how to strip IP address of trailing string - regex

Given an IP address 192.168.10.21.somebody.com.br
I need to extract just 192.168.10.21 I tried CUT below, it gives "cut: invalid byte or field list".
cut -d'.' -f-4

$ echo "192.168.10.21.somebody.com.br" | cut -d'.' -f -4
192.168.10.21
works for me!

All three of the following assume you have the domain name stored in a parameter
dom_name=192.168.10.21.somebody.com.br
More efficient than using cut, assuming the first label to remove doesn't start with a number:
echo "${dom_name%%.[[:alpha:]]*}"
If the first label could start with a number, these are still more efficient than cut, but uglier and much longer to type:
# Match one more dot than necessary to shorten the regular expression;
# then trim that dot when echoing
[[ $dn =~ (([0-9]+\.){4}) ]]
echo "${BASH_REMATCH[1]%.}"
or
# Split the string into an array, then output the
# first four fields rejoined by dots.
IFS=. read -a labels <<< "$dom_name"
(IFS=.; echo "${labels[*]:0:4}")

Related

Get multiple values in an xml file

<!-- someotherline -->
<add name="core" connectionString="user id=value1;password=value2;Data Source=datasource1.comapany.com;Database=databasename_compny" />
I need to grab the values in userid , password, source, database. Not all lines are in the same format.My desired result would be (username=value1,password=value2, DataSource=datasource1.comapany.com,Database=databasename_compny)
This regex seems little bit more complicated as it is more complicated. Please, explain your answer if possible.
I realised its better to loop through each line. Code I wrote so far
while read p || [[ -n $p ]]; do
#echo $p
if [[ $p =~ .*connectionString.* ]]; then
echo $p
fi
done <a.config
Now inside the if I have to grab the values.
For this solution I am considering:
Some lines can contain no data
No semi-colon ; is inside the data itself (nor field names)
No equal sign = is inside the data itself (nor field names)
A possible solution for you problem would be:
#!/bin/bash
while read p || [[ -n $p ]]; do
# 1. Only keep what is between the quotes after connectionString=
filteredLine=`echo $p | sed -n -e 's/^.*connectionString="\(.\+\)".*$/\1/p'`;
# 2. Ignore empty lines (that do not contain the expected data)
if [ -z "$filteredLine" ]; then
continue;
fi;
# 3. split each field on a line
oneFieldByLine=`echo $filteredLine | sed -e 's/;/\r\n/g'`;
# 4. For each field
while IFS= read -r field; do
# extract field name + field value
fieldName=`echo $field | sed 's/=.*$//'`;
fieldValue=`echo $field | sed 's/^[^=]*=//' | sed 's/[\r\n]//'`;
# do stuff with it
echo "'$fieldName' => '$fieldValue'";
done < <(printf '%s\n' "$oneFieldByLine")
done <a.xml
Explanations
General sed replacement syntax :
sed 's/a/b/' will replace what matches the regex a by the content of b
Step 1
-n argument tells sed not to output if no match is found. In this case this is useful to ignore useless lines.
^.* - anything at the beginning of the line
connectionString=" - literally connectionString="
\(.\+\)" - capturing group to store anything in before the closing quote "
.*$" - anything until the end of the line
\1 tells sed to replace the whole match with only the capturing group (which contains only the data between the quotes)
p tells sed to print out the replacement
Step 3
Replace ; by \r\n ; it is equivalent to splitting by semi-colon because bash can loop over line breaks
Step 4 - field name
Replaces literal = and the rest of the line with nothing (it removes it)
Step 4 - field value
Replaces all the characters at the beginning that are not = ([^=] matches all but what is after the '^' symbol) until the equal symbol by nothing.
Another sed command removes the line breaks by replacing it with nothing.

Swap Strings within a line in Bash

I'm parsing a document with a bash script and output different parts of it. At one point i need find and reformat text in the form of:
(foo)[X]
[Y]
(bar)[Z]
to something like:
X->foo
Y
Z->bar
Now, I'm able to grep the parts I want with RegEx, but I'm having trouble swapping the two elements in one line and handling the fact that the text in parentheses is optional. Is this even possible with a combination of sed and grep?
Thank You for your time.
You can use sed:
sed -e 's/(\([^)]*\))\[\([^]]*\)]/\2->\1/' -e 's/\[\([^]]*\)]/\1/' file
This works for your given input example:
X->foo
Y
Z->bar
You might need to make the patterns more strict if you have more kinds of input to handle.
You can use awk:
awk -F '[][()]+' '{print (NF>3 ? $3 "->" $2 : $2)}' file
X->foo
Y
Z->bar
You can even do it in bash itself, although it's not pretty.
# Three capture groups:
# 1. The optional paranthesized text
# 2. The contents of the parentheses
# 3. The contents of the square brackets
regex="(\((.*)\))?\[(.*)\]"
while IFS= read -r str; do
[[ "$str" =~ $regex ]]
# If the 2nd array element is not empty, print -> followed by the
# non-empty value.
echo "${BASH_REMATCH[3]}${BASH_REMATCH[2]:+->${BASH_REMATCH[2]}}"
done < file.txt

Get digit from filename immediately preceeding file extension, with other digits in filename

I'm trying to extract the last number before a file extension in a bash script. So the format varies but it'll be some combination of numbers and letters, and the last character will always be a digit. I need to pull those digits and store them in a variable.
The format is generally:
sdflkej10_sdlkei450_sdlekr_1.txt
I want to store just the final digit 1 into a variable.
I'll be using this to loop through a large number of files, and the last number will get into double and triple digits.
So for this file:
kej10_sdlkei450_sdlekr_310.txt
I'd need to return 310.
The number of alphanumeric characters and underscores varies with each file, but the number I want always is immediately before the .txt extension and immediately after an underscore.
I tried:
bname=${f%%.*}
number=$(echo $bname | tr -cd '[[:digit:]]')
but this returns all digits.
If I try
number = $(echo $(bname -2) it changes the number it returns.
The problem i'm having is mostly related to the variability, and the fact that I've been asked to do it in bash. Any help would really be appreciated.
regex='([0-9]+)\.[^.]*$'
[[ $file =~ $regex ]] && number=${BASH_REMATCH[1]}
This uses bash's underappreciated =~ regex operator which stores matches in an array named BASH_REMATCH.
You could do this using parameter substitution
var=kej10_sdlkei450_sdlekr_310.txt
var=${var%.*}
var=${var##*_}
echo $var
310
Use a Series of Bash Shell Expansions
While not the most elegant solution, this one uses a sequence of shell parameter expansions to achieve the desired result without having to define a specific extension. For example, this function uses the length and offset expansions to find the digit after removing filename extensions:
extract_digit() {
local basename=${1%%.*}
echo "${basename:$(( ${#basename} - 1 ))}"
}
Capturing Function Output
You can capture the output in a variable with something like:
$ foo=$(extract_digit sdflkej10_sdlkei450_sdlekr_1.txt)
$ echo $foo
1
Sample Output from Function
$ extract_digit sdflkej10_sdlkei450_sdlekr_1.txt
1
$ extract_digit sdflkej10_sdlkei450_sdlekr_9.txt
9
$ extract_digit sdflkej10_sdlkei450_sdlekr_10.txt
0
This should take care of your situation:
INPUT="some6random7numbers_12345_moreletters_789.txt"
SUBSTRING=`expr match "$INPUT" '.*_\([[:digit:]]*\)'`
echo $SUBSTRING
This will output 789
No need of regex here, you can utilize IFS
var="kej10_sdlkei450_sdlekr_310.txt"
v=$(IFS=[_.] read -ra arr <<< "$var" && echo "${arr[#]:(-2):1}")
echo "$v"
310

Extract a numeric substring and another line as variables from multi-line text file

I'm writing a bash shell script that I hope to ultimately use to automate the naming and 'attachment' of scanned documents to our db. The script OCR's a section of the first page of the pdf and outputs a text file containing three lines; a name, unique id, and a datetime string:
Smith, John
Case #: 234567 ( )
09/04/2013 11:34 AM
What I'd like to do is end up with two seperate strings as variables, "Smith, John" and "234567". I'm looking for help using regex with sed/awk/etc to extract this number. One issue is that the OCR will rarely output strings like:
"Case #2 234567 ( )"
or
"Ca$e # 2234567 ( 7"
So I'm thinking to take the only last 6-digits in the string, since only maybe 1 in 10,000+ of these ever get the last 6-digits read incorrectly. This unique ID is only 6 digits, and is always between 200000-999999. I'm learning regex, but it's slow going. Any help is greatly appreciated.
Edit:
For now I am using:
casename="$(cat test.txt | sed '1!d')"
casenum="$(cat test.txt | sed -n -r 's/.*([0-9]{6}).*/\1/p')"
echo ${casenum} ${casename}
234567 Smith, John
Any input for why this might not be a good way to do it, or what could be improved is (very) welcomed.
you might try this Regex (BRE):
[2-9][0-9]\{5\}\>
You could use the following regex for the second line:
^.*(\d{6})[^\d].*$
Here, the first named sub-group would denote the digits of interest.
For example, using Notepad++,
Orginal text:
Replace options:
Resulting text:
The regex should remain more or less the same across environments. You might need to simply change the way the named subexpression ($1 here) is referenced.
You could probably use something like this untested but somewhat-syntactically-valid snippet:
shopt -s extglob
declare -a cases
for casefile in casefiles/*
do
name=""
while read l
do
if [[ -z "$name" ]]
then
[[ "$l" == #(*, *) ]] && name=$l
elif [[ "$l" == +([0-9]) ]]
then
after=${l#*[2-9][0-9][0-9][0-9][0-9][0-9]}
l=${l%$after}
l=${l#${l%[2-9][0-9][0-9][0-9][0-9][0-9]}}
if [[ "$l" == #([2-9][0-9][0-9][0-9][0-9][0-9]) ]]
then
cases[$l]=$name
fi
name=""
fi
done < $casefile
done
The "hard part" prunes the first 6-digit number in your range and everything after it, then removes what's left (the stuff before the number) from the line. Then it removes the number from the beginning of the string, and removes what's left (the part after the number) from the end. If what's left is a 6-digit number in your range, it uses that as the index and the case name as the value in an array which you can later iterate over.
The rest should be pretty straightforward. :) If this doesn't quite work as expected, I blame the fact that I mostly use ksh, not bash. ;)

In GNU Grep or another standard bash command, is it possible to get a resultset from regex?

