This question already has answers here:
Extract number from a line with awk/sed
(2 answers)
Closed 7 years ago.
I have a cookie file,I just want to extract the number before the .jpg extension using grep command ..how can i do it
USERSTAFFPHOTO 09480106177557.jpg
I want to copy 09480106177557
echo "USERSTAFFPHOTO 09480106177557.jpg" | grep -oE '[0-9]{1,}'
You should chain 2 grep commands, the first one to get the file name along with the .jpg extension and the second one to get only the number from the file name:
grep -oE '[0-9]+\.jpg' | grep -oE '[0-9]+'
Otherwise, you would get all the numbers that exist in the file with or without the .jpg extension.
Why restrict yourself to a particular tool? Do you want the number, or do you want to use grep? If the format is fixed, I would do this with tr:
echo USERSTAFFPHOTO 09480106177557.jpg | tr -cd [0-9]
If this is one line in a file and you need to filter it out, try awk:
awk '/USERSTAFFPHOTO/{print $2}' FS=' |\\.' input
Related
This question already has answers here:
How can I remove the extension of a filename in a shell script?
(15 answers)
Closed 11 months ago.
I have a filename with the format yyyymmdd.txt. How can I output only yyyymmdd without the .txt extension?
Example
20220414.txt (before output)
20220414 (after the output)
basename has an option to remove a suffix:
basename -s .txt 20220414.txt
gives:
20220414
Or, if your filename is stored in a variable, bash can help:
a=20220414.txt
echo ${a%.*}
gives:
20220414
You can user awk with flag -F to specify the separator . and then print the first part with $1
echo "20220414.txt" | awk -F "." ' {print $1}'
output
20220414
grep doesn't manipulate anything, it shows what you have in a file. So, you can't modify that file using grep, but you can modify what it shows, using the -o switch, as you can see here:
Prompt> echo "20220414.txt" | grep -o "[0-9]*"
20220414
The [0-9]* means "a list of integers, going from character '0' to character '9'.
This question already has an answer here:
Using regex in Grep for Windows command line
(1 answer)
Closed 2 years ago.
Trying to find all occurrences of a word in a range of different .txt files in a directory.
I'm looking for commands that will give me an exact count if possible.
so far I have tried:
$ grep -w 'string' *
and:
$ grep --include=\*.{txt} -rnw desktop/testfiles/ -e "string"
The first outputs its entire contents and the second doesn't seem to do anything.
Any ideas?
simply :
grep -w "string" *.txt | wc -l
also GNU awk allows it to be done in single command
awk -v w="string" '$1==w{n++} END{print n}' RS=' |\n' *.txt
This question already has answers here:
Remove empty lines in a text file via grep
(11 answers)
Closed 4 years ago.
Exists way to remove empty lines with cat myfile | grep -w #something ?
I looking for simple way for remove empty lines from my output like in the way the presented above.
This really belongs on the codegolfing stackexchange because it's not related to how anyone would ever write a script. However, you can do it like this:
cat myfile | grep -w '.*..*'
It's equivalent to the more canonical grep ., but adds explicit .*s on either side so that it will always match the complete line, thereby satisfying the word boundary conditions imposed by -w
You can pipe your output to awk to easily remove empty lines
cat myfile | grep -w #something | awk NF
EDIT: so... you just want cat myfile | awk NF?
if you have to use grep, you can do grep myfile -v '^[[:blank:]]*$'
This question already has answers here:
Extract filename and extension in Bash
(38 answers)
Closed 7 years ago.
I have files with "multiple" extension , for better manipulation I would like to create new folder for each last extension but first I need to retrieve the last extension.
Just for example lets assume i have file called info.tar.tbz2 how could I get "tbz2" ?
One way that comes to my mind is using cut -d "." but in this case I would need to specify -f parameter of the last column, which I don't know how to achieve.
What is the fastest way to do it?
You may use awk,
awk -F. '{print $NF}' file
or
sed,
$ echo 'info.tar.tbz2' | awk -F. '{print $NF}'
tbz2
$ echo 'info.tar.tbz2' | sed 's/.*\.//'
tbz2
This question already has answers here:
How to split a string into an array in Bash?
(24 answers)
Closed 7 years ago.
Suppose I have a file with a name ABC_DE_FGHI_10_JK_LMN.csv. I want to extract the ID from the file-name i.e. 10 with the help of ID position and file-name separator. I have following two inputs
File-name_ID_Position=4; [since 10 is at fourth position in file-name]
File-name_Delimiter="_";
Here ID can be numeric or alpha-numeric. So how extract the 10 from above file with the help of above two inputs. How to achieve this in bash?
Instead of writing a regex in bash, I would do it with awk:
echo 'ABC_DE_FGHI_10_JK_LMN.csv' | awk -F_ -v pos=4 '{print $pos}'
or if you want the dot to also be a delimiter (requires GNU awk):
echo 'ABC_DE_FGHI_10_JK_LMN.csv' | awk -F'[_.]' -v pos=4 '{print $pos}'