I have the following format for filenames: filename_1234.svg
How can I retrieve the numbers preceded by an underscore and followed by a dot. There can be between one to four numbers before the .svg
I have tried:
width=${fileName//[^0-9]/}
but if the fileName contains a number as well, it will return all numbers in the filename, e.g.
file6name_1234.svg
I found solutions for two underscores (and splitting it into an array), but I am looking for a way to check for the underscore as well as the dot.
You can use simple parameter expansion with substring removal to simply trim from the right up to, and including, the '.', then trim from the left up to, and including, the '_', leaving the number you desire, e.g.
$ width=filename_1234.svg; val="${width%.*}"; val="${val##*_}"; echo $val
1234
note: # trims from left to first-occurrence while ## trims to last-occurrence. % and %% work the same way from the right.
Explained:
width=filename_1234.svg - width holds your filename
val="${width%.*}" - val holds filename_1234
val="${val##*_}" - finally val holds 1234
Of course, there is no need to use a temporary value like val if your intent is that width should hold the width. I just used a temp to protect against changing the original contents of width. If you want the resulting number in width, just replace val with width everywhere above and operate directly on width.
note 2: using shell capabilities like parameter expansion prevents creating a separate subshell and spawning a separate process that occurs when using a utility like sed, grep or awk (or anything that isn't part of the shell for that matter).
Try the following code :
filename="filename_6_1234.svg"
if [[ "$filename" =~ ^(.*)_([^.]*)\..*$ ]];
then
echo "${BASH_REMATCH[0]}" #will display 'filename_6_1234.svg'
echo "${BASH_REMATCH[1]}" #will display 'filename_6'
echo "${BASH_REMATCH[2]}" #will display '1234'
fi
Explanation :
=~ : bash operator for regex comparison
^(.*)_([^.])\..*$ : we look for any character, followed by an underscore, followed by any character, followed by a dot and an extension. We create 2 capture groups, one for before the last underscore, one for after
BASH_REMATCH : array containing the captured groups
Some more way
[akshay#localhost tmp]$ filename=file1b2aname_1234.svg
[akshay#localhost tmp]$ after=${filename##*_}
[akshay#localhost tmp]$ echo ${after//[^0-9]}
1234
Using awk
[akshay#localhost tmp]$ awk -F'[_.]' '{print $2}' <<< "$filename"
1234
I would use
sed 's!_! !g' | awk '{print "_" $NF}'
to get from filename_1234.svg to _1234.svg then
sed 's!svg!!g'
to get rid of the extension.
If you set IFS, you can use Bash's build-in read.
This splits the filename by underscores and dots and stores the result in the array a.
IFS='_.' read -a a <<<'file1b2aname_1234.svg'
And this takes the second last element from the array.
echo ${a[-2]}
There's a solution using cut:
name="file6name_1234.svg"
num=$(echo "$name" | cut -d '_' -f 2 | cut -d '.' -f 1)
echo "$num"
-d is for specifying a delimiter.
-f refers to the desired field.
I don't know anything about performance but it's simple to understand and simple to maintain.
Related
I have a path ./test/test1 and I need to extract the test1 part.
I can do that with
cut -d '/' -f 3
But I may also have a path like ./test/test1/test1a in which case I need to extract the test1a part.
I can do this in a similar manner by switching 2, 3, 4 to suit my needs.
But how can I achieve this if I have a list which contains some paths.
./test/test1
./test/test1/test1a/
./test/test1/test1a/example
How can I always make sure I extract the last part of the string after the last / delimiter? How do I start cutting from the last string up till the delimiter?
EDIT: Expected output:
test1
test1a
example
You can easily cut after the last delimiter, using awk, as you can see here:
cat conf.conf.txt | awk -F "/" '{ print $NF}'
(For your information: NF in awk stands for "Number of Fields".)
