Sed - How to switch two words in a line [duplicate] - linux

This question already has answers here:
exchange two words using sed
(5 answers)
Closed 5 years ago.
I'm trying to write a shell script that switches the first and third words in a line. In this case only strings that contain letters (both upper- and lowercase) count as words, everything else (numbers, punctuation, whitespace) is considered whitespace.
For example:
abc123def. ghi...jkl
would turn into:
ghi123def. abc...jkl
I tried the following, but it doesn't work:
sed 's/\([a-zA-Z][a-zA-Z]*\)[^A-Z^a-z]\([a-zA-Z][a-zA-Z]*\)[^A-Z^a-z]\([a-zA-Z][a-zA-Z]*\)/\3 \2 \1/' input.txt


With sed:
$ echo "abc123def. ghi...jkl" | sed -r 's/([A-Za-z]*)([^A-Za-z]*[A-Za-z]*[^A-Za-z]*)([A-Za-z]*)(.*)/\3\2\1\4/g'
$ ghi123def. abc...jkl

Related

Print new line as string literal in unix or shell [duplicate]

This question already has answers here:
How to replace one character with two characters using tr
(5 answers)
Closed 3 years ago.
Hi I have a shell script that has
variable="apple banana monkey"
I want it to be
apple\nbanana\nmonkey
But when I try and execute
echo $variable | tr ' ' '\n'
It results to
apple
banana
monkey
I want to get the actual literal of new line and not the evaluated value.
I have tried echo -e or echo -n or even put numerous escapes \\ but to no avail.
Please help. Thanks

tr command translates chars into chars by performing a 1 to 1 mapping. You are asking the tool to translate a space into two chars, which is something that cannot be done with tr.
If you accept a command switch, you can try with sed:
echo "$variable" | sed 's/ /\\n/g'

how replace null characters in UNIX? [duplicate]

This question already has answers here:
Identifying and removing null characters in UNIX
(9 answers)
Closed 3 years ago.
I had this string and to show the special characters and then replace
0|A0140017511|1|DEMÁS TEMAS|�|�|2014-08-01
I want this
0|A0140017511|1|DEMÁS TEMAS|#|#|2014-08-01
$ xxd -p file.txt
307c41303134303031373531317c317c44454dc381532054454d41537c00
7c007c323031342d30382d3031

From xxd's output I can tell that you're trying to replace NULs with something else. To replace them with a single character, for example -, use tr:
$ tr '\0' '-' < file
0|A0140017511|1|DEMÁS TEMAS|-|-|2014-08-01
Or if you have GNU sed, you can use a string as well:
$ sed 's/\x0/^#/g' file
0|A0140017511|1|DEMÁS TEMAS|^#|^#|2014-08-01

How to use sed or awk or something similar to replace every odd occurrence of character? [duplicate]

This question already has answers here:
Replace every n'th occurrence in huge line in a loop
(4 answers)
Closed 4 years ago.
I have the following string:
"1,0,2,0,3,0,4,0,5,0,6,0,13,05,24233,55".
How to use awk, or sed to get
"1.0,2.0,3.0,4.0,5.0,6.0,13.05,24233.55"?
I tried to use
sed 's/,/./g' <<< "1,0,2,0,3,0,4,0,5,0,6,0,13,05,24233,55"
1.0.2.0.3.0.4.0.5.0.6.0.13.05.24233.55
and also
sed 's/,/./2' <<< "1,0,2,0,3,0,4,0,5,0,6,0,13,05,24233,55"
1,0.2,0,3,0,4,0,5,0,6,0,13,05,24233,55
Which replaced the second item only. I need every odd occurrence changed.
For future, what would be the code the replace every odd occurrence of, by . ?
Thanks for your help

With any sed that supports EREs via -E, e.g. GNU sed and OSX/BSD sed:
$ echo "1,0,2,0,3,0,4,0,5,0,6,0,13,05,24233,55" | sed -E 's/,([^,]+(,|$))/.\1/g'
1.0,2.0,3.0,4.0,5.0,6.0,13.05,24233.55
The above was inspired by #PesaThe's comment to my original answer.

try this:
for the end:
sed 's/[,]$/?/' YourFile
putting the , between [] allow you to remove most of the regex behavior taking litteral value (not for some char like ^ that need to be manage another way
putting the $ is telling to refere to end of string
the g in your test mean change every occurence, you only wanted 1 and at the end
for the internal:
sed -e 's/,/./1;p' \
-e ':a' \
-e 's/^\(\([^.]*[.][^,]*,\)*\)\([^,]*\),\([^,]*\)/\1\3.\4/
/[^,]*,[^,.]*,/ ta' YourFile
you need a loop and a special test due to alternance existing

Remove the first 6 columns using Linux [duplicate]

This question already has answers here:
how to remove the first two columns in a file using shell (awk, sed, whatever)
(9 answers)
Closed 4 years ago.
From a text file, I want to remove the first 6 columns. I tried sed as follows, but I have to do it six times (one for each column). Is there any efficient way to do it (or pass the 6 columns at once for sed)?
sed -i -r 's/(\s+)?\S+//1' file
Thanks!

You could do this within regex and quantifying braces if a column consists of non-whitespace characters with optional leading spaces:
sed -i -r 's/^((\s+)?\S+){6} *//' file

I have a requirement of searching a pattern from a file and displaying the pattern only in the screen,not the whole line .How can I do it in linux? [duplicate]

This question already has answers here:
Can grep show only words that match search pattern?
(15 answers)
Closed 5 years ago.
I have a requirement of searching a pattern like x=<followed by any values> from a file and displaying the pattern i.e x=<followed by any values>, only in the screen, not the whole line. How can I do it in Linux?

I have 3 answers, from simple (but with caveats) to complex (but foolproof):
1) If your pattern never appears more than once per line, you could do this (assuming your shell is
PATTERN="x="
sed "s/.*\($PATTERN\).*/\1/g" your_file | grep "$PATTERN"
2) If your pattern can appear more than once per line, it's a bit harder. One easy but hacky way to do this is to use a special characters that will not appear on any line that has your pattern, eg, "#":
PATTERN="x="
SPECIAL="#"
grep "$PATTERN" your_file | sed "s/$PATTERN/$SPECIAL/g" \
| sed "s/[^$SPECIAL]//g" | sed "s/$SPECIAL/$PATTERN/g"
(This won't separate the output pattern per line, eg. you'll see x=x=x= if a source line had 3 times "x=", this is easy to fix by adding a space in the last sed)
3) Something that always works no matter what:
PATTERN="x="
awk "NF>1{for(i=1;i<NF;i++) printf FS; print \"\"}" \
FS="$PATTERN" your_file

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