Bash: Sed replacing all occurrences except one - bash

I need to replace all patterns of either AllowUsers, #AllowUsers or # AllowUsers in a file.
I've got that part covered with sed -e 's/^\s*#\?AllowUsers.*//'
But the thing I'm having trouble with is that it should leave one occurrence in the file and remove everything else.
Let me know, thank you in advance!

awk '/AllowUsers||#AllowUsers||# AllowUsers/&&c++ {next} 1' did the job!

Related

Removing all the characters from a string after pattern+2

I am trying to remove all the characters from a string after a specific pattern +2 in bash.
In this case I have for example:
3434.586909
3434.58690932454
3434.5869093232r3353
I'd like to keep just 3434.58
I tried with awk and a wildcard but my test haven't worked yet.
You can use sed:
sed 's/\(\...\).*/\1/'
It means "remembering a dot and two following characters, replace them and everything that follows with the remembered part".
How about using floating point logic?
awk '{printf("%.02f\n",$0)}' Input_file
awk '{print substr($0,1,7)}' file
3434.58
3434.58
3434.58

How to insert a specific character at a specific line of a file using sed or awk?

I want to use command to edit the specific line of a file instead of using vi. This is the thing. If there is a # starting with the line, then replace the # to make it uncomment. Otherwise, add the # to make it comment. I'd like to use sed or awk. But it won't work as expected.
This is the file.
what are you doing now?
what are you gonna do? stab me?
this is interesting.
This is a test.
go big
don't be rude.
For example, I just want to add the # at the beginning of the the line 4 This is a test if it doesn't start with #. And if it starts with #, then remove the #.
I've already tried via sed & gawk (awk)
gawk -i inplace '$1!="#" {print "#",$0;next};{print substr($0,3,length-1)}' file
sed -i /test/s/^#// file # make it uncomment
sed -i /test/s/^/#/ file # make it comment
I don't know how to use if else to make sed work. I could only make it with a single command, then use another regex to make the opposite.
Using gawk, it works as the main line. But it will mess the rest of the code up.
This might work for you (GNU sed):
sed '4{s/^/#/;s/^##//}' file
On line 4 prepend a # to the line and if there 2 #'s remove them.
Could also be written:
sed '4s/^/#/;4s/^##//' file
This will remove # from the start of line 4 or add it if it wasn't already there:
sed -i '4s/^#/\n/; 4s/^[^\n]/#&/; 4s/^\n//' File
The above assume GNU sed. If you have BSD/MacOS sed, some minor changes will be required.
When sed reads a new line, the one thing that we know for sure about the new line is that it does not contain \n. (If it did, it would be two lines, not one.) Using this knowledge, the script works by:
s/^#/\n/
If the fourth line starts with #, replace # with \n. (The \n serves as a notice that the line had originally been commented out.)
4s/^[^\n]/#&/
If the fourth line now starts with anything other than \n (meaning that it was not originally commented), put a # in front.
4s/^\n//
If the fourth line now starts with \n, remove it.
Alternative: Modifying lines that contain test
To comment/uncomment lines that contain test:
sed '/test/{s/^#/\n/; s/^[^\n]/#&/; s/^\n//}' File
Alternative: using awk
The exact same logic can be applied using awk. If we want to comment/uncomment line 4:
awk 'NR==4 {sub(/^#/, "\n"); sub(/^[^\n]/, "#&"); sub(/^\n/, "")} 1' File
If we want to comment/uncomment any line containing test:
awk '/test/ {sub(/^#/, "\n"); sub(/^[^\n]/, "#&"); sub(/^\n/, "")} 1' File
Alternative: using sed but without newlines
To comment/uncomment any line containing test:
sed '/test/{s/^#//; t; s/^/#/; }' File
How it works:
s/^#//; t
If the line begins with #, then remove it.
t tells sed that, if the substitution succeeded, then it should skip the rest of the commands.
s/^/#/
If we get to this command, that means that the substitution did not succeed (meaning the line was not originally commented out), so we insert #.
If you end up on a system with a sed that doesn't support in-place editing, you can fall back to its uncle ed:
ed -s file 2>/dev/null <<EOF
4 s/^/#/
s/^##//
w
q
EOF
(Standard error is redirected to /dev/null because in ed, unlike sed, it's an error if s doesn't replace anything and a question mark is thus printed to standard error.)
$ awk 'NR==4{$0=(sub(/^#/,"") ? "" : "#") $0} 1' file
what are you doing now?
what are you gonna do? stab me?
this is interesting.
#This is a test.
go big
don't be rude.
$ awk 'NR==4{$0=(sub(/^#/,"") ? "" : "#") $0} 1' file |
awk 'NR==4{$0=(sub(/^#/,"") ? "" : "#") $0} 1'
what are you doing now?
what are you gonna do? stab me?
this is interesting.
This is a test.
go big
don't be rude.

