I am looking for two consecutive lines matching a certain pattern, say containing word 'pat' using sed and have noticed that I am able to detect it sometimes with this command:
sed -n 'N; /.*pat.*\n.*pat.*/p'
but this command fails if the line numbers for the duplicates are not of the same parity and I assume it's because we're searching lines 1+2, 3+4, 5+6 etc.. if this is the case, what would be the correct way to do this?
Why does it need to be sed? May I suggest awk?
awk '{/pat/?f++:f=0} f==2' file
If pat is found, increment f with 1
If pat is not found, reset f to 0
If f==2 print the line.
This might work for you (GNU sed):
sed '$!N;/pattern.*\n.*pattern/p;D' file
This keeps 2 lines in the pattern space and prints both of them out if the regexp matches.
Related
I have a plain text file:
line1_text
line2_text
I need to add a number of whitespaces between the two lines.
Adding 10 whitespaces is easy.
But say I need to add 10000 whitespaces, how would I achieve that using sed?
P.S. This is for experimental purposes
There undoubtedly is a sed method to do this but, since sed does not have any natural understanding of arithmetic, it is not a natural choice for this problem. By contrast, awk understands arithmetic and can readily, for example, print an empty line n times for any integer value of n.
As an example, consider this input file:
$ cat infile
line1_text
line2_text
This code will add as many blank lines as you like before any line that contains the string line2_text:
$ awk -v n=5 '/line2_text/{for (i=1;i<=n;i++)print""} 1' infile
line1_text
line2_text
If you want 10,000 blank lines instead of 5, then replace n=5 with n=10000.
How it works
-v n=5
This defines an awk variable n with value 5.
/line2_text/{for (i=1;i<=n;i++)print""}
Every time that a line matches the regex line2_text, then a for loop is performed with prints an empty line n times.
1
This is awk's shorthand for print-the-line and it causes every line from input to be printed to the output.
This might work for you (GNU sed):
sed -r '/line1_text/{x;s/.*/ /;:a;ta;s/ /&\n/10000;tb;s/^[^\n]*/&&/;ta;:b;s/\n.*//;x;G}' file
This appends the hold space to the first line. The hold space is manipulated to hold the required number of spaces by a looping mechanism based on powers of 2. This may produce more than necessary and the remainder are chopped off using a linefeed as a delimiter.
To change spaces to newlines, use:
sed -r '/line1_text/{x;s/.*/ /;:a;ta;s/ /\n&/10000;tb;s/^[^\n]*/&&/;ta;:b;s/\n.*//;s/ /\n/g;x;G}' file
In essence the same can be achieved using this (however it is very slow for large numbers):
sed -r '/line1_text/{x;:a;/ {20}/bb;s/^/ /;ta;:b;x;G}' file
I am in the learning phase of sed and awk commands, trying some complicated logic but couldn't get solution for the below.
File contents:
This is apple,apple.com 443,apple2.com 80,apple3.com 232,
We talk on 1 banana,banana.com 80,banannna.com 23,
take 5 grape,grape5.com 23,
When I try with
$ cat sample.txt | sed -e 's/[[:space:]][^,]*,/,/g'
,apple.com,apple2.com,apple3.com,
,banana.com,banannna.com,
,grape5.com,
is ok but I want to skip this sed for the first comma in each line, so expected output is
This is apple,apple.com,apple2.com,apple3.com,
We talk on 1 banana,banana.com,banannna.com,
take 5 grape,grape5.com,
Any help is appreciated.
If you are using GNU sed, you can do something like
sed -e 's/[[:space:]][^,]*,/,/2g' file
where the 2g specifies something like start the substitution from the 2nd occurrence and g for doing it subsequently to the rest of the occurrences.
