I am trying to grep out the lines in a file where the third field matches certain criteria.
I tried using grep but had no luck in filtering out by a field in the file.
I have a file full of records like this:
12794357382;0;219;215
12795287063;0;220;215
12795432063;0;215;220
I need to grep only the lines where the third field is equal to 215 (in this case, only the third line)
Thanks a lot in advance for your help!
Put down the hammer.
$ awk -F ";" '$3 == 215 { print $0 }' <<< $'12794357382;0;219;215\n12795287063;0;220;215\n12795432063;0;215;220'
12795432063;0;215;220
grep:
grep -E "[^;]*;[^;]*;215;.*" yourFile
in this case, awk would be easier:
awk -F';' '$3==215' yourFile
A solution in pure bash for the pre-processing, still needing a grep:
while read line; do
OLF_IFS=$IFS; IFS=";"
line_array=( $line )
IFS=$OLD_IFS
test "${line_array[2]}" = 215 && echo "$line"
done < file | grep _your_pattern_
Simple egrep (=grep -E)
egrep ';215;[0-d][0-d][0-d]$' /path/to/file
or
egrep ';215;[[:digit:]]{3}$' /path/to/file
How about something like this:
cat your_file | while read line; do
if [ `echo "$line" | cut -d ";" -f 3` == "215" ]; then
# This is the line you want
fi
done
Here is the sed version to grep for lines where 3rd field is 215:
sed -n '/^[^;]*;[^;]*;215;/p' file.txt
Simplify your problem by putting the 3rd field at the beginning of the line:
cut -d ";" -f 3 file | paste -d ";" - file
then grep for the lines matching the 3rd field and remove the 3rd field at the beginning:
grep "^215;" | cut -d ";" -f 2-
and then you can grep for whatever you want. So the complete solution is:
cut -d ";" -f 3 file | paste -d ";" - file | grep "^215;" | cut -d ";" -f 2- | grep _your_pattern_
Advantage: Easy to understand; drawback: many processes.
Related
I try to retrieve a line from a file without using a loop.
myFile.txt
val1;a;b;c
val2;b;d;e
val3;c;r;f
I would like to get the line where the second column is b.
If I do grep "b" myFile.txt then both first and second line will be outputed.
If I do cat myFile.txt | cut -d ';' -f2 | grep "b" then the output will just be b whereas I'd like to get the full line val2;b;d;e.
Is there a way of reaching the desired results without using a loop as below ? My file being huge it wouldn't be nice looping through it again and again.
while read line; do
if [ `echo $line | cut -d ';' -f2` = "b" ]; then
echo $line
fi
done < myFile.txt
Given your input file, The below one-liner should work:
awk -F";" '$2 == "b" {print}' myFile.txt
Explanation:
awk -F";" ##Field Separator as ";"
'$2 == "b" ##Searches for "b" in the second column($2)
{print}' ##prints the searched line
Using:
grep:
grep '^[^;]*;b;' myFile.txt
sed:
sed '/^[^;]*;b;/!d' myFile.txt
Output is the same for both:
val2;b;d;e
Here is my code
cd /bin/
echo *xyz?2* | cut -f 1 -d '.'
Please, how can i change this command to display files without extension ?
Bests.
Dump the filenames into an array and then use parameter expansion:
$ arr=(xyz?2*); echo "${arr[*]%.*}"
xyz32281 xyz32406 xyz32459 xyz3252 xyz7214 xyz8286
Assuming your filenames don't have any whitespace or glob characters.
You can just use printf '%s\n' instead of echo in your command:
printf '%s\n' *xyz?2* | cut -f 1 -d '.'
xyz32281
xyz32406
xyz32459
xyz3252
xyz7214
xyz8286
If you must use echo then use awk as this:
echo *xyz?2* | awk '{for(i=1; i<=NF; i++) print (split($i, a, /\./)==2 ? a[1] : $i)}'
xyz32281
xyz32406
xyz32459
xyz3252
xyz7214
xyz8286
This awk command iterated through each filename matched by glob pattern and splits each name by dot. If dot is found then first part is printed otherwise full filename is printed.
