I'm trying to extract the size (in kb) from a file. Trying to do so as follows:
textA=$(du a)
sizeA=$(expr match "$textA" '\(^[^\s]*\)')
textB=$(du b)
sizeB=$(expr match "$textB" '\(^[^\s]*\)')
echo $textA
echo $sizeA
echo $textB
echo $sizeB
[[ $sizeA == $sizeB ]] && echo "eq"
But this just prints in console textA and textB. Both are like:
30745 a
Can someone please explain why is not the regex matching? I've tried to test the regex against the text in many sites, just to make sure, and it appears to capture the correct text.
I've also tried changing it to:
'^\([^\s]*\)'
But this way it will capture all the text. Any thoughts?
My expr match does not understand \s or other extended regexps. Try '\([0-9]*\)' instead.
But as others mentioned already, using regexp for getting "the first word" is a little overkill. I'd use du s | { read a b; echo $a; }, but you could also use the awk version or solutions using cut.
Not a direct answer, but I would do it like this:
sizeA=$(du a | awk '{print $1}')
size=$(wc -c < file)
If you want to use du, I would use the bash builtin read:
read size filename < <(du file)
Note that you can't say du file | read size filename because in bash, components of a pipeline are executed in subshells, so the variables will disappear when the subshell exits.
Do not parse the output of du, if available you can e.g. use stat to get the size of a file in bytes:
sizeA=$(stat -c%s "${fileA}")
Related
row1=$('+00 00:30:07.880000')
rowX=$('row1 | tr -dc '0-9')
I basically want to filter out all the special characters and space.
I wish to have a output as follows.
echo $'row1' = 003007.880000
You don't need regular expressions or external commands like tr for this. Bash's built-in parameter expansion can do it:
row1='+00 00:30:07.880000'
row1=${row1//[^0-9.]/}
echo "row1=$row1"
outputs row1=00003007.880000.
The output has two leading zeros that are not in the output suggested in the question. Maybe there's an unstated requirement to remove prefixes delimited by spaces. If that is the case, possible code is:
row1='+00 00:30:07.880000'
row1=${row1##* }
row1=${row1//[^0-9.]/}
echo "row1=$row1"
That outputs row1=003007.880000.
See How do I do string manipulations in bash? for explanations of ${row1//[^0-9.]/} and ${row1##* }.
This is the easiest way to do that :
$ echo '+00 00:30:07.880000' | tr -dc '[0-9].'
00003007880000
Regards!
I am trying to extract a part of the filename - everything before the date and suffix. I am not sure the best way to do it in bashscript. Regex?
The names are part of the filename. I am trying to store it in a shellscript variable. The prefixes will not contain strange characters. The suffix will be the same. The files are stored in a directory - I will use loop to extract the portion of the filename for each file.
Expected input files:
EXAMPLE_FILE_2017-09-12.out
EXAMPLE_FILE_2_2017-10-12.out
Expected Extract:
EXAMPLE_FILE
EXAMPLE_FILE_2
Attempt:
filename=$(basename "$file")
folder=sed '^s/_[^_]*$//)' $filename
echo 'Filename:' $filename
echo 'Foldername:' $folder
$ cat file.txt
EXAMPLE_FILE_2017-09-12.out
EXAMPLE_FILE_2_2017-10-12.out
$
$ cat file.txt | sed 's/_[0-9]*-[0-9]*-[0-9]*\.out$//'
EXAMPLE_FILE
EXAMPLE_FILE_2
$
No need for useless use of cat, expensive forks and pipes. The shell can cut strings just fine:
$ file=EXAMPLE_FILE_2_2017-10-12.out
$ echo ${file%%_????-??-??.out}
EXAMPLE_FILE_2
Read all about how to use the %%, %, ## and # operators in your friendly shell manual.
Bash itself has regex capability so you do not need to run a utility. Example:
for fn in *.out; do
[[ $fn =~ ^(.*)_[[:digit:]]{4}-[[:digit:]]{2}-[[:digit:]]{2} ]]
cap="${BASH_REMATCH[1]}"
printf "%s => %s\n" "$fn" "$cap"
done
With the example files, output is:
EXAMPLE_FILE_2017-09-12.out => EXAMPLE_FILE
EXAMPLE_FILE_2_2017-10-12.out => EXAMPLE_FILE_2
Using Bash itself will be faster, more efficient than spawning sed, awk, etc for each file name.
Of course in use, you would want to test for a successful match:
for fn in *.out; do
if [[ $fn =~ ^(.*)_[[:digit:]]{4}-[[:digit:]]{2}-[[:digit:]]{2} ]]; then
cap="${BASH_REMATCH[1]}"
printf "%s => %s\n" "$fn" "$cap"
else
echo "$fn no match"
fi
done
As a side note, you can use Bash parameter expansion rather than a regex if you only need to trim the string after the last _ in the file name:
for fn in *.out; do
cap="${fn%_*}"
printf "%s => %s\n" "$fn" "$cap"
done
And then test $cap against $fn. If they are equal, the parameter expansion did not trim the file name after _ because it was not present.
The regex allows a test that a date-like string \d\d\d\d-\d\d-\d\d is after the _. Up to you which you need.
Code
See this code in use here
^\w+(?=_)
Results
Input
EXAMPLE_FILE_2017-09-12.out
EXAMPLE_FILE_2_2017-10-12.out
Output
EXAMPLE_FILE
EXAMPLE_FILE_2
Explanation
^ Assert position at start of line
\w+ Match any word character (a-zA-Z0-9_) between 1 and unlimited times
(?=_) Positive lookahead ensuring what follows is an underscore _ character
Simply with sed:
sed 's/_[^_]*$//' file
The output:
EXAMPLE_FILE
EXAMPLE_FILE_2
----------
In case of iterating through the list of files with extension .out - bash solution:
for f in *.out; do echo "${f%_*}"; done
awk -F_ 'NF-=1' OFS=_ file
EXAMPLE_FILE
EXAMPLE_FILE_2
Could you please try awk solution too, which will take care of all the .out files, note this has ben written and tested in GNU awk.
awk --re-interval 'FNR==1{if(val){close(val)};split(FILENAME, array,"_[0-9]{4}-[0-9]{2}-[0-9]{2}");print array[1];val=FILENAME;nextfile}' *.out
Also my awk version is old so I am using --re-interval, if you have latest version of awk you may need not to use it then.
Explanation and Non-one liner fom of solution: Adding a non-one liner form of solution too here with explanation.
awk --re-interval '##Using --re-interval for supporting ERE in my OLD awk version, if OP has new version of awk it could be removed.
FNR==1{ ##Checking here condition that when very first line of any Input_file is being read then do following actions.
if(val){ ##Checking here if variable named val value is NOT NULL then do following.
close(val) ##close the Input_file named which is stored in variable val, so that we will NOT face problem of TOO MANY FILES OPENED, so it will be like one file read close it in background then.
};
split(FILENAME, array,"_[0-9]{4}-[0-9]{2}-[0-9]{2}");##Splitting FILENAME(which will have Input_file name in it) into array named array only, whose separator is a 4 digits-2 digits- then 2 digits, actually this will take care of YYYY-MM-DD format in Input_file(s) and it will be easier for us to get the file name part.
print array[1]; ##Printing array 1st element here.
val=FILENAME; ##Storing FILENAME variable value which will have current Input_file name in it to variable named val, so that we could close it in background.
nextfile ##nextfile as it name suggests it will skip all the lines in current line and jump onto the next file to save some cpu cycles of our system.
}
' *.out ##Mentioning all *.out Input_file(s) here.
I have a bunch of daily printer logs in CSV format and I'm writing a script to keep track of how much paper is being used and save the info to a database, but I've come across a small problem
Essentially, some of the document names in the logs include commas in them (which are all enclosed within double quotes), and since it's in comma separated format, my code is messing up and pushing everything one column to the right for certain records.
From what I've been reading, it seems like the best way to go about fixing this would be using awk or sed, but I'm unsure which is the best option for my situation, and how exactly I'm supposed to implement it.
Here's a sample of my input data:
2015-03-23 08:50:22,Jogn.Doe,1,1,Ineo 4000p,"MicrosoftWordDocument1",COMSYRWS14,A4,PCL6,,,NOT DUPLEX,GRAYSCALE,35kb,
And here's what I have so far:
#!/bin/bash
#Get today's file name
yearprefix="20"
currentdate=$(date +"%m-%d-%y");
year=${currentdate:6};
year="$yearprefix$year"
month=${currentdate:0:2};
day=${currentdate:3:2};
filename="papercut-print-log-$year-$month-$day.csv"
echo "The filename is: $filename"
# Remove commas in between quotes.
#Loop through CSV file
OLDIFS=$IFS
IFS=,
[ ! -f $filename ] && { echo "$Input file not found"; exit 99; }
while read time user pages copies printer document client size pcl blank1 blank2 duplex greyscale filesize blank3
do
#Remove headers
if [ "$user" != "" ] && [ "$user" != "User" ]
then
#Remove any file name with an apostrophe
if [[ "$document" =~ "'" ]];
then
document="REDACTED"; # Lazy. Need to figure out a proper solution later.
fi
echo "$time"
#Save results to database
mysql -u username -p -h localhost -e "USE printerusage; INSERT INTO printerlogs (time, username, pages, copies, printer, document, client, size, pcl, duplex, greyscale, filesize) VALUES ('$time', '$user', '$pages', '$copies', '$printer', '$document', '$client', '$size', '$pcl', '$duplex', '$greyscale', '$filesize');"
fi
done < $filename
IFS=$OLDIFS
Which option is more suitable for this task? Will I have to create a second temporary file to get this done?
