My file have text in multiple lines in following format
number:characters
example
123:dfgd
3:hjdfg
23423:fdlgj
I want to extract all lines where number part has only one digit using following sed
sed '/^\d:/p' file.txt
But this is printing all the lines. Please point out the issue in the command.
You are pretty close. You can use this sed command:
sed -n '/^[0-9]:/p' file
3:hjdfg
-n is to suppress regular output
[0-9] is to match a digit (\d isn't supported by sed regex even in extended regex mode)
Related
I have two regular expressions:
$ grep -E '\-\- .*$' *.sql
$ sed -E '\-\- .*$' *.sql
(I am trying to grep lines in sql files that have comments and remove lines in sql files that have comments)
The grep command works using this regex; however, the sed returns the following error:
sed: -e expression #1, char 7: unterminated address regex
What am I doing incorrectly with sed?
(The space after the two hyphens is required for sql comments if you are unfamiliar with MySql comments of this type)
You're trying to use:
sed -E '\-\- .*$' *.sql
Here sed command is not correct because you're not really telling sed to do something.
It should be:
sed -n '/-- /p' *.sql
and equivalent grep would be:
grep -- '-- ' *.sql
or even better with a fixed string search:
grep -F -- '-- ' *.sql
Using -- to separate pattern and arguments in grep command.
There is no need to escape - in a regex if it is outside bracket expression (or character class) i.e. [...].
Based on comments below it seems OP's intent is to remove commented section in all *.sql files that start with 2 hyphens.
You may use this sed for that:
sed -i 's/-- .*//g' *.sql
The problem here is not the regex, the problem is that sed requires a command. The equivalent of your grep would be:
sed -n '/\-\- .*$/p'
You suppress output for non-matching lines -n ... you search (wrap your regex in slashes) and you print p (after the last slash).
P.S.: As Anub pointed out, escaping the hyphens - inside the regex is unnecessary.
You are trying to use sed's \cregexpc syntax where with \-<...> you are telling sed the delimiter character you want use is a dash -, but you didn't terminate it where it should be: \-<...>- also add d command to delete those lines.
sed '\-\-\-.*$-d' infile
see man sed about that:
\cregexpc
Match lines matching the regular expression regexp. The c may be any character.
if default / was used this was not required so:
sed '/--.*$/d' infile
or simply:
sed '/^--/d' infile
and more accurately:
sed '/^[[:blank:]]*--/d' infile
I am trying to add 5 blank line spaces in a text file (text.txt) before and after string pattern matches. I used the following to get spaces after the 'string' match which worked for me-
sed '/string/{G;G;G;G;G;}' text.txt
I want to apply the same sed command to obtain 5 blank lines before the 'string' Here I don't want spaces, but rather blank lines before and after them. Any suggestions?
sed -r 's/(^.*)(string)(.*$)/\1\n\n\n\n\n\2\n\n\n\n\n\3/' text.txt
Use -r or -E to allow regular expressions, split likes into three sections and then substitute the line for the first section, 5 new lines, the second section, 5 new lines and then finally the third section.
Use this Perl one-liner:
perl -pe 's/string/\n\n\n\n\n$&\n\n\n\n\n/' text.txt
The Perl one-liner uses these command line flags:
-e : Tells Perl to look for code in-line, instead of in a file.
-p : Loop over the input one line at a time, assigning it to $_ by default. Add print $_ after each loop iteration.
s/PATTERN/REPLACEMENT/ : change PATTERN to REPLACEMENT.
$& : matched pattern.
\n : newline character.
SEE ALSO:
perldoc perlrun: how to execute the Perl interpreter: command line switches
perldoc perlrequick: Perl regular expressions quick start
For a single string match:
$ sed -e '/string/{ s/^/\n\n\n\n\n/; s/$/\n\n\n\n\n/ }' text.txt
For multiple strings, assuming same requirements:
$ sed -E '/(string1|string2|string3)/{ s/^/\n\n\n\n\n/; s/$/\n\n\n\n\n/ }' text.txt
This might work for you:
sed '/string/{G;s/\(string\)\(.*\)\(.\)/\3\3\3\3\3\1\3\3\3\3\3\2/}' file
Match on string, append an empty line, pattern match using the newline to separate the match by 5 lines either side.
