sed pattern matching, stop at first found character - sed

Take the string "hello_world 1 2 3"
I want the output to be "hello_world"
My attempt is "s/\(.*\) .*/\1/g"
But I get "hello_world 1 2"
Instead of stopping at the first space after the sequence, it gets the last space on the line.
I want to take any length of characters \(.*\) followed by a space ' ' and remove anything that comes after it .*
How can I do it?

Could you please try following.
echo "hello_world 1 2 3" | sed 's/\([^ ]*\).*/\1/'
Explanation of above:
Using sed's capability of storing matched regex into a temp buffer. Which could be later accessed by variables like 1, 2 and so on(depending upon number of buffers you are mentioning).
In here we are capturing everything till occurrence of first space into 1st temp buffer and then keeping everything as it is .*. While substituting we are mentioning \1 here which means substitute whole line's value with first matched/stored value of 1st temp buffer(which is hello_world).
Why OP's code not working: Because OP using .* which is a greedy matched regex and capturing all the line in 1st buffer itself that's why when its used \1 its actually printing whole line there.

This might work for you (GNU sed):
sed 's/\s.*//' file
Matches the first white space character and everything thereafter and removes it, leaving whatever is in front of i.e. all non-white space characters.
Same as:
sed 's/^(\S+).*/\1/' -E file

Related

how to type the beginning or end in sed multiple-lines mode?

As we all knew,the "\‘" and "\’"
indicates the beginning or end respectively in multiple-lines mode.But under ASCII(or input-in-english) only "'" exists.
How to type the beginning?
This might work for you (GNU sed):
seq 3 | sed -n 'p;H;1h;$!d;g;l0
s/^.*$/ALL/mgp
s/\`.*$/START/mp
s/^.*\'\''/END/mp'
1
2
3
1\n2\n3$
ALL
ALL
ALL
START
ALL
ALL
START
ALL
END
The command seq generates a file of three consecutive integers.
The sed uses the -n option to turn off implicit printing and then slurps the three integers into hold space. Printing each integer as it is read.
The first substitution, replace all lines with the literal ALL.
The second substitution, replaces the first line with START.
The third substitution, replace the last line with END.
N.B. The use of the m(multiline), g(global) and p(print) substitution flags. Lastly, if the -z option is in use, these zero width anchors work with respect to null characters not newlines.

getting the first letter of an filtered part in sed

I have a filename e.g. 15736--1_brand-new-image.jpg
My goal is to get the first letter after the _ in this case the b.
With s/\(.*\)\_\(.*\)$/\2/ I am able to extract brand-new-image.jpg
which is partly based on the info found on https://www.oncrashreboot.com/use-sed-to-split-path-into-filename-extension-and-directory
I've already found get first letter of words using sed but fail to combine the two.
To validate my sed statement I've used https://sed.js.org/
How can I combina a new sed statement on the part I've filtered to get the first letter?
With your shown samples could you please try following.
echo "15736--1_brand-new-image.jpg" | sed 's/[^_]*_\(.\).*/\1/'
Explanation: Simply using substitution operation of sed, then looking till 1st occurrence of _ then saving next 1 char into back reference and mentioning .* will cover everything after it, while substituting simply substituting everything with 1st back reference value which will be after 1st _ in this case its b.
Explanation: Following is only for explanation purposes.
sed ' ##Starting sed program from here.
s/ ##using s to tell sed to perform substitution operation.
[^_]*_\(.\).* ##using regex to match till 1st occurrence of _ then using back reference \(.\) to catch value in temp buffer memory here.
/\1/ ##Substituting whole line with 1st back reference value here which is b in this case.
'
Using a . or \w could also match _ in case there are 2 consecutive __
If you want to match the first word character without matching the _ you could also use
echo "15736--1_brand-new-image.jpg" | sed 's/[^_]*_\([[:alnum:]]\).*/\1/'
Output
b
This might work for you (GNU sed):
sed -nE 's/^[^_]*_[^[:alpha:]]*([[:alpha:]]).*/\1/p' file
Since this a filtering type operation use the -n option to print only when there is a positive match.
Match the first _ from the start of the line and then discard any non-alpha characters until an alpha character and finally discard any other characters.
Print the result if there is a match.
N.B. Anchoring the match to the start of the line, prevents the result containing more than one character i.e. consider the string 123_456_abc might otherwise result in 4 or 123_a.

A way to append the beginning of every line before a pattern to the end of each same line?

I am trying to copy the beginning of every line in a text file before a certain character to the end of the same line.
I've tried duplicating each line to the end of itself, and then deleting everything after the character, but the trouble is I haven't been able to figure out how to skip the first instance of the character so the result is that the duplicated text gets deleted as well as everything beyond the first instance of the character.
I've tried things like
sed '/S/ s/$/ append text/' sample.txt > cleaned.txt
but this only adds a fixed text. I've also tried using:
s/\(.*\)/\1 \1/
to duplicate the line, and then deleting everything after the S, but I can't figure out how to get it to go to the 3rd S not the 1st to start deleting.
What I have to start with:
dog 50_50_S5_Scale
cat 10_RV_S76_Scale
mouse 15_SQ_S81_Scale
What I'm trying to get:
dog 50_50_S5_Scale dog 50_50_
cat 10_RV_17_S76_Scale cat 10_RV_17_
mouse 15_EQ_S81_Scale mouse 15_EQ_
Where everything before the first S gets copied to the end of the line.
You may use
sed 's/\([^S]*\)S.*/& \1/' file
See the online demo
Details
\([^S]*\) - Capturing group 1 (\1): any 0+ chars other than S
S.* - S and the rest of the string (actually, line, since sed processes line by line by default).
The replacement is the concatenation of the whole match (&), space and Group 1 value.
You could try:
awk '{print $0 " " substr($0, 0, index($0,"S") - 1)}' file
We take the substring from the first character up to but not including the first occurance of "S".

