I am trying to delete all empty lines after line number 3, till the end of the file:
cat ${File}
1
2
3
4
5
5
6
7
8
9
sed -e '3,${s~^$~~g}' ${File}
1
2
3
4
5
5
6
7
8
9
Observation: No change in the output.
Desired Output
1
2
3
4
5
5
6
7
8
9
try:
sed '4,$ {/^$/d}' infile
start from the 4th line until end of the file, delete the empty lines.
The problem with your command is that you replace empty line again with empty string in s~^$~~g (which is same as s~^$~~) and you are not deleting it.
Note: also since you use different delimiter other than default slash /, to use this style ~^$~d you need to escape the first ~ to tell sed that is not part of your regex:
sed -e '4,${ \~^$~d }' infile
see man sed under "Addresses" about it:
\cregexpc Match lines matching the regular expression regexp. The c may be any character.
In case you wanted to delete empty lines as well as the lines containing only whitespaces (Tabs/Spaces), you can do:
sed '4,${ /^[[:blank:]]*$/d }' infile
TextWithPipe="Alpha|Beta|Gamma"; ReplacementTextWithPipe="Delta"; echo "${TextWithPipe}" | sed "s|Beta|${ReplacementTextWithPipe}|g" gives desired output Alpha|Delta|Gamma. But according to you, it should not work as I am using a differnt delimiter?
s command, which is different from what I said. /regex/ is for regex match and anything other than / for that you must escape the first character
Another awk solution:
awk 'NR<=3||NF' file
This will print all lines where either the line number is smaller or equal 3, or which have at least one "field", defined by default as at least one non-blank character. This way, you can also exclude lines that contain no "text" but are only "visually" empty, i.e. that still contain space or tabs.
sed '1,3b;/./!d'
Branch in range 1,3 or delete if empty.
You were almost there. Just needed to add the following steps.
sed -ne '4,${s~^$~~;t};p' < "${File}"
4,$s/^$//s/// succeeds, skip to next record t};pan awk solution:
awk '((NR > 3 && $0!~/^$/) || NR <= 3) { print }' file
awk 'NR<3 || /./' can be simplified to this.