Delete all lines before last case of a string

How would I go about deleting all the lines before the last occurrence of a string. Like if I had a file that looked like
Icecream is good
And
Chocolate is good
And
They have lots of sugar
If I want all lines after and including the last occurrence of "And" what's the cleanest way to do this? Specifically, I want
And
They have lots of sugar
I was doing sed -n -E -e '/And/,$p' file but I see this gives me the first occurrence.
This might work for you (GNU sed):
sed -n '/And/h;//!H;$!d;x;//p' file
Replace anything in the hold space by the line containing And. Append all other lines to the hold space. At the end of the file, swap the pattern space for the hold space and print out the result as long it matches the required string And.
I know that you asked for sed and that Potong provided a good sed solution. But, for comparison, here is an awk solution:
$ awk 's{s=s"\n"$0;} /And/{s=$0;} END{print s;}' file
And
They have lots of sugar
How it works:
s{s=s"\n"$0;}
If the variable s is not empty, then add to it the current line, $0.
/And/{s=$0;}
If the current line contains And, then set s to the current line, $0.
END{print s;}
After we have reached the end of the file, print s.
$ tac file | awk '!f; /And/{f=1}' | tac
And
They have lots of sugar
$ awk 'NR==FNR{if(/And/)nr=NR;next} FNR>=nr' file file
And
They have lots of sugar

SED bash script Assistance

I'm trying to follow who my friend is following (all 1,522 of them)
and a got a text file with from his twitter page and I want to see just the last word of a line that begins with #.
Example:
Podcaster, broadcaster and tech pundit. The Tech Guy on the Premiere Radio
Networks. Live at live.twit.tv For my link feed follow #links_for_twit
(Line-wrapped to remove hateful horizontal scrollbar.)
I want that to turn into #links_for_twit.
Use awk instead:
awk '$NF ~ /^#/ {print $NF}'
You mean, like:
grep -o '#[a-zA-Z_0-9]*$' tweets.txt
?
If you're wanting to use sed, try this:
sed -n 's/.*\(#.*\)/\1/p'
-n: don't print anything unless asked
s/.*\(#.*\): capture everything after the last '#' in the line
/\1/: replace the whole line with the captured bit
p: print if a substitution was made
Hope that helps
EDIT: I just saw the complaint below about email addresses. you can add \s just before the # to ensure there's a space: sed -n 's/.*\s(#.*\)/\1/p'
If you have GNU grep, you could use a Perl-flavoured regex to ensure the # is at the start of a word:
grep -Po '(?<=^|\s)#\w+' filename

Remove a line from a csv file bash, sed, bash

I'm looking for a way to remove lines within multiple csv files, in bash using sed, awk or anything appropriate where the file ends in 0.
So there are multiple csv files, their format is:
EXAMPLEfoo,60,6
EXAMPLEbar,30,10
EXAMPLElong,60,0
EXAMPLEcon,120,6
EXAMPLEdev,60,0
EXAMPLErandom,30,6
So the file will be amended to:
EXAMPLEfoo,60,6
EXAMPLEbar,30,10
EXAMPLEcon,120,6
EXAMPLErandom,30,6
A problem which I can see arising is distinguishing between double digits that end in zero and 0 itself.
So any ideas?
Using your file, something like this?
$ sed '/,0$/d' test.txt
EXAMPLEfoo,60,6
EXAMPLEbar,30,10
EXAMPLEcon,120,6
EXAMPLErandom,30,6
For this particular problem, sed is perfect, as the others have pointed out. However, awk is more flexible, i.e. you can filter on an arbitrary column:
awk -F, '$3!=0' test.csv
This will print the entire line is column 3 is not 0.
use sed to only remove lines ending with ",0":
sed '/,0$/d'
you can also use awk,
$ awk -F"," '$NF!=0' file
EXAMPLEfoo,60,6
EXAMPLEbar,30,10
EXAMPLEcon,120,6
EXAMPLErandom,30,6
this just says check the last field for 0 and don't print if its found.
sed '/,[ \t]*0$/d' file
I would tend to sed, but there is an egrep (or: grep -e) -solution too:
egrep -v ",0$" example.csv

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