sed使用shell脚本变量将选定的行删除到文件

Question

I have shell script variable var="7,8,9" These are the line number use to delete to file using sed.

Here I tried: sed -i "$var"'d' test_file.txt

But i got error `sed: -e expression #1, char 4: unknown command: ,'

Is there any other way to remove the line?

Answer 1

sed command doesn't accept comma delimited line numbers.

You can use this awk command that uses a bit if BASH string manipulation to form a regex with the given comma separated line numbers:

awk -v var="^(${var//,/|})$" 'NR !~ var' test_file.txt

This will set awk variable var as this regex:

^(7|8|9)$

And then condition NR !~ var ensures that we print only those lines that don't match above regex.

For inline editing, if you gnu-awk with version > 4.0 then use:

awk -i inplace -v var="^(${var//,/|})$" 'NR !~ var' test_file.txt

Or for older awk use:

awk -v var="^(${var//,/|})$" 'NR !~ var' test_file.txt > $$.tmp && mv $$.tmp test_file.txt

Answer 2

I like sed, you were close to it. You just need to split each line number into a separate command. How about this:

sed -e "$(echo 1,3,4 | tr ',' '\n' | while read N; do printf '%dd;' $N; done)"

Answer 3

这样做：

sed -i "`echo $var|sed 's/,/d;/g'`d;" file

Answer 4

Another option to consider would be ed , with printf '%s\\n' to put commands onto separate lines:

lines=( 9 8 7 )
printf '%s\n' "${lines[@]/%/d}" w | ed -s file

The array lines contains the line numbers to be deleted; it's important to put these in descending order ! The expansion ${lines[@]/%/d} adds a d (delete) command to each line number and w writes to the file at the end. You can change this to ,p instead, to check the output before overwriting your file.

As an aside, for this example, you could also just use 7,9 as a single entry in the array.