I have a large file that looks like this
(something,something1,something2),(something,something1,something2)
how do I use sed and find ),( and replace it with );( or add a newline between the parentheses that has a comma character.
I did try sed 's/),(/),\n(/g' filename.txt but for some reason it does not work
for those who come here and want to know how this work without getting a lot of stackoverflow "greetings"
since I was on Mac os x you need to replace your \n with \'$'\n''
so to find ),( and add a new line between the parentheses this is the command I used
sed 's/;/\'$'\n''/g' testdone.txt > testdone2.txt
ES
echo "(something,something1,something2),(something,something1,something2)" | sed "s|),(|);(|"
This prints the below for me.
(something,something1,something2);(something,something1,something2)
For new line
echo "(something,something1,something2),(something,something1,something2)" | sed "s|),(|)\n(|"
And the above prints the below.
(something,something1,something2)
(something,something1,something2)
Related
I want to substring the File name in unix using sed command.
File name : Test_Test1_Test2_10082019_030013.csv.20191008-075740
I want the characters after the 3rd underscore or (all the characters after Test2 ) i need to be printed .
Can this be done using sed command?
I have tried this command
sed 's/^.*_\([^_]*\)$/\1/' <<< 'Test_Test1_Test2_10082019_030013.csv.20191008-075740'
but this is giving result as 030013.csv.20191008-075740
I need it from 10082019_030013.csv.20191008-075740
Thanks
Neha
To remove from the beginning up to including the 3rd underscore you can use
sed 's/^\([^_]*_\)\{3\}//' <<< 'Test_Test1_Test2_10082019_030013.csv.20191008-075740'
This removes the initial part that consists of 3 groups of (any number of non-underscore characters followed by an underscore). The result is
10082019_030013.csv.20191008-075740
If you use GNU sed you can switch it to extended regular expressions and omit the backslashes.
sed -r 's/^([^_]*_){3}//' <<< 'Test_Test1_Test2_10082019_030013.csv.20191008-075740'
Could you please try following.
sed 's/\([^_]*\)_\([^_]*\)_\([^_]*\)_\(.*\)/\4/' Input_file
Or as per Bodo's nice suggestion:
sed 's/[^_]*_[^_]*_[^_]_\(.*\)/\1/' Input_file
This might work for you (GNU sed):
sed 's/_/\n/3;s/.*\n//;t;s/Test2/\n/;s/.*\n//;t;d' file
Replace the third _ by a newline and then remove everything upto and including the first newline. If this succeeds, bail out and print the result. Otherwise, try the same method with Test2 and if this fails delete the entire line.
I'm trying to come up with a sed script to take all lines containing a pattern and move them to the end of the output. This is an exercise in learning hold vs pattern space and I'm struggling to come up with it (though I feel close).
I'm here:
$ echo -e "hi\nfoo1\nbar\nsomething\nfoo2\nyo" | sed -E '/foo/H; //d; $G'
hi
bar
something
yo
foo1
foo2
But I want the output to be:
hi
bar
something
yo
foo1
foo2
I understand why this is happening. It is because the first time we find foo the hold space is empty so the H appends \n to the blank hold space and then the first foo, which I suppose is fine. But then the $G does it again, namely another append which appends \n plus what is in the hold space to the pattern space.
I tried a final delete command with /^$/d but that didn't remove the blank line (I think this is because this pattern is being matched not against the last line, but against the, now, multiline pattern space which has a \n\n in it.
I'm sure the sed gurus have a fix for me.
This might work for you (GNU sed):
sed '/foo/H;//!p;$!d;x;//s/.//p;d' file
If the line contains the required string append it to the hold space (HS) otherwise print it as normal. If it is not the last line delete it otherwise swap the HS for the pattern space (PS). If the required string(s) is now in the PS (what was the HS); since all such patterns were appended, the first character will be a newline, delete the first character and print. Delete whatever is left.
An alternative, using the -n flag:
sed -n '/foo/H;//!p;$!b;x;//s/.//p' file
N.B. When the d or b (without a parameter) command is performed no further sed commands are, a new line is read into the PS and the sed script begins with the first command i.e. the sed commands do not resume following the previous d command.
Why? Stuff like this is absolutely trivial in awk, awk is available everywhere that sed is, and the resulting awk script will be simpler, more portable, faster and better in almost every other way than a sed script to do the same task. All that hold space stuff was necessary in sed before the mid-1970s when awk was invented but there's absolutely no use for it now other than as a mental exercise.
$ echo -e "hi\nfoo1\nbar\nsomething\nfoo2\nyo" |
awk '/foo/{buf = buf $0 RS;next} {print} END{printf "%s",buf}'
hi
bar
something
yo
foo1
foo2
The above will work as-is in every awk on every UNIX installation and I bet you can figure out how it works very easily.