However, as the second line ends with a slash, the second result will be empty, as you see here:
test1
example
Is that what you want?
path=./foo/bar/baz
basename "$path"
# or pure shell:
echo "${path##*/}"
Both return baz. The counterpart, dirname, returns ./foo/bar.
It's not entirely clear what you mean by a "list". Perhaps you have the paths in an array, or just a space separated string. In either case, you can use basename, but the way you will use it depends on the data. If you have a space separated string, you can just use:
$ cat a.sh
#!/bin/sh
list='./test/test1
./test/test1/test1a/
./test/test1/test1a/example'
basename -a $list
$ ./a.sh
test1
test1a
example
That form will fail if there are characters in IFS in any of the names. If you have the names in an array, it is slightly easier to deal with that issue:
#!/bin/sh
list=('./test/with space/test1'
./test/test1/test1a/
./test/test1/test1a/example)
basename -a "${list[#]}"
Clean one-liner solution :
<<<"${test2}" mawk -F/ '$!_=$-_=$(NF-=_==$NF)'
test1
test1a
example
Tested and confirmed working on mawk-1.3.4, mawk 1.996, macos nawk, and gawk 5.1.1,
including invocation flags of -c/-P/-t/-S
—- The 4Chan Teller
I want to extract version string (1_4_5) from my-app-1_4_5.img and then convert into dot version (1.4.5) without filename. Version string will have three (1_4_5) or four (1_4_5_7) segments.
Have this one liner working ls my-app-1_4_5.img | cut -d'-' -f 3 | cut -d'.' -f 1 | tr _ .
Would like to know if there is any better way rather than piping output from cut.
Here's an attempt with parameter expansion. I'm assuming you have a wildcard pattern you want to loop over.
for file in *-*.img; do
base=${file%.img}
ver=${base##*-}
echo "${ver//_/.}"
done
The construct ${var%pattern} returns the variable var with any suffix matching pattern trimmed off. Similarly, ${var#pattern} trims any prefix which matches pattern. In both cases, doubling the operator switches to trimming the longest possible match instead of the shortest. (These are POSIX-compatible pattenr expansion, i.e. not strictly Bash only.) The construct ${var/pattern/replacement} replaces the first match in var on pattern with replacement; doubling the first slash causes every match to be replaced. (This is Bash only.)
You can do it with sed:
sed -E "s/.*([0-9]+)_([0-9]+)_([0-9]+).*/\1.\2.\3/" <<< my-app-1_4_5.img
Assuming the version number will always be between the last dash and the file extension, you can use something like this in pure Bash:
name="file-name-x-1_2_3_4_5.ext"
version=${name##*-}
version=${version%%.ext}
version=${version//_/.}
echo $version
The code above will result in:
1.2.3.4.5
For a complete explanation about the brace expansions used above, please take a look at Bash Reference Manual: 3.5.1 Brace Expansion.
Remove everything but 0 to 9, _ and newline and then replace all _ with .:
echo "my-app-1_4_5.img" | tr -cd '0-9_\n' | tr '_' '.'
Output:
1.4.5
With bash and a regex:
echo "my-app-1_4_5.img" | while IFS= read -r line; do [[ "$line" =~ [^0-9]([0-9_]+)[^0-9] ]] && echo "${BASH_REMATCH[1]//_/.}"; done
Output:
1.4.5
A slightly shorter variant
name=my-app-1_4_5.img
vers=${name//[!0-9_]}
$ echo ${vers//_/.}
1.4.5
I have a string like this:
/var/cpanel/users/joebloggs:DNS9=domain.example
I need to extract the username (joebloggs) from this string and store it in a variable.
The format of the string will always be the same with exception of joebloggs and domain.example so I am thinking the string can be split twice using cut?
The first split would split by : and we would store the first part in a variable to pass to the second split function.
The second split would split by / and store the last word (joebloggs) into a variable
I know how to do this in PHP using arrays and splits but I am a bit lost in bash.
To extract joebloggs from this string in bash using parameter expansion without any extra processes...
