How can I turn this:
aaa
bbb
ccc
into this:
aaa,
bbb,
ccc
using sed?
Note how all lines end a comma, except the last one.
In my real problem I also do some regex substitutions on the lines. Is there a solution that doesn't duplicate them?
You could use:
sed '$q;s/$/,/'
If you want to apply a different substitution on the last line, you can still use the $ address:
sed '$s/$/;/;$q;s/$/,/'
The above will replace the end of the line with ; if it's the last line, otherwise it will use ,.
$s/$/;/ = at the last line, replace the end of the line with ;
$q = at the last line, quit
s/$/,/ = replace the end of the line with ,
The last s command will run for each line, but not for the last line in which the q command at 2. tells it to quit.
See:
Restricting to a line number
Ranges by line number
The q or quit command
Related
I am on Mac, I want to find a pattern in lines, replace it with something, then append the resulting string to the end of the original line. Here is what I tried:
echo "test='123'" | sed -E '/([^a-z])/ s/$/ \1/'
sed: 1: "/([^a-z])/ s/$/ \1/": \1 not defined in the RE
What do I need to define \1? I thought I did it with ([^a-z]). No?
Edit: Perhaps this code will represent better what I want:
1) echo "test='123'" | sed 's/[a-zA-Z0-9]//g'
2) I want the new line = original line + line #1 above
In other words:
Before (what I get): test='123'
After (what I want): test='123' =''
You can edit this command this way:
echo "test='123'" | sed -E 'h;s/([a-zA-Z0-9])//g;G;s/(.*)\n(.*)/\2\1/'
For readability, the script, line by line, reads
h
s/([a-zA-Z0-9])//g
G
s/(.*)\n(.*)/\2\1/
h stores the current line in the hold space,
your s command does what it does
G appends the content of the hold space, i.e. the original line, to the pattern space, i.e. the current line as you have edited it, putting a newline \n in between.
another s command reorders the two pieces, also removing the \n that the G command inserted.
Comments
Your original attempt sed -E '/([^a-z])/ s/$/ \1/' could not work because \1 refers to what is captured by the leftmost (…) group in the search portion of the s command, it does not "remember" the group(s) you used to address the line.
Once you print the pattern space with p, a newline comes with it, and once it's been printed, there's no way you can remove it within the same sed program.
I have a text file that looks like:
#filelists.txt
a
# aaa
b
#bbb
c #ccc
I want to delete parts of lines starting with '#' and afterwards, if line starts with #, then to delete whole line.
So I use 'sed' command in my shell:
sed -e "s/#*//g" -e "/^$/d" filelists.txt
I wish its result is:
a
b
c
but actually result is:
filelists.txt
a
aaa
b
bbb
c ccc
What's wrong in my "sed" command?
I know '*' which means "any", so I think that '#*' means string after "#".
Isn't it?
You may use
sed 's/#.*//;/^$/d' file > outfile
The s/#.*// removes # and all the rest of the line and /^$/d drops empty lines.
See an online test:
s="#filelists.txt
a
# aaa
b
#bbb
c #ccc"
sed 's/#.*//;/^$/d' <<< "$s"
Output:
a
b
c
Another idea: match lines having #, then remove # and the rest of the line there and drop if the line is empty:
sed '/#/{s/#.*//;/^$/d}' file > outfile
See another online demo.
This way, you keep the original empty lines.
* does not mean "any" (at least not in regular expression context). * means "zero or more of the preceding pattern element". Which means you are deleting "zero or more #". Since you only have one #, you delete it, and the rest of the line is intact.
You need s/#.*//: "delete # followed by zero or more of any character".
EDIT: was suggesting grep -v, but didn't notice the third example (# in the middle of the line).
How can I replace the final character of a file with a ]?
For example, if the document content looked like this:
This is the first line with a full stop.
This is the second line with a full stop.
This is the third line with square bracket.
I would want it to look like this afterwards:
This is the first line with a full stop.