Your problem is that all files of echo *xyz?2* are shown in one line. When the filenames are without spaces/newlines, you can fix this by moving them to different lines and joining theem again when finished.
echo *xyz?2* | tr ' ' '\n' | cut -f 1 -d '.' | tr '\n' ' '| sed '$s/ $/\n/'
You can do this a lot easier with sed:
echo *xyz?2* | sed 's/[.][^. ]*//g'
I'm working on a shell script.
OUT=$1
here, the OUT variable is my filename.
I'm using grep search as follows:
l=`grep "$pattern " -A 15 $OUT | grep -w $i | awk '{print $8}'|tail -1 | tr '\n' ','`
The issue is that the filename parameter I must pass is test.log.However, I have the folder structure :
test.log
test.log.001
test.log.002
I would ideally like to pass the filename as test.log and would like it to search it in all log files.I know the usual way to do is by using test.log.* in command line, but I'm facing difficulty replicating the same in shell script.
My efforts:
var-$'.*'
l=`grep "$pattern " -A 15 $OUT$var | grep -w $i | awk '{print $8}'|tail -1 | tr '\n' ','`
However, I did not get the desired result.
Hopefully this will get you closer:
#!/bin/bash
for f in "${1}*"; do
grep "$pattern" -A15 "$f"
done | grep -w $i | awk 'END{print $8}'
I have a script which iterates through a file and finds matches in another file. How to I get the process to stop once I've found a match.
For example:
I take the first line in name.txt, and then try to find a match for it in file.txt.
name.txt:
7,7,FRESH,98,135,
65,10,OLD,56,45,
file.txt:
7,7,Dave,S
8,10,Frank,S
31,7,Gregg
45,5,Jake,S
Script:
while read line
do
name_id=`echo $line | cut -f1,2 -d ','`
identiferOne=`echo $name_id | cut -f1 -d ','`
identiferTwo=`echo $name_id | cut -f2 -d ','`
while IFS= read line
do
CHECK=`echo $line | cut -f4 -d','`
if [ $CHECK = "S" ]
then
symbolName=`echo $line | cut -f3 -d ','`
numberOne=`echo $line | awk -F',' '{print $1}'`
numberTwo=`echo $line | cut -f2 -d ','`
if [ "$numberOne" == $identiferOne ] && [ "$numberTwo" == $identifierTwo ]
then
echo "WE HAVE A MATCH with $symbolName"
break
fi
fi
done < /tmp/file.txt
done < /tmp/name.txt
My question is - how do I stop the script from iterating through file.txt once it has found an initial match, and then set that matched record into a variable, stop the if statement, then do some other stuff within the loop using that variable. I tried using break; but that exits the loop, which is not what I want.
You can tell grep different things:
Stop searching after the first match (option -m 1).
Read the searchkeys from a file (option -f file).
Pretend that the output of a command is a file (not really grep, bash helps here) with <(cmmnd).
Combining these will give you
grep -m1 -f <(cut -d"," -f1-2 name.txt) file.txt
Close, but not what you want. The substrings given by cut -d"," -f1-2 name.txt will match everywhere in the line, and you want to match the first two fields. Matching at the start of the line is done with ^, so we use sed to make strings like ^field1,field2 :
grep -m1 -f <(sed 's/\([^,]*,[^,]*,\).*/^\1/' name.txt) file.txt
I have an application (myapp) that gives me a multiline output
result:
abc|myparam1|def
ghi|myparam2|jkl
mno|myparam3|pqr
stu|myparam4|vwx
With grep and sed I can get my parameters as below
myapp | grep '|' | sed -e 's/^[^|]*//' | sed -e 's/|.*//'
But then want these myparamx values as paramaters of a script to be executed for each parameter.
myscript.sh myparam1
myscript.sh myparam2
etc.
Any help greatly appreciated
Please see xargs. For example:
myapp | grep '|' | sed -e 's/^[^|]*//' | sed -e 's/|.*//' | xargs -n 1 myscript.sh
May be this can help -
myapp | awk -F"|" '{ print $2 }' | while read -r line; do /path/to/script/ "$line"; done
I like the xargs -n 1 solution from Dark Falcon, and while read is a classical tool for such kind of things, but just for completeness:
myapp | awk -F'|' '{print "myscript.sh", $2}' | bash
As a side note, speaking about extraction of 2nd field, you could use cut:
myapp | cut -d'|' -f 1 # -f 1 => second field, starting from 0