MYVAR="/var/cpanel/users/joebloggs:DNS9=domain.example"
NAME=${MYVAR%:*} # retain the part before the colon
NAME=${NAME##*/} # retain the part after the last slash
echo $NAME
Doesn't depend on joebloggs being at a particular depth in the path.
Summary
An overview of a few parameter expansion modes, for reference...
${MYVAR#pattern} # delete shortest match of pattern from the beginning
${MYVAR##pattern} # delete longest match of pattern from the beginning
${MYVAR%pattern} # delete shortest match of pattern from the end
${MYVAR%%pattern} # delete longest match of pattern from the end
So # means match from the beginning (think of a comment line) and % means from the end. One instance means shortest and two instances means longest.
You can get substrings based on position using numbers:
${MYVAR:3} # Remove the first three chars (leaving 4..end)
${MYVAR::3} # Return the first three characters
${MYVAR:3:5} # The next five characters after removing the first 3 (chars 4-9)
You can also replace particular strings or patterns using:
${MYVAR/search/replace}
The pattern is in the same format as file-name matching, so * (any characters) is common, often followed by a particular symbol like / or .
Examples:
Given a variable like
MYVAR="users/joebloggs/domain.example"
Remove the path leaving file name (all characters up to a slash):
echo ${MYVAR##*/}
domain.example
Remove the file name, leaving the path (delete shortest match after last /):
echo ${MYVAR%/*}
users/joebloggs
Get just the file extension (remove all before last period):
echo ${MYVAR##*.}
example
NOTE: To do two operations, you can't combine them, but have to assign to an intermediate variable. So to get the file name without path or extension:
NAME=${MYVAR##*/} # remove part before last slash
echo ${NAME%.*} # from the new var remove the part after the last period
domain
Define a function like this:
getUserName() {
echo $1 | cut -d : -f 1 | xargs basename
}
And pass the string as a parameter:
userName=$(getUserName "/var/cpanel/users/joebloggs:DNS9=domain.example")
echo $userName
What about sed? That will work in a single command:
sed 's#.*/\([^:]*\).*#\1#' <<<$string
The # are being used for regex dividers instead of / since the string has / in it.
.*/ grabs the string up to the last backslash.
\( .. \) marks a capture group. This is \([^:]*\).
The [^:] says any character _except a colon, and the * means zero or more.
.* means the rest of the line.
\1 means substitute what was found in the first (and only) capture group. This is the name.
Here's the breakdown matching the string with the regular expression:
/var/cpanel/users/ joebloggs :DNS9=domain.example joebloggs
sed 's#.*/ \([^:]*\) .* #\1 #'
Using a single Awk:
... | awk -F '[/:]' '{print $5}'
That is, using as field separator either / or :, the username is always in field 5.
To store it in a variable:
username=$(... | awk -F '[/:]' '{print $5}')
A more flexible implementation with sed that doesn't require username to be field 5:
... | sed -e s/:.*// -e s?.*/??
That is, delete everything from : and beyond, and then delete everything up until the last /. sed is probably faster too than awk, so this alternative is definitely better.
Using a single sed
echo "/var/cpanel/users/joebloggs:DNS9=domain.example" | sed 's/.*\/\(.*\):.*/\1/'
I like to chain together awk using different delimitators set with the -F argument. First, split the string on /users/ and then on :
txt="/var/cpanel/users/joebloggs:DNS9=domain.com"
echo $txt | awk -F"/users/" '{print$2}' | awk -F: '{print $1}'
$2 gives the text after the delim, $1 the text before it.
I know I'm a little late to the party and there's already good answers, but here's my method of doing something like this.
DIR="/var/cpanel/users/joebloggs:DNS9=domain.example"
echo ${DIR} | rev | cut -d'/' -f 1 | rev | cut -d':' -f1
I had this
string="dontcare noone &11111-&1111-&C00 noone"
and I had to extract the substring &11111-&1111-&C00
from the first & to the first blank
I've tried some index and some sed without any luck.
