Hello I have an output from grep
cc.log.1:<Operation adaptorMethod="search" adaptorName="Search" status="Success"/>
cc.log.12:<Operation adaptorMethod="getOrderId" adaptorName="PersistenceAdaptor" status="Success"/>
I need to cut off file name from beginning and leave only tag
I tried several variants of awk '/<Operation/,/>$/' sed -n '/^<Ope/,/>$/p' and etc
But have no success. Could anyone help me?
You don't need sed, awk or cut you just want the -h option of grep:
-h, --no-filename
Suppress the prefixing of file names on output. This is the default when
there is only one file (or only standard input) to search.
Related
I have tried to scan through the other posts in stack overflow for this, but couldn't get my code work, hence I am posting a new question.
Below is the content of file temp.
<?xml version="1.0" encoding="UTF-8"?>
<env:Envelope xmlns:env="http://schemas.xmlsoap.org/soap/envelope/<env:Body><dp:response xmlns:dp="http://www.datapower.com/schemas/management"><dp:timestamp>2015-01-
22T13:38:04Z</dp:timestamp><dp:file name="temporary://test.txt">XJzLXJlc3VsdHMtYWN0aW9uX18i</dp:file><dp:file name="temporary://test1.txt">lc3VsdHMtYWN0aW9uX18i</dp:file></dp:response></env:Body></env:Envelope>
This file contains the base64 encoded contents of two files names test.txt and test1.txt. I want to extract the base64 encoded content of each file to seperate files test.txt and text1.txt respectively.
To achieve this, I have to remove the xml tags around the base64 contents. I am trying below commands to achieve this. However, it is not working as expected.
sed -n '/test.txt"\>/,/\<\/dp:file\>/p' temp | perl -p -e 's#<dp:file name="temporary://test.txt">##g'|perl -p -e 's#</dp:file>##g' > test.txt
sed -n '/test1.txt"\>/,/\<\/dp:file\>/p' temp | perl -p -e 's#<dp:file name="temporary://test1.txt">##g'|perl -p -e 's#</dp:file></dp:response></env:Body></env:Envelope>##g' > test1.txt
Below command:
sed -n '/test.txt"\>/,/\<\/dp:file\>/p' temp | perl -p -e 's#<dp:file name="temporary://test.txt">##g'|perl -p -e 's#</dp:file>##g'
produces output:
XJzLXJlc3VsdHMtYWN0aW9uX18i
<dp:file name="temporary://test1.txt">lc3VsdHMtYWN0aW9uX18i</dp:response> </env:Body></env:Envelope>`
Howeveer, in the output I am expecting only first line XJzLXJlc3VsdHMtYWN0aW9uX18i. Where I am commiting mistake?
When i run below command, I am getting expected output:
sed -n '/test1.txt"\>/,/\<\/dp:file\>/p' temp | perl -p -e 's#<dp:file name="temporary://test1.txt">##g'|perl -p -e 's#</dp:file></dp:response></env:Body></env:Envelope>##g'
It produces below string
lc3VsdHMtYWN0aW9uX18i
I can then easily route this to test1.txt file.
UPDATE
I have edited the question by updating the source file content. The source file doesn't contain any newline character. The current solution will not work in that case, I have tried it and failed. wc -l temp must output to 1.
OS: solaris 10
Shell: bash
sed -n 's_<dp:file name="\([^"]*\)">\([^<]*\).*_\1 -> \2_p' temp
I add \1 -> to show link from file name to content but for content only, just remove this part
posix version so on GNU sed use --posix
assuming that base64 encoded contents is on the same line as the tag around (and not spread on several lines, that need some modification in this case)
Thanks to JID for full explaination below
How it works
sed -n
The -n means no printing so unless explicitly told to print, then there will be no output from sed
's_
This is to substitute the following regex using _ to separate regex from the replacement.
<dp:file name=
Regular text
"\([^"]*\)"
The brackets are a capture group and must be escaped unless the -r option is used( -r is not available on posix). Everything inside the brackets is captured. [^"]* means 0 or more occurrences of any character that is not a quote. So really this just captures anything between the two quotes.
>\([^<]*\)<
Again uses the capture group this time to capture everything between the > and <
.*
Everything else on the line
_\1 -> \2
This is the replacement, so replace everything in the regex before with the first capture group then a -> and then the second capture group.
_p
Means print the line
Resources
http://unixhelp.ed.ac.uk/CGI/man-cgi?sed
http://www.grymoire.com/Unix/Sed.html
/usr/xpg4/bin/sed works well here.
/usr/bin/sed is not working as expected in case if the file contains just 1 line.
below command works for a file containing only single line.
/usr/xpg4/bin/sed -n 's_<env:Envelope\(.*\)<dp:file name="temporary://BackUpDir/backupmanifest.xml">\([^>]*\)</dp:file>\(.*\)_\2_p' securebackup.xml 2>/dev/null
Without 2>/dev/null this sed command outputs the warning sed: Missing newline at end of file.
This because of the below reason:
Solaris default sed ignores the last line not to break existing scripts because a line was required to be terminated by a new line in the original Unix implementation.
GNU sed has a more relaxed behavior and the POSIX implementation accept the fact but outputs a warning.
I have a big file which is composed of alot of different lines which only have one commen keyword, storaged.
PROC:storage123:0702:2108:0,1,2,3,4,5:storage:vers:storaged:storage123:Storage
123:storage123:-R /etc/orc/storage123 -e emr123#localhost -p Xxx::
PROC:storageabc:0606:2108:0,1,2,3,4,5:storage:vers:storaged:storageabc:Storage
abc:storageabc: -e emabc#localhost -R /etc/orc/storageabc -p 654::
What i need to do is grep for the path that can be found on all storaged keywords that comes after -R. But I only want the path, nothing after that. -R can be found on different places so there is no pattern to it.
I created one espressionen which seemed to work, but I think I made it much for complex (and not 100% sure to match) than it should have to be.
[root:~/scripts/] <conf.txt grep -o 'R *[^ ]*' | grep -o '[^ ]*$' | sed 's/.*R\///'
/etc/orc/storage123
/etc/orc/storagerabc
The espression also is hard to implement in a bash script so something simpler would be great. I need these paths in the script later on.
Cheers
Your attempt is nice, but you can simplify it by using a look-behind:
$ grep -Po '(?<=-R )[^ ]*' file
/etc/orc/storage123
/etc/orc/storageabc
Basically it looks for the string -R (note the space) and from that, it prints everything up to a space.
$ sed 's/.*-R \([^ ]*\).*/\1/' file
/etc/orc/storage123
/etc/orc/storageabc
i have several files in which i want to replace a certain word with the name of the file itself..
for example i have 2 files named test1.txt and test2.txt
both files are equal and look like
bla1,bla2,temp
bla2,bla3,temp
with the sed i want to replace the word temp with the name of the file itself
so after the sed operation i have 2 different files
test1.txt , which looks like :
bla1,bla2,test1
bla2,bla3,test1
test2.txt, which looks like :
bla1,bla2,test2
bla2,bla3,test2
so my question ... how do i use the actual name of the input file itself as part of the replace command?
sed "s/temp/ ??filename??/ ??? " *.txt
thanks for your suggestions
I'm not sure you can reference the filename using sed although I could be wrong. You would probably use a shell hack. A better aproach to substitute all occurrences of temp with the filename would be the following awk script:
$ awk '{gsub(/temp/,FILENAME)}1' file
use awk, awk has FILENAME variable:
awk '{sub(/temp/,FILENAME)}7' yourfile
awk 'BEGIN{FS=OFS=","} {$NF=FILENAME}1' file
The difference between this and the sub() solutions is that this will work even if the word "temp" exists elsewhere in your file, e.g. if "bla1" contains the word "temperature".
If you need to strip ".txt" from the file name as it appears from your posted desired output, tweak it to:
awk 'BEGIN{FS=OFS=","} {t=FILENAME; sub(/\.txt$/,"",t); $NF=t}1' file
You can probably edit FILENAME itself but I find it best not to mess with the builtin variables if you don't have to.
You could do it with a little bit of bash to help you out, if that's available.
find . -name "test*.txt" -type f | awk -F '/' '{print $2;}' | while read file; do sed -i "s|temp|$file|" ./$file; done
That's a kind of hacky adaptation of a script I have to do something similar. It can undoubtedly be shortened.
no sed internal variable for the file name so you need some previous batch command for a generic process
for FileName in MyFileShellFilter
do
cat <> ${FileName} | sed "s|,temp$|,${FileName}|"
done
just be carrefull with file name used, they normaly don't have \ but could have & that are s// special meaning. I use | as separator to allow / in file name but for this reason, no unescaped | are allowed in file name (normaly not)
with xargs:
printf "%s\n" *.txt | xargs -I FILE -L 1 sed 's/temp/FILE/' FILE
The filename cannot have: newlines, slashes, ampersand, single quote.
I've got a file which contains lot of strings like below input.
Need to extract the below output and process it further.
