I need to delete "" in file
"CITFFUSKD-E0"
I have tried sed 's/\"//.
Result is:
CITFFUSKD-E0"
How I can delete both ?
Also I need to delete everything behind first word but input can be this one:
"CITFFUSKD-E0"
"CITFFUSKD_E0"
"CITFFUSKD E0"
Result I want it:
CITFFUSKD
You may use
sed 's/"//g' file | sed 's/[^[:alnum:]].*//' > newfile
Or, contract the two sed commands into one sed call as #Wiimm suggests:
sed 's/"//g;s/[^[:alnum:]].*//' file > newfile
If you want to replace inline, see sed edit file in place.
Explanation:
sed 's/"//g' file - removes all " chars from the file
sed 's/[^[:alnum:]].*//' > newfile - also removes all chars from a line starting from the first non-alphanumeric char and saves the result into a newfile.
Could you please try following.
awk 'match($0,/[a-zA-Z]+[^a-zA-Z]*/){val=substr($0,RSTART,RLENGTH);gsub(/[^a-zA-Z]+/,"",val);print val}' Input_file
delete everything behind first word
sed 's/^"\([[:alpha:]]*\)[^[:alpha:]]*.*/\1/'
Match the first ". Then match a sequence of alphabetic characters. Match until you find non-alphabetic character ^[:alpha:]. Then match the rest. Substitute it all for \1 - it is a backreference for the part inside \( ... \), ie. the first word.
I need delete two “ ” with sed command
Remove all possible ":
sed 's/"//g'
Extract the string between ":
sed 's/"\([^"]*\)"/\1/'
Remove everything except alphanumeric characters (numbers + a-z + a-Z, ie. [0-9a-zA-z]):
sed 's/[^[:alnum:]]//g'
This should do all in one go, remove the ", print the first part:
awk -F\" '{split($2,a,"-| |_");print a[1]}' file
CITFFUSKD
CITFFUSKD
CITFFUSKD
When you have 1 line, you can use
grep -Eo "(\w)*" file | head -1
For normal files (starting with a double quote on each line)
, try this
tr -c [^[:alnum:]] '"' < file | cut -d'"' -f2
Many legitimate ways to solve this.
I favor using what you know about your data to simplify solutions -- this is usually an option. If everything in your file follows the same pattern, you can simply extract the first set of capitalized letters encountered:
sed 's/"\([A-Z]\+\).*$/\1/' file
awk '{gsub(/^.|....$/,"")}NR==1' file
CITFFUSKD
Related
I want to delete three words with a special character on a line such as
Input:
\cf4 \cb6 1749,1789 \cb3 \
Output:
1749,1789
I have tried a couple sed and grep statements but so far none have worked, mainly due to the character \.
My unsuccessful attempt:
sed -i 's/ [.\c ] //g' inputfile.ext >output file.ext
Awk accepts a regex Field Separator (in this case, comma or space):
$ awk -F'[ ,]' '$0 = $3 "." $4' <<< '\cf4 \cb6 1749,1789 \cb3 \'
1749.1789
-F'[ ,]' - Use a single character from the set space/comma as Field Separator
$0 = $3 "." $4 - If we can set the entire line $0 to Field 3 $4 followed by a literal period "." followed by Field 4 $4, do the default behavior (print entire line)
Replace <<< 'input' with file if every line of that file has the same delimeters (spaces/comma) and number of fields. If your input file is more complex than the sample you shared, please edit your question to show actual input.
The backslash is a special meta-character that confuses bash.
We treat it like any other meta-character, by escaping it, with--you guessed it--a backslash!
But first, we need to grep this pattern out of our file
grep '\\... \\... [0-9]+,[0-9]+ \\... \\' our_file # Close enough!
Now, just sed out those pesky backslashes
| sed -e 's/\\//g' # Don't forget the g, otherwise it'll only strip out 1 backlash
Now, finally, sed out the clusters of 2 alpha followed by a number and a space!
| sed -e 's/[a-z][a-z][0-9] //g'
And, finally....
grep '\\... \\... [0-9]+,[0-9]+ \\... \\' our_file | sed -e 's/\\//g' | sed -e 's/[a-z][a-z][0-9] //g'
Output:
1749,1789
My guess is you are having trouble because you have backslashes in input and can't figure out how to get backslashes into your regex. Since backslashes are escape characters to shell and regex you end up having to type four backslashes to get one into your regex.
Ben Van Camp already posted an answer that uses single quotes to make the escaping a little easier; however I shall now post an answer that simply avoids the problem altogether.
grep -o '[0-9]*,[0-9]*' | tr , .
Locks on to the comma and selects the digits on either side and outputs the number. Alternately if comma is not guaranteed we can do it this way:
egrep -o ' [0-9,]*|^[0-9,]*' | tr , . | tr -d ' '
Both of these assume there's only one usable number per line.
$ awk '{sub(/,/,".",$3); print $3}' file
1749.1789
$ sed 's/\([^ ]* \)\{2\}\([^ ]*\).*/\2/; s/,/./' file
1749.1789
I am having trouble figuring out how to grep the characters between two single quotes .
I have this in a file
version: '8.x-1.0-alpha1'
and I like to have the output like this (the version numbers can be various):
8.x-1.0-alpha1
I wrote the following but it does not work:
cat myfile.txt | grep -e 'version' | sed 's/.*\?'\(.*?\)'.*//g'
Thank you for your help.
Addition:
I used the sed command sed -n "s#version:\s*'\(.*\)'#\1#p"
I also like to remove 8.x- which I edited to sed -n "s#version:\s*'8.x-\(.*\)'#\1#p".
This command only works on linux and it does not work on MAC. How to change this command to make it works on MAC?
sed -n "s#version:\s*'8.x-\(.*\)'#\1#p"
If you just want to have that information from the file, and only that you can quickly do:
awk -F"'" '/version/{print $2}' file
Example:
$ echo "version: '8.x-1.0-alpha1'" | awk -F"'" '/version/{print $2}'
8.x-1.0-alpha1
How does this work?
An awk program is a series of pattern-action pairs, written as:
condition { action }
condition { action }
...
where condition is typically an expression and action a series of commands.
-F "'": Here we tell awk to define the field separator FS to be a <single quote> '. This means the all lines will be split in fields $1, $2, ... ,$NF and between each field there is a '. We can now reference these fields by using $1 for the first field, $2 for the second ... etc and this till $NF where NF is the total number of fields per line.
/version/{print $2}: This is the condition-action pair.
condition: /version/:: The condition reads: If a substring in the current record/line matches the regular expression /version/ then do action. Here, this is simply translated as if the current line contains a substring version
action: {print $2}:: If the previous condition is satisfied, then print the second field. In this case, the second field would be what the OP requests.
There are now several things that can be done.
Improve the condition to be /^version :/ && NF==3 which reads _If the current line starts with the substring version : and the current line has 3 fields then do action
If you only want the first occurance, you can tell the system to exit immediately after the find by updating the action to {print $2; exit}
I'd use GNU grep with pcre regexes:
grep -oP "version: '\\K.*(?=')" file
where we are looking for "version: '" and then the \K directive will forget what it just saw, leaving .*(?=') to match up to the last single quote.
Try something like this: sed -n "s#version:\s*'\(.*\)'#\1#p" myfile.txt. This avoids the redundant cat and grep by finding the "version" line and extracting the contents between the single quotes.
Explanation:
the -n flag tells sed not to print lines automatically. We then use the p command at the end of our sed pattern to explicitly print when we've found the version line.
Search for pattern: version:\s*'\(.*\)'
version:\s* Match "version:" followed by any amount of whitespace
'\(.*\)' Match a single ', then capture everything until the next '
Replace with: \1; This is the first (and only) capture group above, containing contents between single quotes.
When your only want to look at he quotes, you can use cut.
grep -e 'version' myfile.txt | cut -d "'" -f2
grep can almost do this alone:
grep -o "'.*'" file.txt
But this may also print lines you don't want to: it will print all lines with 2 single quotes (') in them. And the output still has the single quotes (') around it:
'8.x-1.0-alpha1'
But sed alone can do it properly:
sed -rn "s/^version: +'([^']+)'.*/\1/p" file.txt
I have a file with a format like this:
First Last UID
First Middle Last UID
Basically, some names have middle names (and sometimes more than one middle name). I just want a file that only as UIDs.
Is there a sed or awk command I can run that removes everything before the last space?
awk
Print the last field of each line using awk.
The last field is indexed using the NF variable which contains the number of fields for each line. We index it using a dollar sign, the resulting one-liner is easy.
awk '{ print $NF }' file
rs, cat & tail
Another way is to transpose the content of the file, then grab the last line and transpose again (this is fairly easy to see).
The resulting pipe is:
cat file | rs -T | tail -n1 | rs -T
cut & rev
Using cut and rev we could also achieve this goal by reversing the lines, cutting the first field and then reverse it again.
rev file | cut -d ' ' -f1 | rev
sed
Using sed we simply remove all chars until a space is found with the regex ^.* [^ ]*$. This regex means match the beginning of the line ^, followed by any sequence of chars .* and a space . The rest is a sequence of non spaces [^ ]* until the end of the line $. The sed one-liner is:
sed 's/^.* \([^ ]*\)$/\1/' file
Where we capture the last part (in between \( and \)) and sub it back in for the entire line. \1 means the first group caught, which is the last field.
Notes
As Ed Norton cleverly pointed out we could simply not catch the group and remove the former part of the regex. This can be as easily achieved as
sed 's/.* //' file
Which is remarkably less complicated and more elegant.
For more information see man sed and man awk.
Using grep:
$ grep -o '[^[:blank:]]*$' file
UID
UID
-o tells grep to print only the matching part. The regex [^[:blank:]]*$ matches the last word on the line.
Say I have txt file with characters as follows:
abcd|123|kds|Name|Place|Phone
ldkdsd|323|jkds|Name1|Place1|Phone1
I want to remove all the characters in each line that exist within first 3 occurences of | character in each line. I want my output as:
Name|Place|Phone
Name1|Place1|Phone1
Could anyone help me figure this out? How can I achieve this using sed?
This would be a typical task for cut
cut -d'|' -f4- file
output:
Name|Place|Phone
Name1|Place1|Phone1
the -f4- means you want from the forth field till the end. Adjust the 4 if you have a different requirement.
You could try the below sed commad,
$ sed -r 's/^(\s*)[^|]*\|[^|]*\|[^|]*\|/\1/g' file
Name|Place|Phone
Name1|Place1|Phone1
^(\s*) captures all the spaces which are at the start.
[^|]*\|[^|]*\|[^|]*\| Matches upto the third |. So this abcd|123|kds| will be matched.
All the matched characters are replaced by the chars which are present inside the first captured group.
This might work for you (GNU sed):
sed 's/^\([^|]*|\)\{3\}//' file
or more readably:
sed -r 's/^([^|]*\|){3}//' file
sed 's/\(\([^|]*|\)\{3\}\)//' YourFile
this is a posix version, on GNU sed force --posix due to the use of | that is interpreted as "OR" and not in posix version.
Explaination
Replace the 3 first occurence (\{3\}) of [ any charcater but | followed by | (\([^|]*|\)) ] by nothing (// that is an empty pattern)
You can print the last 3 fields:
awk '{print $(NF-2),$(NF-1),$NF}' FS=\| OFS=\| file
Name|Place|Phone
Name1|Place1|Phone1
I have a file as show below
1.2.3.4.ask
sanma.nam.sam
c.d.b.test
I want to remove the last field from each line, the delimiter is . and the number of fields are not constant.
Can anybody help me with an awk or sed to find out the solution. I can't use perl here.
Both these sed and awk solutions work independent of the number of fields.
Using sed:
$ sed -r 's/(.*)\..*/\1/' file
1.2.3.4
sanma.nam
c.d.b
Note: -r is the flag for extended regexp, it could be -E so check with man sed. If your version of sed doesn't have a flag for this then just escape the brackets:
sed 's/\(.*\)\..*/\1/' file
1.2.3.4
sanma.nam
c.d.b
The sed solution is doing a greedy match up to the last . and capturing everything before it, it replaces the whole line with only the matched part (n-1 fields). Use the -i option if you want the changes to be stored back to the files.
Using awk:
$ awk 'BEGIN{FS=OFS="."}{NF--; print}' file
1.2.3.4
sanma.nam
c.d.b
The awk solution just simply prints n-1 fields, to store the changes back to the file use redirection:
$ awk 'BEGIN{FS=OFS="."}{NF--; print}' file > tmp && mv tmp file
Reverse, cut, reverse back.
rev file | cut -d. -f2- | rev >newfile
Or, replace from last dot to end with nothing:
sed 's/\.[^.]*$//' file >newfile
The regex [^.] matches one character which is not dot (or newline). You need to exclude the dot because the repetition operator * is "greedy"; it will select the leftmost, longest possible match.
With cut on the reversed string
cat youFile | rev |cut -d "." -f 2- | rev
If you want to keep the "." use below:
awk '{gsub(/[^\.]*$/,"");print}' your_file