Extract filename from multiple lines in unix - sed

I'm trying to extract the name of the file name that has been generated by a Java program. This Java program spits out multiple lines and I know exactly what the format of the file name is going to be. The information text that the Java program is spitting out is as follows:
ABCASJASLEKJASDFALDSF
Generated file YANNANI-0008876_17.xml.
TDSFALSFJLSDJF;
I'm capturing the output in a variable and then applying a sed operator in the following format:
sed -n 's/.*\(YANNANI.\([[:digit:]]\).\([xml]\)*\)/\1/p'
The result set is:
YANNANI-0008876_17.xml.
However, my problem is that want the extraction of the filename to stop at .xml. The last dot should never be extracted.
Is there a way to do this using sed?

Let's look at what your capture group actually captures:
$ grep 'YANNANI.\([[:digit:]]\).\([xml]\)*' infile
Generated file YANNANI-0008876_17.xml.
That's probably not what you intended:
\([[:digit:]]\) captures just a single digit (and the capture group around it doesn't do anything)
\([xml]\)* is "any of x, m or l, 0 or more times", so it matches the empty string (as above – or the line wouldn't match at all!), x, xx, lll, mxxxxxmmmmlxlxmxlmxlm, xml, ...
There is no way the final period is removed because you don't match anything after the capture groups
What would make sense instead:
Match "digits or underscores, 0 or more": [[:digit:]_]*
Match .xml, literally (escape the period): \.xml
Make sure the rest of the line (just the period, in this case) is matched by adding .* after the capture group
So the regex for the string you'd like to extract becomes
$ grep 'YANNANI.[[:digit:]_]*\.xml' infile
Generated file YANNANI-0008876_17.xml.
and to remove everything else on the line using sed, we surround regex with .*\( ... \).*:
$ sed -n 's/.*\(YANNANI.[[:digit:]_]*\.xml\).*/\1/p' infile
YANNANI-0008876_17.xml
This assumes you really meant . after YANNANI (any character).

You can call sed twice: first in printing and then in replacement mode:
sed -n 's/.*\(YANNANI.\([[:digit:]]\).\([xml]\)*\)/\1/p' | sed 's/\.$//g'
the last sed will remove all the last . at the end of all the lines fetched by your first sed
or you can go for a awk solution as you prefer:
awk '/.*YANNANI.[0-9]+.[0-9]+.xml/{print substr($NF,1,length($NF)-1)}'
this will print the last field (and truncate the last char of it using substr) of all the lines that do match your regex.

Related

sed match first word replace full line

I know this should be straight forward but I'm stuck, sorry.
I have two files both contain the same parameters but with different values. I'm trying to read one file line at a time, get the parameter name, use this to match in the second file and replace the whole line with that from file 1.
e.g. rw_2.core.fvbCore.Param.isEnable 1 (FVB_Params)
becomes
rw_2.core.fvbCore.Param.isEnable true (FVB_Boolean)
The lines are not always the same length but I always want to replace the whole line.
The code I have is as follows but it doesn't make the substitutions and I can't work out why not.
while read line; do
ParamName=`awk '{print $1}'`
sed -i 's/$ParamName.*/$line/g' FVB_Params.txt
done < FVB_Boolean.txt
You need your sed command within double quotes if you want those variables to be replaced with their values. You have single quotes, so sed is actually looking for strings with dollar signs to replace with the string '$line', not whatever your shell has in the $line variable.
In short, sed's not seeing the values you want. Switch to double quotes.

How to use sed to isolate only the first part of a file

I'm running Windows and have the GnuWin32 toolkit, which includes sed. Specifically:
C:\TEMP>sed --version
GNU sed version 4.2.1
I have a text file with two sections: A fixed part I want to preserve, and a part that's appended after running a job.
In the file is a unique string that identifies the start of the part that's added, and I'd like to use Gnu sed to isolate only the part of the file that's before the unique string - i.e., so I can append different data to the fixed part each time the job is run.
I know I could keep the fixed portion in a separate file, but that adds complexity and it would be more elegant if I could just reuse the data at the start of the same file.
A long time ago I knew how to set up sed scripts, and I'm sure this can be done with sed, but I've slept since then. :)
Can you please describe how to use sed to display the lines of text in a file up to and not including a specific string?
Example:
line 1 of fixed portion
line 2 of fixed portion
unique string
line 1 of appended portion
line 2 of appended portion
line 3 of appended portion
What I'd like is to see as output:
line 1 of fixed portion
line 2 of fixed portion
I've gotten as far as:
sed -r -n -e "0,/unique string/p"
but that prints the unique string as well.
Thanks in advance.
-Noel
This should work for you:
sed -n '/unique string/q;p' file
It quits processing at unique string. Other lines get printed.
An alternative might be to use a range address like this:
sed -n '1,/unique string/{/unique string/!p}' file
Note that sed includes the range border. We need to exclude unique string from printing.
Furthermore I'm using the -n option which makes sed suppress the output of input lines by default.
One thing, if unique string can contain characters which are also syntax characters in the regex like ...
test*
... sed might not be the right tool for the job any more since it can only match regular expressions but not fixed strings.
In that case awk might be the tool of choice:
awk 'index("*unique string*"){exit}1' file
index("string") returns a non zero value (the position) if the string has been found. We cancel further processing of input lines in that case and don't print that line as well.
The trailing 1 always evaluates to true and makes awk print all the lines until the previous condition applies.

Printing all words that start with "#" using sed in BASH

I have a file with a lot of text, but I want to print only words that contain "#" at the beginning. Ex:
My name is #Laura and I live in #London. Name=#Laura. City=#London
How can I print all words that start with #?.I did this the following and it worked, but I want to do it using sed. I tried several patters, but I cannot make it print anything.
grep -o -E "#\w+" file.txt
Thanks
Use this sed command:
sed 's/[^#]*\(#[^ .]*\)/\1\n/g' file.txt
Explanation: we invoke the substitution command of sed. This has following structure: sed 's/regex/replace/options'. We will search for a regex and replace it using the g option. g makes sure the match is made multiple times per line.
We look for a series of non at chars followed by an # and a number of non-spaces #[^ ]*. We put this last part in a group \(\) and sub it during the replacement \1.
Note that we add a newline at the end of each match, you can also get the output on a single line by omitting the \n.

Using command line to remove text?

I have a huge file that contains lines that follow this format:
New-England-Center-For-Children-L0000392290
Southboro-Housing-Authority-L0000392464
Crew-Star-Inc-L0000391998
Saxony-Ii-Barber-Shop-L0000392491
Test-L0000392334
What I'm trying to do is narrow it down to just this:
New-England-Center-For-Children
Southboro-Housing-Authority
Crew-Star-Inc
Test
Can anyone help with this?
Using GNU awk:
awk -F\- 'NF--' OFS=\- file
New-England-Center-For-Children
Southboro-Housing-Authority
Crew-Star-Inc
Saxony-Ii-Barber-Shop
Test
Set the input and output field separator to -.
NF contains number of fields. Reduce it by 1 to remove the last field.
Using sed:
sed 's/\(.*\)-.*/\1/' file
New-England-Center-For-Children
Southboro-Housing-Authority
Crew-Star-Inc
Saxony-Ii-Barber-Shop
Test
Simple greedy regex to match up to the last hyphen.
In replacement use the captured group and discard the rest.
Version 1 of the Question
The first version of the input was in the form of HTML and parts had to be removed both before and after the desired text:
$ sed -r 's|.*[A-Z]/([a-zA-Z-]+)-L0.*|\1|' input
Special-Restaurant
Eliot-Cleaning
Kennedy-Plumbing
Version 2 of the Question
In the revised question, it is only necessary to remove the text that starts with -L00:
$ sed 's|-L00.*||' input2
New-England-Center-For-Children
Southboro-Housing-Authority
Crew-Star-Inc
Saxony-Ii-Barber-Shop
Test
Both of these commands use a single "substitute" command. The command has the form s|old|new|.
The perl code for this would be: perl -nle'print $1 if(m{-.*?/(.*?-.*?)-})
We can break the Regex down to matching the following:
- for that's between the city and state
.*? match the smallest set of character(s) that makes the Regex work, i.e. the State
/ matches the slash between the State and the data you want
( starts the capture of the data you are interested in
.*?-.*? will match the data you care about
) will close out the capture
- will match the dash before the L####### to give the regex something to match after your data. This will prevent the minimal Regex from matching 0 characters.
Then the print statement will print out what was captured (your data).
awk likes these things:
$ awk -F[/-] -v OFS="-" '{print $(NF-3), $(NF-2)}' file
Special-Restaurant
Eliot-Cleaning
Kennedy-Plumbing
This sets / and - as possible field separators. Based on them, it prints the last_field-3 and last_field-2 separated by the delimiter -. Note that $NF stands for last parameter, hence $(NF-1) is the penultimate, etc.
This sed is also helpful:
$ sed -r 's#.*/(\w*-\w*)-\w*\.\w*</loc>$#\1#' file
Special-Restaurant
Eliot-Cleaning
Kennedy-Plumbing
It selects the block word-word after a slash / and followed with word.word</loc> + end_of_line. Then, it prints back this block.
Update
Based on your new input, this can make it:
$ sed -r 's/(.*)-L\w*$/\1/' file
New-England-Center-For-Children
Southboro-Housing-Authority
Crew-Star-Inc
Saxony-Ii-Barber-Shop
Test
It selects everything up to the block -L + something + end of line, and prints it back.
You can use also another trick:
rev file | cut -d- -f2- | rev
As what you want is every slice of - separated fields, let's get all of them but last one. How? By reversing the line, getting all of them from the 2nd one and then reversing back.
Here's how I'd do it with Perl:
perl -nle 'm{example[.]com/bp/(.*?)/(.*?)-L\d+[.]htm} && print $2' filename
Note: the original question was matching input lines like this:
<loc>http://www.example.com/bp/Lowell-MA/Special-Restaurant-L0000423916.htm</loc>
<loc>http://www.example.com/bp/Houston-TX/Eliot-Cleaning-L0000422797.htm</loc>
<loc>http://www.example.com/bp/New-Orleans-LA/Kennedy-Plumbing-L0000423121.htm</loc>
The -n option tells Perl to loop over every line of the file (but not print them out).
The -l option adds a newline onto the end of every print
The -e 'perl-code' option executes perl-code for each line of input
The pattern:
/regex/ && print
Will only print if the regex matches. If the regex contains capture parentheses you can refer to the first captured section as $1, the second as $2 etc.
If your regex contains slashes, it may be cleaner to use a different regex delimiter ('m' stands for 'match'):
m{regex} && print
If you have a modern Perl, you can use -E to enable modern feature and use say instead of print to print with a newline appended:
perl -nE 'm{example[.]com/bp/(.*?)/(.*?)-L\d+[.]htm} && say $2' filename
This is very concise in Perl
perl -i.bak -lpe's/-[^-]+$//' myfile
Note that this will modify the input file in-place but will keep a backup of the original data in called myfile.bak

Extract CentOS mirror domain names using sed

I'm trying to extract a list of CentOS domain names only from http://mirrorlist.centos.org/?release=6.4&arch=x86_64&repo=os
Truncating prefix "http://" and "ftp://" to the first "/" character only resulting a list of
yum.phx.singlehop.com
mirror.nyi.net
bay.uchicago.edu
centos.mirror.constant.com
mirror.teklinks.com
centos.mirror.netriplex.com
centos.someimage.com
mirror.sanctuaryhost.com
mirrors.cat.pdx.edu
mirrors.tummy.com
I searched stackoverflow for the sed method but I'm still having trouble.
I tried doing this with sed
curl "http://mirrorlist.centos.org/?release=6.4&arch=x86_64&repo=os" | sed '/:\/\//,/\//p'
but doesn't look like it is doing anything. Can you give me some advice?
Here you go:
curl "http://mirrorlist.centos.org/?release=6.4&arch=x86_64&repo=os" | sed -e 's?.*://??' -e 's?/.*??'
Your sed was completely wrong:
/x/,/y/ is a range. It selects multiple lines, from a line matching /x/ until a line matching /y/
The p command prints the selected range
Since all lines match both the start and end pattern you used, you effectively selected all lines. And, since sed echoes the input by default, the p command results in duplicated lines (all lines printed twice).
In my fix:
I used s??? instead of s/// because this way I didn't need to escape all the / in the patterns, so it's a bit more readable this way
I used two expressions with the -e flag:
s?.*://?? matches everything up until :// and replaces it with nothing
s?/.*?? matches everything from / until the end replaces it with nothing
The two expressions are executed in the given order
In modern versions of sed you can omit -e and separate the two expressions with ;. I stick to using -e because it's more portable.