be interpreted as a string continuation character. The consequence is that neither the backslash or the newline are contained in the string and so not seen by sed.
$ printf %s "hello\
world" | hexdump -C
The string is continued without backslash and without newline:
00000000 68 65 6c 6c 6f 77 6f 72 6c 64 |helloworld|
0000000a
EDIT: