How can I use grep to find an exact word inside of a file entered by the user as string?
For example I need to select the word I want to find and the file I want to find it in. I've been told I am really close but something is not working as it should be. I'm using bash shell under Linux.
Here's what I've done so far:
#!/bin/bash
echo "Find the file you want to search the word in?"
read filename
echo "Enter the word you want to find."
read word1
grep $word1 $filename
How can I use grep to find an exact word inside of a file entered by
the user as string.
Try using -F option. Type man grep on shell for more details.
grep -F "$word1" "$filename"
It's recommended to enclose search string and file name with quotes to avoid unexpected output because of white spaces.
Not sure why you have fi on the last line. It's not needed.
Try:
grep -R WORD ./ to search the entire current directory, or grep WORD ./path/to/file.ext to search inside a specific file.
#!/bin/bash/
echo "Find the file you want to search the word in?"
read filename
echo "Enter the word you want to find."
read word
cat $filename | grep "$word"
This works fine to find the exact word match in a file.
If you want an exact word match within lines use
grep "\b$word\b"
Related
I have a file with partial matches to lines in another file. In order to do this I was looking to generate a while loop with read and substituting a variable for each line of partial matches into a grep command to search a database files with a partial match but for some reason, I am not getting an output (an empty outputfile.txt).
Here is my current script
while read -r line; do
grep $line /path/to/databasefile >> /path/to/folder/outputfile.txt
done < "/partial_matches.txt"
the database has multiple lines with a sequence name then DNA sequence after:
>transcript_ab
AGTCAGTCATGTC
>transcript_ac
AGTCAGTCATGTC
>transctipt_ad
AGTCAGTCATGTC
and the partial matching search file has lines of text:
ab
ac
and I'm looking for a return of:
>transcript_ab
>transcript_ac
any help would be appreciated. Thanks.
If you are using GNU grep, then its -f option is what you are looking for:
grep -f /partial_matches.txt /path/to/databasefile
(if you don't have any pattern in partial_matches.txt but only strings, then use grep -F instead of grep)
you can use a for loop instead:
for i in $(cat partial_matches.txt); do
grep $i /path/to/databasefile >> /path/to/folder/outputfile.txt
done
Also, check if you have a typo:
"/partial_matches.txt" -> "./partial_matches.txt"
iam trining to find the longest word in a text file.
i tried it and find out the no of characters in the longest word in a file
by using the command
wc -L
i need to print the longest word By using this number and grep command .
If you must use the two commands give, I'd suggest:
grep -E ".{$(wc -L < test.txt)}" test.txt
The command substitution is used to build the correct brace expression to match the line(s) with exactly the given number of characters. -E is needed to enable extended regular expression support; otherwise, the braces need to be escaped: grep ".\{...\}" test.txt.
Using an awk command that makes a single pass through the file may be faster.
I have a file named testfile.It contains some information like
methun:x:500:500:comment:/home/methun:bin/bash
salahuddin:x:501:500:comment:/home/methun:bin/bash
Now implemented a following shell program:
echo "Enter a Name:"
read username
users='cat /mypractice/myfiles/testfile | awk -F ':' '{print $1}''
for user in $users
do
if [ "$user" == "$username" ]; then
echo "Name found and Enter a new name to change."
read newUsername
#need code to change text on my file --->testfile
fi
done
Now suppose I need to change methun to Moin. Comment to newcomment. I used
sed -i 's/"$user"/"$newuser"/g' /mypractice/myfiles/testfile
But it not working here. I test with it in my testfile singly it change and replace all.But i need to change only that position i want .
I also tried with usermod but it will not works here..
Can anyone give me the solution or correct my code...Thanks
You are using g flag in your sed command which means global substitution (will change all occurrences). Also, the variables although quoted are wrapped inside single quotes and hence are not interpolated.
Try this:
sed -i "s/^$user/$newuser/" /mypractice/myfiles/testfile
I have placed a ^ anchor in the substitution part which means only substitute if the word is at the beginning of the line. This protects you from making a change if the name of the user is not at the start but somewhere in the middle.
I have a text file with many links which aren't in separate lines.
I want to save in another file probably, all the links which contains a specific word.
How can I do this with grep?
EDIT:
To become more specifique, I have a messy txt file with many links. I want to copy in onother file all links starting with https:://, ending with .jpg and contains anywhere "10x10" string for example
You can get all the lines containing a specific word from the file like this:
LINKS=$(cat myfile.txt | grep MYWORD)
Then with LINKS, you can use a delimiter to create an array of links, which you can print to another file.
# Using a space as the delimeter
while IFS=' 'read -ra ind_link
do
echo $ind_link >> mynewfile.txt
done <<< "$LINKS"
Something along those lines I think is what you are looking for no?
Also if you need to refine your search, you can use the grep options such as -w to get more specific.
Hope it helps.
Could you give us the specific word and an example of input file ?
You could try to use egrep or/and sed like this (for example) :
egrep -o "href=\".*\.html\"" file|sed "s/\"\([^\"]*\)/\1/g"
Another exemple for all kind of http/https ressources links (whithout spaces in the URL) :
$ echo "<a href=http://titi/toto.jpg >"|egrep -o "https?:\/\/[^\ ]*"
http://titi/toto.jpg
$ echo "<a href=https://titi/toto.htm >"|egrep -o "https?:\/\/[^\ ]*"
https://titi/toto.htm
You have to customize the regexp according to your needs.
I have long file with the following list:
/drivers/isdn/hardware/eicon/message.c//add_b1()
/drivers/media/video/saa7134/saa7134-dvb.c//dvb_init()
/sound/pci/ac97/ac97_codec.c//snd_ac97_mixer_build()
/drivers/s390/char/tape_34xx.c//tape_34xx_unit_check()
(PROBLEM)/drivers/video/sis/init301.c//SiS_GetCRT2Data301()
/drivers/scsi/sg.c//sg_ioctl()
/fs/ntfs/file.c//ntfs_prepare_pages_for_non_resident_write()
/drivers/net/tg3.c//tg3_reset_hw()
/arch/cris/arch-v32/drivers/cryptocop.c//cryptocop_setup_dma_list()
/drivers/media/video/pvrusb2/pvrusb2-v4l2.c//pvr2_v4l2_do_ioctl()
/drivers/video/aty/atyfb_base.c//aty_init()
/block/compat_ioctl.c//compat_blkdev_driver_ioctl()
....
It contains all the functions in the kernel code. The notation is file//function.
I want to copy some 100 files from the kernel directory to another directory, so I want to strip every line from the function name, leaving just the filename.
It's super-easy in python, any idea how to write a 1-liner in the bash prompt that does the trick?
Thanks,
Udi
cat "func_list" | sed "s#//.*##" > "file_list"
Didn't run it :)
You can use pure Bash:
while read -r line; do echo "${line%//*}"; done < funclist.txt
Edit:
The syntax of the echo command is doing the same thing as the sed command in Eugene's answer: deleting the "//" and everything that comes after.
Broken down:
"echo ${line}" is the same as "echo $line"
the "%" deletes the pattern that follows it if it matches the trailing portion of the parameter
"%" makes the shortest possible match, "%%" makes the longest possible
"//*" is the pattern to match, "*" is similar to sed's ".*"
See the Parameter Expansion section of the Bash man page for more information, including:
using ${parameter#word} for matching the beginning of a parameter
${parameter/pattern/string} to do sed-style replacements
${parameter:offset:length} to retrieve substrings
etc.
here's a one liner in (g)awk
awk -F"//" '{print $1}' file
Here's one using cut and rev
cat file | rev | cut -d'/' -f2-| rev