SED search and replace substring in a database file

To all,
I have spent alot of time searching for a solution to this but cannot find it.
Just for a background, I have a text database with thousands of records. Each record is delineated by :
"0 #nnnnnn# Xnnn" // no quotes
The records have many fields on a line of their own, but the field I am interested in to search and replace a substring (notice spaces) :
" 1 X94 User1.faculty.ventura.ca" // no quotes
I want to use sed to change the substring ".faculty.ventura.ca" to ".students.moorpark.ut", changing nothing else on the line, globally for ALL records.
I have tested many things with negative results.
How can this be done ?
Thank You for the assistance.
Bob Perez (robertperez1957#gmail.com)
If I understand you correctly, you want this:
sed 's/1 X94 \(.*\).faculty.ventura.ca/1 X94 \1.students.moorpark.ut/' mydatabase.file
This will replace all records of the form 1 X94 XXXXXX.faculty.ventura.ca with 1 X94 XXXXX.students.moorpark.ut.
Here's details on what it all does:
The '' let you have spaces and other messes in your script.
s/ means substitute
1 X94 \(.*\).faculty.ventura.ca is what you'll be substituting. The \(.*\) stores anything in that regular expression for use in the replacement
1 X94 \1.students.moorpark.ut is what to replace the thing you found with. \1 is filled in with the first thing that matched \(.*\). (You can have multiple of those in one line, and the next one would then be \2.)
The final / just tells sed that you're done. If your database doesn't have linefeeds to separate its records, you'll want to end with /g, to make this change multiple times per line.
mydatabase.file should be the filename of your database.
Note that this will output to standard out. You'll probably want to add
> mynewdatabasefile.name
to the end of your line, to save all the output in a file. (It won't do you much good on your terminal.)
Edit, per your comments
If you want to replace 1 F94 bperez.students.Napvil.NCC to 1 F94 bperez.JohnSmith.customer, you can use another set of \(.*\), as:
sed 's/1 X94 \(.*\).\(.*\).Napvil.NCC/1 X94 \1.JohnSmith.customer/' 251-2.txt
This is similar to the above, except that it matches two stored parameters. In this example, \1 evaluates to bperez and \2 evaluates to students. We match \2, but don't use it in the replace part of the expression.
You can do this with any number of stored parameters. (Sed probably has some limit, but I've never hit a sufficiently complicated string to hit it.) For example, we could make the sed script be '\(.\) \(...\) \(.*\).\(.*\).\(.*\).\(.*\)/\1 \2 \3.JohnSmith.customer/', and this would make \1 = 1, \2 = X94, \3 = bperez, \4 = Napvil and \5 = NCC, and we'd ignore \4 and \5. This is actually not the best answer though - just showing it can be done. It's not the best because it's uglier, and also because it's more accepting. It would then do a find and replace on a line like 2 Z12 bperez.a.b.c, which is presumably not what you want. The find query I put in the edit is as specific as possible while still being general enough to suit your tasks.
Another edit!
You know how I said "be as specific as possible"? Due to the . character being special, I wasn't. In fact, I was very generic. The . means "match any character at all," instead of "match a period". Regular expressions are "greedy", matching the most they could, so \(.*\).\(.*\) will always fill the first \(.*\) (which says, "take 0 to many of any character and save it as a match for later") as far as it can.
Try using:
sed 's/1 X94 \(.*\)\.\(.*\).Napvil.NCC/1 X94 \1.JohnSmith.customer/' 251-2.txt
That extra \ acts as an escape sequence, and changes the . from "any character" to "just the period". FYI, since I don't (but should) escape the other periods, technically sed would consider 1 X94 XXXX.StdntZNapvilQNCC as a valid match. Since . means any character, a Z or a Q there would be considered a fit.
The following tutorial helped me
sed - replace substring in file
try the same using a -i prefix to replace in the file directly
sed -i 's/unix/linux/' file.txt

Can I use the sed command to replace multiple empty line with one empty line?

I know there is a similar question in SO How can I replace mutliple empty lines with a single empty line in bash?. But my question is can this be implemented by just using the sed command?
Thanks
Give this a try:
sed '/^$/N;/^\n$/D' inputfile
Explanation:
/^$/N - match an empty line and append it to pattern space.
; - command delimiter, allows multiple commands on one line, can be used instead of separating commands into multiple -e clauses for versions of sed that support it.
/^\n$/D - if the pattern space contains only a newline in addition to the one at the end of the pattern space, in other words a sequence of more than one newline, then delete the first newline (more generally, the beginning of pattern space up to and including the first included newline)
You can do this by removing empty lines first and appending line space with G command:
sed '/^$/d;G' text.txt
Edit2: the above command will add empty lines between each paragraph, if this is not desired, you could do:
sed -n '1{/^$/p};{/./,/^$/p}'
Or, if you don't mind that all leading empty lines will be stripped, it may be written as:
sed -n '/./,/^$/p'
since the first expression just evaluates the first line, and prints it if it is blank.
Here: -n option suppresses pattern space auto-printing, /./,/^$/ defines the range between at least one character and none character (i.e. empty space between newlines) and p tells to print this range.