I have a file, including some strings and variables, like:
${cat.mouse.dog}
bird://localhost:${xfire.port}/${plfservice.url}
bird://localhost:${xfire.port}/${spkservice.synch.url}
bird://localhost:${xfire.port}/${spkservice.asynch.request.url}
${soabp.protocol}://${hpc.reward113.host}:${hpc.reward113.port}
${configtool.store.folder}/config/hpctemplates.htb
I want to print all the strings between "{}". In some lines there are more than one such string and in this case they should remain in the same line. The output should be:
cat.mouse.dog
xfire.port plfservice.url
xfire.port spkservice.synch.url
xfire.port spkservice.asynch.request.url
soabp.protocol hpc.reward113.host hpc.reward113.port
configtool.store.folder
I tried the following:
sed -n 's/.*{//;s/}.*//p' filename
but it printed only the last occurrence of each line. How can I get all the occurrences, remaining in the same line, as in the original file?
This might work for you (GNU sed):
sed -n 's/${/\n/g;T;s/[^\n]*\n\([^}]*\)}[^\n]*/\1 /g;s/ $//p' file
Replace all ${ by newlines and if there are non then move on as there is nothing to process. If there are newlines then remove non-newline characters to the left and non-newline characters to the right of the next } globally. To finish off remove the extra space introduced in the RHS of the global substitution.
If you're not against awk, you can try the following:
awk -v RS='{|}' -v ORS=' ' '/\n/{printf "\n"} (NR+1)%2' file
The record separator RS is set to either { or }. This splits the wanted pattern from the rest.
The script then displays 1 record out of 2 with the statement (NR+1)%2.
In order to keep the alignment as expected, the output record separator is set to a space ORS=' ' and everytime a newline is encountered this statement /\n/{printf "\n"} inserts one.
Related
I have string like this:
20.07.2010|Berlin|id 100|bd-22.10.94|Marry Scott Robinson|msc#gmail.com
I need to replace whitespaces only between "Marry Scott Robinson" with "|". So to have bd-22.10.94|Marry|Scott|Robinson|
There many of such rows, so problem is in replace whitespace only between "bd-" and vertical line after name.
I'll assume that the name is always on the fifth column :
awk 'BEGIN{FS=OFS="|"}{gsub(/ /,OFS,$5)}1' file
If it is not the case, you can do :
awk 'BEGIN{FS=OFS="|"}{for(i=1;i<=NF;i++){if($i ~ /bd-/){break}};gsub(/ /,OFS,$(i+1))}1' file
Returns :
20.07.2010|Berlin|id 100|bd-22.10.94|Marry|Scott|Robinson|msc#gmail.com
Perl to the rescue!
perl -lne '($before, $change, $after) = /(.*\|bd-.*?\|)(.*?)(\|.*)/;
print $before, $change =~ s/ /|/gr, $after' -- file
-n reads the input line by line, running the code for each line
-l removes newlines from input and adds them to output
the first line populates three variables by values captured from the line. $before contains verything up to the first | after bd-; $change contains what follows up to the next |, and $after contains the rest.
s/ /|/gr replaces spaces by pipes (/g for "all of them") and returns (/r) the result.
This might work for you (GNU sed):
sed 's/[^|]*/\n&\n/5;:a;s/\(\n[^\n ]*\) /\1\|/;ta;s/\n//g' file
Sometimes to fix a problem we must erect scaffolding, then fix the original problem and finally remove the scaffolding.
Here we need to isolate the field by surrounding it by newlines.
Remove the spaces between the newlines by looping until failure.
Finally, remove the scaffolding i.e. the introduced newlines.
Another perl version:
$ perl -F'\|' -ne '$F[4] =~ tr/ /|/; print join("|", #F)' foo.txt
20.07.2010|Berlin|id 100|bd-22.10.94|Marry|Scott|Robinson|msc#gmail.com
Same basic idea as Corentin's first awk example. Split each line into columns based on |, replace spaces in the 5th one with |'s, print the re-joined lines.
I have a requirement to read through all the files and look for <double quotes> (") and copy the whole line to a different file. The challenge is here that to identify the whole line when there is a new character in the line.
The file format is like this - values are separated with delimiter |*| and end with |##|.
In the attached (image), the highlighted in green should go to new file, Logic would be check for " and if it finds read line starting from (line after |##| to until next |##| )
10338|*|BVL-O-G-01020-R4|*||*|BVL|*||*|Y|*|Y|*||*|CFC6E82284990A7AE040800AA5644B19|*|jmorlan|*|2011.12.21 15:52:01|##|
10358|*|BI-MED-CDMA-MCS-90-118-EXAM|*|Exam for 001-MCS-90-118:
Planning, Conducting and Reporting Post Marketing Surveillance "Studies and Safety Reporting from Non Trial Activities |*|GLOBAL_MEDICAL|*||*|Y|*|N|*||*|CFC6E822849A0A7AE040800AA5644B19|*|finke|*|2012.04.30 04:23:27|##|
10342|*|BVL-O-4-01020-R7|*||*|DVL|*||*|Y|*|Y|*||*|RRFC6E82284990A7AE040800AA5644B19|*|sppa|*|2011.12.21 15:52:01|##|
Assuming you mean that the sections between |##| should be considered as newline, next question is does you file contain any real newlines? If not, grep is probably not going to be very efficient as it works on a line-by-line basis. If any real newlines are supposed to be considered part of the text, then definitely, grep is going to be unhappy.
If you really want to do it in 1 go in grep:
grep -Eoz '(^|\|##\|)([^|]|\|[^#]|\|#[^#]|\|##[^|])"([^|]|\|[^#]|\|#[^#]|\|##[^|])(\|##\||$)'
This is looking for any sequence that starts with |##| (or is the start of the file) is followed by some characters, a quote, and some more characters, then ends with |##| (or end of file). By using -z grep will ignore any newlines in the file.
The complex "any characters" ([^|]|\|[^#]|\|#[^#]|\|##[^|])* expression is because grep is greedy. It basically looks for repeating sequences that are not |##|. Perhaps turning off greed is good, but that will depend on the power of the regexp engine in your version of grep.
But much easier, and probably faster, to use sed to break up the records and inject "NULL" line-breaks:
sed 's/\|##\|/\x00/g' | grep -z '"'
This is simply replacing your end of line pattern |##| with the null character, then asking grep to find quote while treating null character as end of line.
This answer provides two solutions a Gnu Awk solution and a POSIX version.
POSIX awk
awk '{r=r ? r "\n" $0 : $0}
/\|##\|$/ { if (r ~ /"/) print r; r=""}' inputfile > outputfile
GNU awk 1
awk 'BEGIN{RS="\\|##\\|\n?";ORS="|##|\n"}/"/' inputfile > outputfile
GNU awk 2
awk 'BEGIN{RS="\\|##\\|\n?"}/"/{printf $0 RT}' inputfile > outputfile
On the sample data provided in the question, all provided solutions give the following output:
10358|*|BI-MED-CDMA-MCS-90-118-EXAM|*|Exam for 001-MCS-90-118:
Planning, Conducting and Reporting Post Marketing Surveillance "Studies and Safety Reporting from Non Trial Activities |*|GLOBAL_MEDICAL|*||*|Y|*|N|*||*|CFC6E822849A0A7AE040800AA5644B19|*|finke|*|2012.04.30 04:23:27|##|
note: It is possible that you are suffering from the Carriage Return problem if the file comes from a Windows machine. Pleas run dos2unix on the file before using it with these tools.
How does this work? (POSIX)
Using a POSIX version of awk we can do
awk '{r=r ? r "\n" $0 : $0}
/\|##\|$/ { if (r ~ /"/) print r; r=""}' inputfile > outputfile
The idea is to build a record r by appending every line to r. If the current line ends with "|##|", then we check if the record r contains a <double quote> ". If this is the case, we print the record r and reset the record r to an empty string. If it does not contain the <double quote>, we just reset it.
How does this work? (GNU)
Using GNU awk you can do this directly using the record separator RS
awk 'BEGIN{RS="\\|##\\|\n?";ORS="|##|\n"}/"/' inputfile > outputfile
The idea here is that the file contains various records. The OP clearly stated that the information of a record is split in fields separated by |*|, but more importantly, the records themselves are separated by |##|. So in the presented example of the OP, the first record is line1 while the second record is spread over line 2 and line 3.
In awk, you can define a record separator by means of the variable RS. In its default state, RS is the <newline> character \n which makes each line a separate record which can be referenced by $0. In POSIX, the record separator can only be a single character which separates the records, while in Gnu awk, this can be a regular expression (see addendum below).
Since the record separator of the OP is the string "|##|" followed all or not by a <newline> character \n, we need to define RS=\\|##\\|\n?. Why so complicated?
the <pipe> | symbol is the OR operation (alternation operator) in a regular expression, so we need to escape it. But since string literals that are used as regular expressions are parsed twice, we also need to escape it twice. So | → \\| (see here)
the \n? is because it seems that the actual record separator is the string "|##|\n", but maybe some records do not have a newline character, especially the last record.
When you print records, using the print statement it automatically appends the output record separator ORS after each line. By default this is again a <newline> character \n. Since the record separator RS is not a part of the record $0 you need to update the value ORS to ORS="|##|\n". This time, not a regex, so you do not need to escape at all.
The statement /"/ is a shorthand for /"/{print $0} which means If the current record $0 contains a <double quote> ", then print the current record $0 followed by the output record separator ORS.
Note: since we actually already use Gnu awk, we can actually reduce the whole thing even further to:
awk 'BEGIN{RS="\\|##\\|\n?"}/"/{printf $0 RT}' inputfile > outputfile
Which makes use of the matched record separator RT that corresponds to the text found by RS. By replacing the print statement by a printf statement, we do not need to ORS anymore and just manually add RT to the record $0.
RS: The input record separator. Its default value is a string containing a single newline character, which means that an input record consists of a single line of text. It can also be the null string, in which case records are separated by runs of blank lines. If it is a regexp, records are separated by matches of the regexp in the input text.
The ability for RS to be a regular expression is a gawk extension. In most other AWK implementations, or if gawk is in compatibility mode (see Options), just the first character of RS’s value is used.
ORS: The output record separator. It is output at the end of every print statement. Its default value is "\n", the newline character.
RT: (GNU AWK specific) The input text that matched the text denoted by RS, the record separator. It is set every time a record is read.
source: GNU AWK manual
I need to align decimal numbers with the "," symbol using only the sed command. The "," should go in the 5th position. For example:
183,7
2346,7
7,999
Should turn into:
183,7
2346,7
7,999
The maximum amount of numbers before the comma is 4. I have tried using this to remove spaces:
sed 's/ //g' input.txt > nospaces.txt
And then I thought about adding spaces depending on the number of digits before the comma, but I don't know how to do this using only sed.
Any help would be appreciated.
Assuming that there is only one number on each line; that there are at most four digits before the ,, and that there is always a ,:
sed 's/[^0-9,]*\([0-9]\+,[0-9]*\).*/ \1/;s/.*\(.....,.*\)/\1/;'
The first s gets rid of everything other than the (first) number on the line, and puts four spaces before it. The second one deletes everything before the fifth character prior to the ,, leaving just enough spaces to right justify the number.
The second s command might mangle input lines which didn't match the first s command. If it is possible that the input contains such lines, you can add a conditional branch to avoid executing the second substitution if the first one failed. With Gnu sed, this is trivial:
sed 's/[^0-9,]*\([0-9]\+,[0-9]*\).*/ \1/;T;s/.*\(.....,.*\)/\1/;'
T jumps to the end of the commands if the previous s failed. Posix standard sed only has a conditional branch on success, so you need to use this circuitous construction:
sed 's/[^0-9,]*\([0-9]\+,[0-9]*\).*/ \1/;ta;b;:a;s/.*\(.....,.*\)/\1/;'
where ta (conditional branch to a on success) is used to skip over a b (unconditional branch to end). :a is the label referred to by the t command.
if you change your mind, here is an awk solution
$ awk -F, 'NF{printf "%5d,%-d\n", $1,$2} !NF' file
183,7
2346,7
7,999
set the delimiter to comma and handle both parts as separate fields
Try with this:
gawk -F, '{ if($0=="") print ; else printf "%5d,%-d\n", $1, $2 }' input.txt
If you are using GNU sed, you could do as below
sed -r 's/([0-9]+),([0-9]+)/printf "%5s,%d" \1 \2/e' input.txt
I have a result file, values separated by ; as below:
137;AJP14028.1_VP35;HLA-A*02:01;MVAKYDFLV;0.79200;0.35000;0.87783;0.99826;0.30;<-E
137;AJP14037.1_VP35;HLA-A*02:01;MVAKYDFLV;0.79200;0.35000;0.87783;0.99826;0.30;<-E
137;AJP14352.1_VP35;HLA-A*02:01;MVAKYDFLV;0.79200;0.35000;0.87783;0.99826;0.30;<-E
137;AJP14846.1_VP35;HLA-A*02:01;MVAKYDFLV;0.79200;0.35000;0.87783;0.99826;0.30;<-E
and I want to change the second value (AJP14028.1_VP35) to only AJP14028, without the ".1_VP35" at the back. So the result will be:
137;AJP14028;HLA-A*02:01;MVAKYDFLV;0.79200;0.35000;0.87783;0.99826;0.30;<-E
137;AJP14037;HLA-A*02:01;MVAKYDFLV;0.79200;0.35000;0.87783;0.99826;0.30;<-E
137;AJP14352;HLA-A*02:01;MVAKYDFLV;0.79200;0.35000;0.87783;0.99826;0.30;<-E
137;AJP14846;HLA-A*02:01;MVAKYDFLV;0.79200;0.35000;0.87783;0.99826;0.30;<-E
Any idea on how to do this? I am trying to solve this using either sed or awk but I am not really familiar with them yet.
With that input, and focusing on the second field, you can use awk:
$ awk 'BEGIN{FS=OFS=";"} {split($2, arr, /\.1/); $2=arr[1]} 1' file
137;AJP14028;HLA-A*02:01;MVAKYDFLV;0.79200;0.35000;0.87783;0.99826;0.30;<-E
137;AJP14037;HLA-A*02:01;MVAKYDFLV;0.79200;0.35000;0.87783;0.99826;0.30;<-E
137;AJP14352;HLA-A*02:01;MVAKYDFLV;0.79200;0.35000;0.87783;0.99826;0.30;<-E
137;AJP14846;HLA-A*02:01;MVAKYDFLV;0.79200;0.35000;0.87783;0.99826;0.30;<-E
Explanation:
BEGIN{FS=OFS=";"} sets FS and OFS to ";". This splits the input on the ; character and set the output field separator to that same character.
{split($2, arr, /\.1/) splits the second field on the pattern of a literal .1 and places the result in an array.
$2=arr[1] is an awk idiom that resets the second field, $2, to the trimmed value. A side effect is the total record, $0 is reset using the output field separator, OFS
1 at the end is another awkism -- print the current record.
If you just have the fixed string .1_VP35 to remove (and you do not care if it is field specific) you can just used sed:
sed 's/\.1_VP35//' file
awk '{sub(/.1_VP35/,"")}1' file
137;AJP14028;HLA-A*02:01;MVAKYDFLV;0.79200;0.35000;0.87783;0.99826;0.30;<-E
137;AJP14037;HLA-A*02:01;MVAKYDFLV;0.79200;0.35000;0.87783;0.99826;0.30;<-E
137;AJP14352;HLA-A*02:01;MVAKYDFLV;0.79200;0.35000;0.87783;0.99826;0.30;<-E
137;AJP14846;HLA-A*02:01;MVAKYDFLV;0.79200;0.35000;0.87783;0.99826;0.30;<-E
sed -r 's/(^[^.]*)(.[^;]*)(.*)/\1\3/g' inputfile
137;AJP14028;HLA-A*02:01;MVAKYDFLV;0.79200;0.35000;0.87783;0.99826;0.30;<-E
137;AJP14037;HLA-A*02:01;MVAKYDFLV;0.79200;0.35000;0.87783;0.99826;0.30;<-E
137;AJP14352;HLA-A*02:01;MVAKYDFLV;0.79200;0.35000;0.87783;0.99826;0.30;<-E
137;AJP14846;HLA-A*02:01;MVAKYDFLV;0.79200;0.35000;0.87783;0.99826;0.30;<-E
Here: back referencing is used to divide the input line into three groups,seprated by `()'. Later they are referred as "\1" and so on.
The first group will match from the start of the line till the first dot.
The second group will match string followed by the first dot till the first semicolon.
The third group will match everything followed by it.
This might work for you (GNU sed):
sed 's/\(;[^.]*\)[^;]*/\1/' file
Make a back reference of the first ; and everything thereafter which is not a . and then remove everything from thereon which is not a ;.
I found several related questions, but none of them fits what I need, and since I am a real beginner, I can't figure it out.
I have a text file with entries like this, separated by a blank line:
example entry &with/ special characters
next line (any characters)
next %*entry
more words
I would like the output merge the lines, put a comma between, and delete empty lines. I.e., the example should look like this:
example entry &with/ special characters, next line (any characters)
next %*entry, more words
I would prefer sed, because I know it a little bit, but am also happy about any other solution on the linux command line.
Improved per Kent's elegant suggestion:
awk 'BEGIN{RS="";FS="\n";OFS=","}{$1=$1}7' file
which allows any number of lines per block, rather than the 2 rigid lines per block I had. Thank you, Kent. Note: The 7 is Kent's trademark... any non-zero expression will cause awk to print the entire record, and he likes 7.
You can do this with awk:
awk 'BEGIN{RS="";FS="\n";OFS=","}{print $1,$2}' file
That sets the record separator to blank lines, the field separator to newlines and the output field separator to a comma.
Output:
example entry &with/ special characters,next line (any characters)
next %*entry,more words
Simple sed command,
sed ':a;N;$!ba;s/\n/, /g;s/, , /\n/g' file
:a;N;$!ba;s/\n/, /g -> According to this answer, this code replaces all the new lines with ,(comma and space).
So After running only the first command, the output would be
example entry &with/ special characters, next line (any characters), , next %*entry, more words
s/, , /\n/g - > Replacing , , with new line in the above output will give you the desired result.
example entry &with/ special characters, next line (any characters)
next %*entry, more words
This might work for you (GNU sed):
sed ':a;$!N;/.\n./s/\n/, /;ta;/^[^\n]/P;D' file
Append the next line to the current line and if there are characters either side of the newline substitute the newline with a comma and a space and then repeat. Eventually an empty line or the end-of-file will be reached, then only print the next line if it is not empty.
Another version but a little more sofisticated (allowing for white space in the empty line) would be:
sed ':a;$!N;/^\s*$/M!s/\n/, /;ta;/\`\s*$/M!P;D' file
sed -n '1h;1!H
$ {x
s/\([^[:cntrl:]]\)\n\([^[:cntrl:]]\)/\1, \2/g
s/\(\n\)\n\{1,\}/\1/g
p
}' YourFile
change all after loading file in buffer. Could be done "on the fly" while reading the file and based on empty line or not.
use -e on GNU sed