Input:
History={ExecAt=[2013-05-03 03:00:20,2013-05-03 03:00:23,2013-05-03 03:00:26],MId=["msgId3","msgId4","msgId5"]};
Output should be:
MId=["msgId3","msgId4","msgId5"]
using (sed 's/^.*,MId=/MId/') command i got the output like MId=["msgId3","msgId4","msgId5"]};
but still wanted the exact output (need to remove last 2 special chars }; here).
This works for me:
sed 's/.*\(MId=.*\)\}.*/\1/'
If your grep supports the -o option, you can use it rather than sed:
grep -o 'MId=\[[^]]\+\]'
Using the same regex in sed works fine, just remove anything before and after:
sed -e 's/.*\(MId=\[[^]]\+\]\).*/\1/'
I have a CSV. I want to edit the 35th field of the CSV and write the change back to the 35th field. This is what I am doing on bash:
awk -F "," '{print $35}' test.csv | sed -i 's/^0/+91/g'
so, I am pulling the 35th entry using awk and then replacing the "0" in the starting position in the string with "+91". This one works perfet and I get desired output on the console.
Now I want this new entry to get written in the file. I am thinking of sed's "in -place" replacement feature but this fetuare needs and input file. In above command, I cannot provide input file because my primary command is awk and sed is taking the input from awk.
Thanks.
You should choose one of the two tools. As for sed, it can be done as follows:
sed -ri 's/^(([^,]*,){34})0([^,]*)/\1+91\3/' test.csv
Not sure about awk, but #shellter's comment might help with that.
The in-place feature of sed is misnamed, as it does not edit the file in place. Instead, it creates a new file with the same name. eg:
$ echo foo > foo
$ ln -f foo bar
$ ls -i foo bar # These are the same file
797325 bar 797325 foo
$ echo new-text > foo # Changes bar
$ cat bar
new-text
$ printf '/new/s//newer\nw\nq\n' | ed foo # Edit foo "in-place"; changes bar
9
newer-text
11
$ cat bar
newer-text
$ ls -i foo bar # Still the same file
797325 bar 797325 foo
$ sed -i s/new/newer/ foo # Does not edit in-place; creates a new file
$ ls -i foo bar
797325 bar 792722 foo
Since sed is not actually editing the file in place, but writing a new file and then renaming it to the old file, you might as well do the same.
awk ... test.csv | sed ... > test.csv.1 && mv test.csv.1 test.csv
There is the misperception that using sed -i somehow avoids the creation of the temporary file. It does not. It just hides the fact from you. Sometimes abstraction is a good thing, but other times it is unnecessary obfuscation. In the case of sed -i, it is the latter. The shell is really good at file manipulation. Use it as intended. If you do need to edit a file in place, don't use the streaming version of ed; just use ed
So, it turned out there are numerous ways to do it. I got it working with sed as below:
sed -i 's/0\([0-9]\{10\}\)/\+91\1/g' test.csv
But this is little tricky as it will edit any entry which matches the criteria. however in my case, It is working fine.
Similar implementation of above logic in perl:
perl -p -i -e 's/\b0(\d{10})\b/\+91$1/g;' test.csv
Again, same caveat as mentioned above.
More precise way of doing it as shown by Lev Levitsky because it will operate specifically on the 35th field
sed -ri 's/^(([^,]*,){34})0([^,]*)/\1+91\3/g' test.csv
For more complex situations, I will have to consider using any of the csv modules of perl.
Thanks everyone for your time and input. I surely know more about sed/awk after reading your replies.
This might work for you:
sed -i 's/[^,]*/+91/35' test.csv
EDIT:
To replace the leading zero in the 35th field:
sed 'h;s/[^,]*/\n&/35;/\n0/!{x;b};s//+91/' test.csv
or more simply:
|sed 's/^\(\([^,]*,\)\{34\}\)0/\1+91/' test.csv
If you have moreutils installed, you can simply use the sponge tool:
awk -F "," '{print $35}' test.csv | sed -i 's/^0/+91/g' | sponge test.csv
sponge soaks up the input, closes the input pipe (stdin) and, only then, opens and writes to the test.csv file.
As of 2015, moreutils is available in package repositories of several major Linux distributions, such as Arch Linux, Debian and Ubuntu.
Another perl solution to edit the 35th field in-place:
perl -i -F, -lane '$F[34] =~ s/^0/+91/; print join ",",#F' test.csv
These command-line options are used:
-i edit the file in-place
-n loop around every line of the input file
-l removes newlines before processing, and adds them back in afterwards
-a autosplit mode – split input lines into the #F array. Defaults to splitting on whitespace.
-e execute the perl code
-F autosplit modifier, in this case splits on ,
#F is the array of words in each line, indexed starting with 0
$F[34] is the 35 element of the array
s/^0/+91/ does the substitution