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I work with Bash script and I want to get line from big text by special text
for example i have these lines
first fffffffffffffffffffffffffff
.................................
second ssssssssssssssssssssssssss
.................................
third ttttttttttttttttttttttttttt
and I want to get ssssssssssssssssssssssssss string .
Can anybody help me?
Is this what you want?
echo "$longstring" | awk '$1 == "second" { print $2 }'
since you seem to not have any criterion as to which line you want to output, i suggest something like:
echo "ssssssssssssssssssssssssss"
this is pretty robust regarding the content of your input, doesn't depend on a "file", and is a fast solution.
cat filename | grep "^second" | cut -d " " -f 2
Or, if you are ALF:
<filename grep "^second" | cut -d " " -f 2
Or
grep "^second" filename | cut -d " " -f 2
Related
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I have a text file which contains a series of same line except at the end.
eg
lesi-1-1500-1
lesi-1-1500-2
lesi-1-1500-3
how can I remove the last number? it goes upto 250
to change in the file itself
sed -i 's/[0-9]\+$//' /path/to/file
or
sed 's/[0-9]\+$//' /path/to/file > /path/to/output
see example
You can do it with Awk by breaking it into fields.
echo "lesi-1-1500-2" > foo.txt
echo "lesi-1-1500-3" >> foo.txt
cat foo.txt | awk -F '-' '{print $1 "-" $2 "-" $3 }'
The -F switch allows us to set the delimiter which is -. Then we just print the first three fields with - for formatting.
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This is from a vhost file. This is the output I get
ServerName uat3-dam-something1.prg-dc.brb.com
Hello,
I'm wondering how to cut from this output so only this part remains
something1.prg-dc.brb.com
Keep in mind that "something" could be "something4141411" or "something23". So length operations won't work. Tried with cut command and AWK, but didn't work. I would be happy receive a tips from the bash experts :)
Like this :
grep -o 'something.*' file
or more specific:
grep -oE 'something[0-9]+\..*' file
Output:
something1.prg-dc.brb.com
Could you please try following, written and tested with provided samples only.
awk -F'uat3-dam-' '{print $NF}' Input_file
Description: Making uat3-dam- as field separator and printing last field of it.
2nd solution:
awk 'match($0,/something.*/){print substr($0,RSTART,RLENGTH)}' Input_file
Using:
echo "ServerName uat3-dam-something1.prg-dc.brb.com" |cut -d\- -f3-4
Will return:
something1.prg-dc.brb.com
And if you change the string (as you mention):
echo "ServerName uat3-dam-something111111.prg-dc.brb.com" |cut -d\- -f3-4
It will keep returning:
something111111.prg-dc.brb.com
$ echo 'ServerName uat3-dam-something1.prg-dc.brb.com' | awk -F- '{sub(".*" $2 FS,"")}1'
something1.prg-dc.brb.com
This will work:
echo "ServerName uat3-dam-something1.prg-dc.brb.com" | sed -E 's/.*(something.*)/\1/'
Or, if the string is in a file named file
sed -E 's/.*(something.*)/\1/' file
Explanation:
-E is for extended regex
.*(something.*) means "any char 0 or more times followed by something and any other char 0 or more times".
\1 is used to print only the matching part inside the brackets.
You could also use :
echo ${test#*dam-}
Example :
test="ServerName uat3-dam-something1.prg-dc.brb.com"
echo ${test#*dam-}
which gives:
something1.prg-dc.brb.com
Note that the opposite version would be echo ${test%something*}
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I have some data in a MyFile.CSV file like this:
id,name,country
100,tom cruise,USA
101,Johnny depp,USA
102,John,India
What will be the shell script to take the above file as input and segregate the data in 2 different files as per the country?
I tried using the FOR loop and then using 2 IFs inside it but I am unable to do so. How to do it using awk?
For LINE in MyFile.CSV
Do
If grep "USA" $LINE >0 Then
$LINE >> Out_USA.csv
Else
$LINE >> Out_India.csv
Done
You can try with this
grep -R "USA" /path/to/file >> Out_USA.csv
grep -R "India" /path/to/file >> Out_India.csv
Many ways to do:
One way:
$ for i in `awk -F"," '{if(NR>1)print $3}' MyFile.csv|uniq|sort`;
do
echo $i;
egrep "${i}|country" MyFile.csv > Out_${i}.csv;
done
This assumes that the country name would not clash with other columns.
If it does, then you can fine tune that by adding additional regex.
For example, it country will be the last field, then you can add $ to the grep
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I'm currently using (diff -q directory1 directory2) to output the files in each directory that are different and printing them to a table in html.
Current output: "Files directory1/file1 and directory2/file2 differ"
What I want: "file1 has changed"
I do not want to use comm or sort the files because other applications are pulling from the files and are sensitive to ordering. Any idea on how to get this done?
you need to grep diff output for file that differ then use awk to print file name with your new format
diff -rq dir1 dir2 | grep "differ" | awk '{print $2 "has changed"}'
Will this work?
diff -q $file1 $file2 | awk '{print $2 " has changed"}'
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Suppose i have one line Script :
Script Name is script1.sh has below line on it -
# sh script.sh #
So how can i take only script.sh name from script1.sh.
What I have done is below but that is not fully fruitful to me get the exact output that I want.
while read line
do
called_script= awk -F ':' '{print $1 }' final_calling_script_name
qwe= grep '*.sh' $called_script
echo $called_script " : $qwe"
done<'file_that_contains_data_of_script1_line_by_line'
Can anybody help me?
If what you want here is basically "the second word" you can use "cut"
echo "sh script.sh" | cut -d ' ' -f 2
The -d ' ' tells cut that the "delimiting character" is a space, the -f 2 tells cut that you want column number 2.
echo "sh script.sh" | { read a b; echo "$b"; }
EDIT:
After you've clarified your requirements in the notes below, I would propose this command:
echo "script1.ksh script2.pig script3.sh" | grep -oe '\w*\.sh'