And an awk version:
awk '{if(/string1|string2|.../){printf "\n\n\n\n\n%s\n\n\n\n\n",$0}else{print}}' file
so I have a text file with multiple instances looking like this:
word. word or words [something:'else]
I need to replace with a new line the double space after every period followed by a sequence of words and then a "[", like so:
word.\nword or words [something:'else]
I thought about using the sed command in bash with extended regex syntax, but nothing has worked so far... I've tried different variations of this:
sed -E 's/(\.)( )(.*)(.\[)/\1\n\3\4/g' old.txt > new.txt
I'm an absolute beginner at this, so I'm not sure at all about what I'm doing 😳
This might work for you (GNU sed):
sed -E 's/\. ((\w+ )+\[)/\.\n\1/g' file
Replace globally a period followed by two spaces and one or more words space separated followed by an opening square bracket by; a period followed by a newline followed by the matching back reference from the regexp.
Your sed command is almost correct (but contains some redundancies)
sed -E 's/(\.)( )(.*)(.\[)/\1\n\3\4/' old.txt > new.txt
# ^
# You forget terminating the s command
But you don't need to capture everything. A simpler one could be
sed -E 's/\. (.*\[)/.\n\1/' old.txt > new.txt
I tried to remove the unwanted symbols
%H1256
*+E1111
*;E2311
+-'E3211
{E4511
DE4513
so I tried by using this command
sed 's/+E[0-9]/E/g
but it won't remove the blank spaces, and the digits need to be preserved.
expected:
H1256
E1111
E2311
E3211
E4511
E4513
EDIT
Special thanks to https://stackoverflow.com/users/3832970/wiktor-stribiżew my days have been saved by him
sed -n 's/.*\([A-Z][0-9]*\).*/\1/p' file or grep -oE '[A-Z][0-9]+' file
You may use either sed:
sed -n 's/.*\([[:upper:]][[:digit:]]*\).*/\1/p' file
or grep:
grep -oE '[[:upper:]][[:digit:]]+' file
See the online demo
Basically, the patterns match an uppercase letter ([[:upper:]]) followed with digits ([[:digit:]]* matches 0 or more digits in the POSIX BRE sed solution and [[:digit:]]+ matches 1+ digits in an POSIX ERE grep solution).
While sed solution will extract a single value (last one) from each line, grep will extract all values it finds from all lines.
This should do the job:
sed -E 's/^[^[:alnum:]]+//' file
Or if it is only the last 5 characters you need
sed -E 's/.*(.{5})$/\1/' file
I need to find out how to delete up to 10 digits that are at the end of the line in my text file using sed.
For example if I have this:
ajsdlfkjasldf1234567890
asdlkjfalskdjf123456
adsf;lkjasldfkjas123
it should become:
ajsdlfkjasldf
asdlkjfalskdjf
adsf;lkjasldfkjas
can anyone help?
I have this, but its not working:
sed 's/[0-9]{10}$//g'
Have you tried this:
sed 's/[0-9]+$//'
Your command would only match and delete exactly 10 digits at the end of line and only, if you enabled extended regular expressions (-E or -r, depending on your version of sed).
You should try
sed -r 's/[0-9]{1,10}$//'
The following should work:
sed 's/[0-9]\{1,10\}$//' file
Regex syntax in sed requires backslashes before the brackets to use them for repetition, unless you use an extended regex option.
A quick look here suggests you should try this:
$ sed 's/[0-9]\{0,10\}$//g'
{ } should be escaped, unless you switch to extended regex syntax:
$ sed -r 's/[0-9]{0,10}$//g'