Someone has some great advice?
You can use bash's string manipulation capabilities:
string="dontcare noone &11111-&1111-&C00 noone"
# remove everything up to the first "&"
string="&${string#*&}"
# remove everything from the end to the earliest blank
string="${string%% *}"
# ta da!
echo $string
&11111-&1111-&C00
$ echo "dontcare noone &11111-&1111-&C00 noone" | grep -o '&[^ ]*'
&11111-&1111-&C00
Bash regular expressions will work as well:
[[ $string =~ (&[[:alnum:]]+-?)+ ]]; echo ${BASH_REMATCH[0]}
The regex matches one or more groups, where a group is an ampersand followed by one or more letters/numbers and an optional final hyphen. If the match is successful (i.e., the [[ ]] command has exist status 0), then the first element of the array BASH_REMATCH contains the text from string that matched the regex.
Use grep.
echo "dontcare noone &11111-&1111-&C00 noone" > somefile.txt
grep somefile.txt -e "&" | cut -d " " > someotherfile.txt
grep will basically "cat" the output and search for the pattern, instantiated with -e , and pull the string with &
The > will push the output into a file that you already have, or simply name there, and it'll create it.
You can then overwrite the somefile.txt anytime you want, or append strings.
To be honest, there's a multitude of ways to do this; grep is one of the better options when searching for specific strings, but even that being said, you can already tell in this question there's many, many ways to grep things.
Pipes, | ; and's, && , and many other options, will provide you with ways to combine grep with other commands to get the exact output you are searching for.
I also see you want it to the first "blank space", which I have just added to the top code. Using "cut" and a delimeter of " ", you can extract only data after spaces. Using cut's field settings, it's possible to switch what's being pulled.
Okay so what I am trying to figure out is how do I count the number of periods in a string and then cut everything up to that point but minus 2. Meaning like this:
string="aaa.bbb.ccc.ddd.google.com"
number_of_periods="5"
number_of_periods=`expr $number_of_periods-2`
string=`echo $string | cut -d"." -f$number_of_periods`
echo $string
result: "aaa.bbb.ccc.ddd"
The way that I was thinking of doing it was sending the string to a text file and then just greping for the number of times like this:
grep -c "." infile
The reason I don't want to do that is because I want to avoid creating another text file for I do not have permission to do so. It would also be simpler for the code I am trying to build right now.
EDIT
I don't think I made it clear but I want to make finding the number of periods more dynamic because the address I will be looking at will change as the script moves forward.
If you don't need to count the dots, but just remove the penultimate dot and everything afterwards, you can use Bash's built-in string manuipulation.
${string%substring}
Deletes shortest match of $substring from back of $string.
Example:
$ string="aaa.bbb.ccc.ddd.google.com"
$ echo ${string%.*.*}
aaa.bbb.ccc.ddd
Nice and simple and no need for sed, awk or cut!
What about this:
echo "aaa.bbb.ccc.ddd.google.com"|awk 'BEGIN{FS=OFS="."}{NF=NF-2}1'
(further shortened by helpful comment from #steve)
gives:
aaa.bbb.ccc.ddd
The awk command:
awk 'BEGIN{FS=OFS="."}{NF=NF-2}1'
works by separating the input line into fields (FS) by ., then joining them as output (OFS) with ., but the number of fields (NF) has been reduced by 2. The final 1 in the command is responsible for the print.
This will reduce a given input line by eliminating the last two period separated items.
This approach is "shell-agnostic" :)
Perhaps this will help:
#!/bin/sh
input="aaa.bbb.ccc.ddd.google.com"
number_of_fields=$(echo $input | tr "." "\n" | wc -l)
interesting_fields=$(($number_of_fields-2))
echo $input | cut -d. -f-${interesting_fields}
grep -o "\." <<<"aaa.bbb.ccc.ddd.google.com" | wc -l
5