a simple question please
i have this code, and it add the word echo for all lines, but i want exclusively to odd lines
i kwow that this code sed -n 1~2p' show me all odd lines, but i can't doing the same in script above
sed 's/.*/echo &/' $startdirectory
thanks
For a literal answer to what you're asking (apply an 's' action to every other line), you want
sed -e '1~2s/.*/echo &/'
This is marginally better than the N way of doing things in that it doesn't interfere with other things you might want to do to other lines in the file.
Actually rather simple:
sed -e 'N;s/^/echo /'
The N reads a second line into the pattern space; the substitute puts 'echo' in front of the first, the implicit print prints both lines and empties the pattern space.
Note that if you have an odd number of lines, it drops the last. Fixing that is an exercise for the reader.
Related
sed 'N; D' testfile
testfile contains:
this is the first line
this is the second line
this is the third line
this is the fourth line
I am using RHEL 6 and the output comes as:
this is the fourth line
As per my understanding, N just pulls in the next line into the pattern space and D deletes just the first line of the pattern space. Therefore, the output should have been:
this is the second line
this is the fourth line
Can someone please explain why the output is coming as mentioned above?
According to the documentation:
D
If pattern space contains no newline, start a normal new cycle as if the d command was issued. Otherwise, delete text in the pattern space up to the first newline, and restart cycle with the resultant pattern space, without reading a new line of input.
(Emphasis mine.)
It sounds like this would restart your sed program from the beginning, reading and deleting lines until it runs out of input, at which point only the last line is left in the buffer.
As already shown using D will move to the beginning of program. You can however use the following to print even lines:
sed -n 'n;p'
and to print odds:
sed 'n;d'
In GNU sed you can also use:
sed '0~2!d' # Odd
sed '1~2!d' # Even
An alternative can be something like:
N;s/^[^\n]*\n//
which will read the next line into the pattern space and then substitute the first away.
One might ask why this is the behavior. One reason is to make things like this possible, working with multiply lines in the pattern space:
$!N;/\npattern$/d;P;D
The above will delete lines matching pattern as well as the line before.
my goal is to sed the 100th line and convert it to a string, then separate the data of the sentence to word
#!/bin/bash
fid=log.txt;
sentence=`expr sed -n '100p' ${fid}`;
for word in $sentence
do
echo $word
done
but apparently this has failed.
expr: syntax error
would somebody please let me know what have I done wrong? previously for number it worked.
The expr does not seem to serve a useful purpose here, and if it did, a sed command would certainly not be a valid or useful thing to pass to it, under most circumstances. You should probably just take it out.
However, the following loop is also problematic. Unquoted variables in shell script are very frequently an error. In this case, you can't quote the thing you pass to the for loop (that would cause the loop to only run once, with the loop variable set to the quoted string) but you also cannot prevent the shell from performing wildcard expansion on the unquoted string. So if the string happened to contain *, the shell will expand that to a list of files in the current directory, for example.
Fortunately, this can all be done in an only slightly more complicated sed script.
sed '100!d;s/[ \t]\+/\n/g;q' "$fid"
That is, if the line number is not 100, delete this line and start over with the next line. Otherwise, we are at line 100; replace runs of whitespace with newlines, (print) and quit.
(The backslash escape codes \t and \n are not universally portable; and \+ for repetition is also an optional extension. I believe there are also sed variants which dislike semicolon as a command separator. Consult your sed manual page, experiment, and if everything else fails, maybe switch to Awk or Perl. Just in case, here is a version which works even on Mac OSX:
sed '100!d
s/[ ][ ]*/\
/g;q' log.txt
The stuff inside the square brackets are a space and a literal tab; in Bash, with default keybindings, type ctrl-V, tab to produce a literal tab.)
Incidentally, this also gets rid of the variable capture antipattern. There are good reasons to capture output to a variable, but if it can be avoided, you often end up with a simpler, more robust and efficient, as well as more idiomatic and elegant script. (I see no reason to put the log file name in a variable, either, in this isolated case; but in a larger script, it might make sense.)
I don't think you need expr command in this case.
expr is used to do calculations. Something like:
expr 1 + 1
Just this one is fine:
sentence=`sed -n '100p' ${fid}`;
#!/bin/bash
fid=log.txt;
sentence=$(sed -n '100p' ${fid});
for word in $sentence
do
echo $word
done
put a dollar sign and parenthesis solve the problem
I have a block of text I'm trying to edit like this, in a script:
First, I tried
VAR2=`echo $VAR | sed 's/\n/\n\t/g'
It removes the newlines, but doesn't add the newline or tab back in.
Is this some stupid mistake? Not escaping something I should?
Two things:
You have to prevent shell expansion of $VAR, or the newlines will be lost before you have a chance to handle them
sed works in a line-based manner. It treats every line individually, and it doesn't see the newlines between them (unless you do special things).
The first can be handled by quoting $VAR, the second problem I would circumvent by reformulating the problem as "insert a tab to the beginning of every line but the first." That leaves us with:
VAR2=$(echo "$VAR" | sed '1!s/^/\t/')
Where the sed code means: Under the condition 1! (which is the case when we're not handling the first line), do s/^/\t/ -- i.e., replace the empty string at the beginning of the line with a tab.