This feels like a hack and I think it should be possible to handle this situation more gracefully. The following works on GNU sed:
echo -e "hi\nfoo1\nbar\nsomething\nfoo2\nyo" | sed -r '/foo/{H;d;}; $G; s/\n\n/\n/g'
However, on OSX/BSD sed, results in this odd output:
hi
bar
something
yonfoo1
foo2
Note the 2 consecutive newlines was replaced with the literal character n
The OSX/BSD vs GNU sed is explained in this article. And the following works (in GNU SED as well):
echo -e "hi\nfoo1\nbar\nsomething\nfoo2\nyo" | sed '/foo/{H;d;}; $G; s/\n\n/\'$'\n''/'
TL;DR; in BSD sed, it does not accept escaped characters in the RHS of the replacement expression and so you either have to put a true LF/newline in there at the command line, or do the above where you split the sed script string where you need the newline on the RHS and put a dollar sign in front of '\n' so the shell interprets it as a line feed.
I want to insert a range of lines from a file, say something like 210,221r before the first occurrence of a pattern in a bunch of other files.
As I am clearly not a GNU sed expert, I cannot figure how to do this.
I tried
sed '0,/pattern/{210,221r file
}' bunch_of_files
But apparently file is read from line 210 to EOF.
Try this:
sed -r 's/(FIND_ME)/PUT_BEFORE\1/' test.text
-r enables extendend regular expressions
the string you are looking for ("FIND_ME") is inside parentheses, which creates a capture group
\1 puts the captured text into the replacement.
About your second question: You can read the replacement from a file like this*:
sed -r 's/(FIND_ME)/`cat REPLACEMENT.TXT`\1/' test.text
If replace special characters inside REPLACEMENT.TXT beforehand with sed you are golden.
*= this depends on your terminal emulator. It works in bash.
In https://stackoverflow.com/a/11246712/4328188 CodeGnome gave some "sed black magic" :
In order to insert text before a pattern, you need to swap the pattern space into the hold space before reading in the file. For example:
sed '/pattern/ {
h
r file
g
N
}' in
However, to read specific lines from file, one may have to use a two-calls solution similar to dummy's answer. I'd enjoy knowing of a one-call solution if it is possible though.
I'm having issues matching strings even if they start with any number of white spaces. It's been very little time since I started using regular expressions, so I need some help
Here is an example. I have a file (file.txt) that contains two lines
#String1='Test One'
String1='Test Two'
Im trying to change the value for the second line, without affecting line 1 so I used this
sed -i "s|String1=.*$|String1='Test Three'|g"
This changes the values for both lines. How can I make sed change only the value of the second string?
Thank you
With gnu sed, you match spaces using \s, while other sed implementations usually work with the [[:space:]] character class. So, pick one of these:
sed 's/^\s*AWord/AnotherWord/'
sed 's/^[[:space:]]*AWord/AnotherWord/'
Since you're using -i, I assume GNU sed. Either way, you probably shouldn't retype your word, as that introduces the chance of a typo. I'd go with:
sed -i "s/^\(\s*String1=\).*/\1'New Value'/" file
Move the \s* outside of the parens if you don't want to preserve the leading whitespace.
There are a couple of solutions you could use to go about your problem
If you want to ignore lines that begin with a comment character such as '#' you could use something like this:
sed -i "/^\s*#/! s|String1=.*$|String1='Test Three'|g" file.txt
which will only operate on lines that do not match the regular expression /.../! that begins ^ with optional whiltespace\s* followed by an octothorp #
The other option is to include the characters before 'String' as part of the substitution. Doing it this way means you'll need to capture \(...\) the group to include it in the output with \1
sed -i "s|^\(\s*\)String1=.*$|\1String1='Test Four'|g" file.txt
With GNU sed, try:
sed -i "s|^\s*String1=.*$|String1='Test Three'|" file
or
sed -i "/^\s*String1=/s/=.*/='Test Three'/" file
Using awk you could do:
awk '/String1/ && f++ {$2="Test Three"}1' FS=\' OFS=\' file
#String1='Test One'
String1='Test Three'
It will ignore first hits of string1 since f is not true.
This should be extremely simple, but for the life of me I just can't get gnu-sed to do it this afternoon.
The file in question has lines that look like this:
PART NUMBER PART NUMBER QUANTITY WEIGHT -999 -4,999 -9,999
w/ UL APPROVAL
MIN-3
I need to prepend every line like the "MIN-3" line with a ">" character, and the only thing specifically differentiating those lines from the others are two things:
The first character is a space " ".
The lines do not contain a comma.
I've tried mostly things like any of the following:
/^ +[^,]+$/ s/^/>/
/^ +[\w\-]+$/ s/^/>/
/^ +(\w|\-)+$/ s/^/>/
I will admit, I am somewhat new to sed. :)
Edit: Answers that use perl, or awk could also be appreciated, though my initial target is sed.
try this:
sed '/^ [^,]*$/s/^/>/'
the output is, only the line with MIN-3 with leading >
sed default uses basic regex. so the + should be \+ in your script. I think that could be the problem killing your time. You could add -r however, to let sed use extended-regex.
According to your description this should do:
sed 's/^\([ ][^,]*\)$/> \1/' input
which matches the complete line if the line starts with a space and then contains anything but a comma until the end.
Here is a simple answer:
sed 's/^ [^,]*$/>&/'