Consider the following:
var="text more text and yet more text"
echo $var | egrep "yet more (text)"
It should be possible to get the result of the regex as the string: text
However, I don't see any way to do this in bash with grep or its siblings at the moment.
In perl, php or similar regex engines:
$output = preg_match('/yet more (text)/', 'text more text yet more text');
$output[1] == "text";
Edit: To elaborate why I can't just multiple-regex, in the end I will have a regex with multiple of these (Pictured below) so I need to be able to get all of them. This also eliminates the option of using lookahead/lookbehind (As they are all variable length)
egrep -i "([0-9]+) +$USER +([0-9]+).+?(/tmp/Flash[0-9a-z]+) "
Example input as requested, straight from lsof (Replace $USER with "j" for this input data):
npviewer. 17875 j 11u REG 8,8 59737848 524264 /tmp/FlashXXu8pvMg (deleted)
npviewer. 17875 j 17u REG 8,8 16037387 524273 /tmp/FlashXXIBH29F (deleted)
The end goal is to cp /proc/$var1/fd/$var2 ~/$var3 for every line, which ends up "Downloading" flash files (Flash used to store in /tmp but they drm'd it up)
So far I've got:
#!/bin/bash
regex="([0-9]+) +j +([0-9]+).+?/tmp/(Flash[0-9a-zA-Z]+)"
echo "npviewer. 17875 j 11u REG 8,8 59737848 524264 /tmp/FlashXXYOvS8S (deleted)" |
sed -r -n -e " s%^.*?$regex.*?\$%\1 \2 \3%p " |
while read -a array
do
echo /proc/${array[0]}/fd/${array[1]} ~/${array[2]}
done
It cuts off the first digits of the first value to return, and I'm not familiar enough with sed to see what's wrong.
End result for downloading flash 10.2+ videos (Including, perhaps, encrypted ones):
#!/bin/bash
lsof | grep "/tmp/Flash" | sed -r -n -e " s%^.+? ([0-9]+) +$USER +([0-9]+).+?/tmp/(Flash[0-9a-zA-Z]+).*?\$%\1 \2 \3%p " |
while read -a array
do
cp /proc/${array[0]}/fd/${array[1]} ~/${array[2]}
done
Edit: look at my other answer for a simpler bash-only solution.
So, here the solution using sed to fetch the right groups and split them up. You later still have to use bash to read them. (And in this way it only works if the groups themselves do not contain any spaces - otherwise we had to use another divider character and patch read by setting $IFS to this value.)
#!/bin/bash
USER=j
regex=" ([0-9]+) +$USER +([0-9]+).+(/tmp/Flash[0-9a-zA-Z]+) "
sed -r -n -e " s%^.*$regex.*\$%\1 \2 \3%p " |
while read -a array
do
cp /proc/${array[0]}/fd/${array[1]} ~/${array[2]}
done
Note that I had to adapt your last regex group to allow uppercase letters, and added a space at the beginning to be sure to capture the whole block of numbers. Alternatively here a \b (word limit) would have worked, too.
Ah, I forget mentioning that you should pipe the text to this script, like this:
./grep-result.sh < grep-result-test.txt
(provided your files are named like this). Instead you can add a < grep-result-test after the sed call (before the |), or prepend the line with cat grep-result-test.txt |.
How does it work?
sed -r -n calls sed in extended-regexp-mode, and without printing anything automatically.
-e " s%^.*$regex.*\$%\1 \2 \3%p " gives the sed program, which consists of a single s command.
I'm using % instead of the normal / as parameter separator, since / appears inside the regex and I don't want to escape it.
The regex to search is prefixed by ^.* and suffixed by .*$ to grab the whole line (and avoid printing parts of the rest of the line).
Note that this .* grabs greedy, so we have to insert a space into our regexp to avoid it grabbing the start of the first digit group too.
The replacement text contains of the three parenthesed groups, separated by spaces.
the p flag at the end of the command says to print out the pattern space after replacement. Since we grabbed the whole line, the pattern space consists of only the replacement text.
So, the output of sed for your example input is this:
5 11 /tmp/FlashXXu8pvMg
5 17 /tmp/FlashXXIBH29F
This is much more friendly for reuse, obviously.
Now we pipe this output as input to the while loop.
read -a array reads a line from standard input (which is the output from sed, due to our pipe), splits it into words (at spaces, tabs and newlines), and puts the words into an array variable.
We could also have written read var1 var2 var3 instead (preferably using better variable names), then the first two words would be put to $var1 and $var2, with $var3 getting the rest.
If read succeeded reading a line (i.e. not end-of-file), the body of the loop is executed:
${array[0]} is expanded to the first element of the array and similarly.
When the input ends, the loop ends, too.
This isn't possible using grep or another tool called from a shell prompt/script because a child process can't modify the environment of its parent process. If you're using bash 3.0 or better, then you can use in-process regular expressions. The syntax is perl-ish (=~) and the match groups are available via $BASH_REMATCH[x], where x is the match group.
After creating my sed-solution, I also wanted to try the pure-bash approach suggested by Mark. It works quite fine, for me.
#!/bin/bash
USER=j
regex=" ([0-9]+) +$USER +([0-9]+).+(/tmp/Flash[0-9a-zA-Z]+) "
while read
do
if [[ $REPLY =~ $regex ]]
then
echo cp /proc/${BASH_REMATCH[1]}/fd/${BASH_REMATCH[2]} ~/${BASH_REMATCH[3]}
fi
done
(If you upvote this, you should think about also upvoting Marks answer, since it is essentially his idea.)
The same as before: pipe the text to be filtered to this script.
How does it work?
As said by Mark, the [[ ... ]] special conditional construct supports the binary operator =~, which interprets his right operand (after parameter expansion) as a extended regular expression (just as we want), and matches the left operand against this. (We have again added a space at front to avoid matching only the last digit.)
When the regex matches, the [[ ... ]] returns 0 (= true), and also puts the parts matched by the individual groups (and the whole expression) into the array variable BASH_REMATCH.
Thus, when the regex matches, we enter the then block, and execute the commands there.
Here again ${BASH_REMATCH[1]} is an array-access to an element of the array, which corresponds to the first matched group. ([0] would be the whole string.)
Another note: Both my scripts accept multi-line input and work on every line which matches. Non-matching lines are simply ignored. If you are inputting only one line, you don't need the loop, a simple if read ; then ... or even read && [[ $REPLY =~ $regex ]] && ... would be enough.
echo "$var" | pcregrep -o "(?<=yet more )text"
Well, for your simple example, you can do this:
var="text more text and yet more text"
echo $var | grep -e "yet more text" | grep -o "text"