Note that to look at the result of the substitution, you'll have to quote it as well, or it'll be shell-expanded, and the inserted whitespaces will be lost. That is to say,
echo "$VAR2"
will show the result you want, while
echo $VAR2
will lose all formatting (and potentially do silly things, if there are special characters such as $ in the paragraph).
Im trying to replace numbers in my textfile by adding one to them. i.e.
sed 's/3/4/g' path.txt
sed 's/2/3/g' path.txt
sed 's/1/2/g' path.txt
Instead of this, Can i automate it, i.e. find a /d and add one to it in the replace.
Something like
sed 's/\([0-8]\)/\1+1/g' path.txt
Also wanted to capture more than one digit i.e. ([0-9])\t([0-9]) and change each one keeping the tab inbetween
Thanks
edited #2
Using the perl example,
I also would like it to work with more digits i.e.
perl -pi~ -e 's/(\d+)\.(\d+)\.(\d+)\.(\d+)/ ($1+1)\.($2+1)\.($3+1)\.($4+1) /ge' output.txt
Any tips on making the above work?
There is no support for arithmetic in sed, but you can easily do this in Perl.
perl -pe 's/(\d+)/ $1+1 /ge'
With the /e option, the replacement expression needs to be valid Perl code. So to handle your final updated example, you need
perl -pi~ -e 's/(\d+)\.(\d+)\.(\d+)\.(\d+)/ $1+1 . "." $2+1 . "." . $3+1 . "." . $4+1 /ge'
where strings are properly quoted and adjacent strings are concatenated together with the . Perl string concatenation operator. (The arithmetic numbers are coerced into strings as well when they are concatenated with a string.)
... Though of course, the first script already does that more elegantly, since with the /g flag it already increments every sequence of digits with one, anywhere in the string.
Triplee's perl solution is the more generic answer, but Michal's sed solution works well for this particular case. However, Michal's sed solution is more easily written:
sed y/12345678/23456789/ path.txt
and is better implemented as
tr 12345678 23456789 < path.txt
This utterly fails to handle 2 digit numbers (as in the edited question).
You can do it with sed but it's not easy, see this thread.
And it's hard with awk too, see this.
I'd rather use perl for this (something like this can be seen in action # ideone):
perl -pe 's/([0-8])/$1+1/e'
(The ideone.com example must have some looping as ideone does not sets -pe by default.)
You can't do addition directly in sed - you could do it in awk by matching numbers using a regex in each line and increasing the value, but it's quite complicated. If do not need to handle arbitrary numbers but a limited set, like only single-digit numbers from 0 to 8, you can just put several replacement commands on a single sed command line by separating them with semicolons:
sed 's/8/9/g ; s/7/8/g; s/6/7/g; s/5/6/g; s/4/5/g; s/3/4/g; s/2/3/g; s/1/2/g; s/0/1/g' path.txt
This might work for you (GNU sed & Bash):
sed 's/[0-9]/$((&+1))/g;s/.*/echo "&"/e' file
This will add one to every individual digit, to increment numbers:
sed 's/[0-9]\+/$((&+1))/g;s/.*/echo "&"/e' file
N.B. This method is fraught with problems and may cause unexpected results.
Wow, this one has really got me. Gonna need some tricky sed skill here I think. Here is the output value of command text I'm trying to replace:
...
fast
n : abstaining from food
The value I'd like to replace it with, is:
...
Noun
: abstaining from food
This turns out to be tricker that I thought. Because 'fast' is listed a number of times and because it is listed in other places at the beginning of the line. So I came up with this to define the range:
sed '/fast/,/^ n : / s/fast/Noun/'
Which I thought would do, but... Unfortunately, this doesn't end the replacement and the rest of the output following this match are replaced with Noun. How to get sed to stop replacement after the match? Even better, can I find a two line pattern match and replace it?
Try this:
sed "h; :b; \$b ; N; /^${1}\n n/ {h;x;s//Noun\n/; bb}; \$b ; P; D"
Unfortunately, Paul's answer reads the whole file in which makes any additional processing you might want to do difficult. This version reads the lines in pairs.
By enclosing the sed script in double quotes instead of single quotes, you can include shell variables such as positional parameters. I would recommend surrounding them with curly braces so they are set apart from the adjacent characters. When using double quotes, you'll have to be careful of the shell wanting to do its various expansions. In this example, I've escaped the dollar signs that signify the last line of the input file for the branch commands. Otherwise the shell will try to substitute the value of a variable $b which is likely to be null thus making sed unhappy.
Another technique would be to use single quotes and close and open them each time you have a shell variable:
sed 'h; :b; $b ; N; /^'${1}'\n n/ {h;x;s//Noun\n/; bb}; $b ; P; D'
# ↑open close↑ ↑open close↑
I'm assuming that the "[/code]" in your expected result is a typo. Let me know if it's not.
This seems to do what you want:
sed -e ':a;N;$!ba;s/fast\n n/Noun\n/'
I essentially stole the answer from here.
This might work for you:
sed '$!N;s/^fast\n\s*n :/Noun\n :/;P;D' file
...
Noun
: abstaining from food