The output for the above command.
sed -e 's/[[:space:]][^,]*,/,/2g' file
This is apple,apple.com,apple2.com,apple3.com,
We talk on 1 banana,banana.com,banannna.com,
take 5 grape,grape5.com,
An excerpt from the man page of GNU sed
g
Apply the replacement to all matches to the regexp, not just the first.
number
Only replace the numberth match of the regexp.
awk '{gsub(/[ ]+/," ")gsub(/com [0-9]+/,"com")}1' file
This is apple,apple.com,apple2.com,apple3.com,
We talk on 1 banana,banana.com,banannna.com,
take 5 grape,grape5.com,
The first gsub removes extra space and the next one takes away unwanted numbers between com and comma.
I want to remove last 2 words in the string which is in a file.
I am using this command first to delete the last word. But I couldn't do it. can someone help me
sed 's/\w*$//' <file name>
my strings are like this
Input:
asbc/jahsf/jhdsflk/jsfh/ -0.001 (exam)
I want to remove both numerical value and the one in brackets.
Output:
asbc/jahsf/jhdsflk/jsfh/
Using GNU sed:
$ sed -r 's/([[:space:]]+[-+.()[:alnum:]]+){2}$//' file
asbc/jahsf/jhdsflk/jsfh/
How it works
[[:space:]]+ matches one or more spaces.
[-+.()[:alnum:]]+ matches the 'words' which are allowed to contain any number of plus or minus signs, periods, parens, or any alphanumeric characters.
Note that, when a period is inside square brackets, [.], it is just a period, not a wildcard: it does not need to be escaped.
([[:space:]]+[-+.()[:alnum:]]+) matches one or more spaces followed by a word.
([[:space:]]+[-+.()[:alnum:]]+){2}$ matches two words and the spaces which precede them.
Note the use of character classes like [:space:] and [:alnum:]. Unlike the old-fashioned classes like [a-zA-Z0-9], these classes are unicode safe.
OSX (BSD) sed
The above was tested on GNU sed. For BSD sed, try:
sed -E 's/([[:space:]][[:space:]]*[-+.()[:alnum:][:alnum:]]*){2}$//' file
To remove everything that follows a number with decimal places
This looks for a decimal number with optional sign and removes it, the spaces which precede it, and everything which follows it:
$ sed -r 's/[[:space:]]+[-+]?[[:digit:]]+[.][[:digit:]]+[[:space:]].*//' file
asbc/jahsf/jhdsflk/jsfh/
How it works:
[[:space:]]+ matches one or more spaces
[-+]? matches zero or one signs.
[[:digit:]]+ matches any number of digits.
[.] matches a decimal point (period).
[[:digit:]]+ matches one or more digits following the decimal point.
[[:space:]] matches a space following the number.
.* matches anything which follows.
It looks like there is a tab between what you want to keep and what you want to get rid of. I don't have linux in front of me but try this.
sed 's/\t.*//'
This is assuming your strings are always formatted similarily which is what I take from your comment.
This might work for you (GNU sed):
sed -r 's/\s+\S+\s+\S+\s*$//' file
or if you prefer:
sed -r 's/(\s+\S+){2}\s*$//' file
This matches and removes: one or more whitespaces followed by one or more non-whitespaces twice followed by zero or more whitespaces at the end of the line.
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.
I want to get a list of lines in a batch file which are greater than 120 characters length. For this I thought of using sed. I tried but I was not successful. How can i achieve this ?
Is there any other way to get a list other than using sed ??
Thanks..
Another way to do this using awk:
cat file | awk 'length($0) > 120'
You can use grep and its repetition quantifier:
grep '.\{120\}' script.sh
Using sed, you have some alternatives:
sed -e '/.\{120\}/!d'
sed -e '/^.\{,119\}$/d'
sed -ne '/.\{120\}/p'
The first option matches lines that don't have (at least) 120 characters (the ! after the expression is to execute the command on lines that don't match the pattern before it), and deletes them (ie. doesn't print them).
The second option matches lines that from start (^) to end ($) have a total of characters from zero to 119. These lines are also deleted.
The third option is to use the -n flag, which tells sed to not print lines by default, and only print something if we tell it to. In this case, we match lines that have (at least) 120 characters, and use p to print them.