This is the second line with a full stop.
This is the third line with square bracket]
You may specify line numbers, ranges or even the last line where you want to perform search and replace operations with sed:
This will replace the final , char with ] only on the last line:
sed '$ s/,$/]/'
Here, the $ char tells sed to only replace on the last line.
The sed '1 s/,$/]/' command will do that only on Line 1, and sed '1,4 s/,$/]/' will do that on lines 1 through 4.
everybody.
I don't understand dollar sign ($) in sed script programming, it is stand for last line of a file or a counter of sed?
I want to reverse order of lines (emulates "tac") of /etc/passwd. like following:
$ cat /etc/passwd | wc -l ----> 52 // line numbers
$ sed '1!G;h;$!d' /etc/passwd | wc -l ----> 52 // working correctly
$ sed '1!G;h;$d' /etc/passwd | wc -l ----> 1326 // no ! followed by $
$ sed '1!G;h;$p' /etc/passwd | wc -l ----> 1430 // instead !d by p
Last two example don't work right, who can tell me what mean does dollar sign stand for?
All the commands "work right." They just do something you don't expect. Let's consider the first version:
sed '1!G;h;$!d
Start with the first two commands:
1!G; h
After these two commands have been executed, the pattern space and the hold space both contain all the lines reads so far but in reverse order.
At this point, if we do nothing, sed would take its default action which is to print the pattern space. So:
After the first line is read, it would print the first line.
After the second line is read, it would print the second line followed by the first line.
After the third line is read, it would print the third line, followed by the second line, followed by the first line.
And so on.
If we are emulating tac, we don't want that. We want it to print only after it has read in the last line. So, that is where the following command comes in:
$!d
$ means the last line. $! means not-the-last-line. $!d means delete if we are not on the last line. Thus, this tells sed to delete the pattern space unless we are on the last line, in which case it will be printed, displaying all lines in reverse order.
With that in mind, consider your second example:
sed '1!G;h;$d'
This prints all the partial tacs except the last one.
Your third example:
sed '1!G;h;$p'
This prints all the partial tacs up through the last one but the last one is printed twice: $p is an explicit print of the pattern space for the last line in addition to the implicit print that would happen anyway.
I want to use sed to delete part of code (paragraph) beginning with a pattern and ending with a semicolon (;).
Now I came across an example to delete a paragraph separated by new lines
sed -e '/./{H;$!d;}' -e 'x;/Pattern/!d'
I'm confused how to use semicolon not as a delimiter but as a pattern instead.
Thanks.
Other option is to use the GNU extension of address range.
Next example means: delete everything from a line which begins with pattern until a line ending with semicolon.
sed '/pattern/,/;$/ d' infile
EDIT to comment of Harsh:
Try next sed command:
sed '/^\s*LOG\s*(.*;\s*$/ d ; /^\s*LOG/,/;\s*$/ d' infile
Explanation:
/^\s*LOG\s*(.*;\s*$/ d # Delete line if begins with 'LOG' and ends with semicolon.
/^\s*LOG/,/;\s*$/ d # Delete range of lines between one that begins with LOG and
# other that ends with semicolon.
This might work for you:
cat <<! >file
> a
> b
> ;
> x
> y
> ;
> !
sed '/^[^;]*$/{H;$!d};x;s/;//;/x/!d' file
x
y
Explanation:
For any line the does not have a single ; in it /^[^;]*$/
Append the above line to the hold space (HS) and delete the pattern space (PS) and begin the next iteration unless it is the last line in the file. {H;$!d}
If a line is empty /^$/ or the last line of the file:
Swap to the HS x
Delete the first ; s/;//
Search for pattern (x) and if not found delete the PS /x/!d
N.B. This finds any pattern /x/ to find the beginning pattern use /^x/.
EDIT:
After having seen your data and expected result, this may work for you:
sed '/^\s*LOG(.*);/d;/^\s*LOG(/,/);/d' file