Thanks in advance!
As I wrote in another answer:
Rather than interfere with what is evidently source data, i.e. the stuff inside the quotes, you might consider replacing the field-separator commas (with say |) instead:
s/,([^,"]*|"[^"]*")(?=(,|$))/|$1/g
And then splitting on | (assuming none of your data has | in it).
Is it possible to write a regular expression that matches a particular pattern and then does a replace with a part of the pattern
There is probably an easier way using sed alone, but this should work. Loop on the file, for each line match the parentheses with grep -o then replace the commas in the line with spaces (or whatever it is you would like to use to get rid of the commas - if you want to preserve the data you can use a non printable and explode it back to commas afterward).
i=1 && IFS=$(echo -en "\n\b") && for a in $(< test.txt); do
var="${a}"
for b in $(sed -n ${i}p test.txt | grep -o '"[^"]*"'); do
repl="$(sed "s/,/ /g" <<< "${b}")"
var="$(sed "s#${b}#${repl}#" <<< "${var}")"
done
let i+=1
echo "${var}"
done
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"
How can I find the index of a substring which matches a regular expression on solaris10?
Assuming that what you want is to find the location of the first match of a wildcard in a string using bash, the following bash function returns just that, or empty if the wildcard doesn't match:
function match_index()
{
local pattern=$1
local string=$2
local result=${string/${pattern}*/}
[ ${#result} = ${#string} ] || echo ${#result}
}
For example:
$ echo $(match_index "a[0-9][0-9]" "This is a a123 test")
10
If you want to allow full-blown regular expressions instead of just wildcards, replace the "local result=" line with
local result=$(echo "$string" | sed 's/'"$pattern"'.*$//')
but then you're exposed to the usual shell quoting issues.
The goto options for me are bash, awk and perl. I'm not sure what you're trying to do, but any of the three would likely work well. For example:
f=somestring
string=$(expr match "$f" '.*\(expression\).*')
echo $string
You tagged the question as bash, so I'm going to assume you're asking how to do this in a bash script. Unfortunately, the built-in regular expression matching doesn't save string indices. However, if you're asking this in order to extract the match substring, you're in luck:
if [[ "$var" =~ "$regex" ]]; then
n=${#BASH_REMATCH[*]}
while [[ $i -lt $n ]]
do
echo "capture[$i]: ${BASH_REMATCH[$i]}"
let i++
done
fi
This snippet will output in turn all of the submatches. The first one (index 0) will be the entire match.
You might like your awk options better, though. There's a function match which gives you the index you want. Documentation can be found here. It'll also store the length of the match in RLENGTH, if you need that. To implement this in a bash script, you could do something like:
match_index=$(echo "$var_to_search" | \
awk '{
where = match($0, '"$regex_to_find"')
if (where)
print where
else
print -1
}')
There are a lot of ways to deal with passing the variables in to awk. This combination of piping output and directly embedding one into the awk one-liner is fairly common. You can also give awk variable values with the -v option (see man awk).
Obviously you can modify this to get the length, the match string, whatever it is you need. You can capture multiple things into an array variable if necessary:
match_data=($( ... awk '{ ... print where,RLENGTH,match_string ... }'))
If you use bash 4.x you can source the oobash. A string lib written in bash with oo-style:
http://sourceforge.net/projects/oobash/
String is the constructor function:
String a abcda
a.indexOf a
0
a.lastIndexOf a
4
a.indexOf da
3
There are many "methods" more to work with strings in your scripts:
-base64Decode -base64Encode -capitalize -center
-charAt -concat -contains -count
-endsWith -equals -equalsIgnoreCase -reverse
-hashCode -indexOf -isAlnum -isAlpha
-isAscii -isDigit -isEmpty -isHexDigit
-isLowerCase -isSpace -isPrintable -isUpperCase
-isVisible -lastIndexOf -length -matches
-replaceAll -replaceFirst -startsWith -substring
-swapCase -toLowerCase -toString -toUpperCase
-trim -zfill