I am running the following command:
for file in 2017120[1-9]/54100_*.zip; do unzip -c "$file" | grep "3613825" | grep '3418665' ; done
This does a grep job of pulling the data that matches my grep parameters, but I can't figure out how to capture which file the results came from.
I have tried adding grep -H but the result comes back with (standard input).
How can I capture the file name?
When I need to do something like this I just add an echo of the file name to the for loop like this:
for file in 2017120[1-9]/54100_*.zip; do echo $file; unzip -c "$file" | grep "3613825" | grep '3418665' ; done
This prints out the list of files, and the grep line that matches will print immediately after the file that the match is in. like this:
file_1
file_2
file_3
matching line
file_4
file_5
another matching line
file_6
...
Thus I know the matching lines occurred in file_3 and file_5.
Related
I have a file testtns.txt which has numbers like below :
123
456
I am then passing the input of the file folder path like "/var/www/batchfiles/files/test.csv" . The test.csv has following records :
1,123,On its way to warehouse,20230131
2,456,On its way to warehouse,20230201
3,777,Pickedup,20230201
4,888,Pickedup,20230202
I have created the script printgrep.bash to read the numbers from the file testtns.txt and then grep the csv files in folder "/var/www/batchfiles/files/". I am running the script using command .\printgrep.bash /var/www/batchfiles/files/* and excepting the output to be in output.txt as follows :
1,123,On its way to warehouse,20230131
2,456,On its way to warehouse,20230201
However when I run the above command, output.txt is empty and doesnt have any results. However if I run the grep command as it is, it does return results as excepted.
Can someone let me know why the grep command not working in the below script printgrep.bash :
#!/bin/bash
cat testtns.txt | while read line
do
grep -i "^ $line" $1 >> output.txt
done
Tried even below grep command but still didnt work :
#!/bin/bash
cat testtns.txt | while read line
do
grep -i "$line" $1 >> output.txt
done
I have a file containing list of 4000 words (A.txt). Now I want to grep lines from another file (sentence_per_line.txt) containing those 4000 words mentioned in the file A.txt.
The shell script I wrote for the above problem is
#!/bin/bash
file="A.txt"
while IFS= read -r line
do
# display $line or do somthing with $line
printf '%s\n' "$line"
grep $line sentence_per_line.txt >> output.txt
# tried printing the grep command to check its working or not
result=$(grep "$line" sentence_per_line.txt >> output.txt)
echo "$result"
done <"$file"
And A.txt looks like this
applicable
available
White
Black
..
The code is neither working nor does it shows any error.
Grep has this built in:
grep -f A.txt sentence_per_line.txt > output.txt
Remarks to your code:
Looping over a file to execute grep/sed/awk on each line is typically an antipattern, see this Q&A.
If your $line parameter contains more than one word, you have to quote it (doesn't hurt anyway), or grep tries to look for the first word in a file named after the second word:
grep "$line" sentence_per_line.txt >> output.txt
If you write output in a loop, don't redirect within the loop, do it outside:
while read -r line; do
grep "$line" sentence_per_line.txt
done < "$file" > output.txt
but remember, it's usually not a good idea in the first place.
If you'd like to write to a file and at the same time see what you're writing, you can use tee:
grep "$line" sentence_per_line.txt | tee output.txt
writes to output.txt and stdout.
If A.txt contains words which you want to match only if the complete word matches, i.e., pattern should not match longerpattern, you can use grep -wf – the -w matches only complete words.
If the words in A.txt aren't regular expressions, but fixed strings, you can use grep -fF – the -F option looks for fixed strings and is faster. These two can be combined: grep -WfF
I am trying to figure out how to search a file for example test.txt for "#include" and print the first instance of it that is found. However if it is not found I want to print out the whole file. I can do the first part with grep "#include" test.txt | head -n 1 but can't seem to figure out how to include the second requirement.
grep -m 1 '#include' "$file" || cat "$file"
You can control grep's output in many ways with grep options.
grep exits non-success status if the match is not found.
I have file.txt with names one per line as shown below:
ABCB8
ABCC12
ABCC3
ABCC4
AHR
ALDH4A1
ALDH5A1
....
I want to grep each of these from an input.txt file.
Manually i do this one at a time as
grep "ABCB8" input.txt > output.txt
Could someone help to automatically grep all the strings in file.txt from input.txt and write it to output.txt.
You can use the -f flag as described in Bash, Linux, Need to remove lines from one file based on matching content from another file
grep -o -f file.txt input.txt > output.txt
Flag
-f FILE, --file=FILE:
Obtain patterns from FILE, one per line. The empty file
contains zero patterns, and therefore matches nothing. (-f is
specified by POSIX.)
-o, --only-matching:
Print only the matched (non-empty) parts of a matching line, with
each such part on a separate output line.
for line in `cat text.txt`; do grep $line input.txt >> output.txt; done
Contents of text.txt:
ABCB8
ABCC12
ABCC3
ABCC4
AHR
ALDH4A1
ALDH5A1
Edit:
A safer solution with while read:
cat text.txt | while read line; do grep "$line" input.txt >> output.txt; done
Edit 2:
Sample text.txt:
ABCB8
ABCB8XY
ABCC12
Sample input.txt:
You were hired to do a job; we expect you to do it.
You were hired because ABCB8 you kick ass;
we expect you to kick ass.
ABCB8XY You were hired because you can commit to a rational deadline and meet it;
ABCC12 we'll expect you to do that too.
You're not someone who needs a middle manager tracking your mouse clicks
If You don't care about the order of lines, the quick workaround would be to pipe the solution through a sort | uniq:
cat text.txt | while read line; do grep "$line" input.txt >> output.txt; done; cat output.txt | sort | uniq > output2.txt
The result is then in output.txt.
Edit 3:
cat text.txt | while read line; do grep "\<${line}\>" input.txt >> output.txt; done
Is that fine?
In a very large file I need to find the position (line number) of a string, then extract the 2 lines above and below that string.
To do this right now - I launch vi, find the string, note it's line number, exit vi, then use sed to extract the lines surrounding that string.
Is there a way to streamline this process... ideally without having to run vi at all.
Maybe using grep like this:
grep -n -2 your_searched_for_string your_large_text_file
Will give you almost what you expect
-n : tells grep to print the line number
-2 : print 2 additional lines (and the wanted string, of course)
You can do
grep -C 2 yourSearch yourFile
To send it in a file, do
grep -C 2 yourSearch yourFile > result.txt
Use grep -n string file to find the line number without opening the file.
you can use cat -n to display the line numbers and then use awk to get the line number after a grep in order to extract line number:
cat -n FILE | grep WORD | awk '{print $1;}'
although grep already does what you mention if you give -C 2 (above/below 2 lines):
grep -C 2 WORD FILE
You can do it with grep -A and -B options, like this:
grep -B 2 -A 2 "searchstring" | sed 3d
grep will find the line and show two lines of context before and after, later remove the third one with sed.
If you want to automate this, simple you can do a Shell Script. You may try the following:
#!/bin/bash
VAL="your_search_keyword"
NUM1=`grep -n "$VAL" file.txt | cut -f1 -d ':'`
echo $NUM1 #show the line number of the matched keyword
MYNUMUP=$["NUM1"-1] #get above keyword
MYNUMDOWN=$["NUM1"+1] #get below keyword
sed -n "$MYNUMUP"p file.txt #display above keyword
sed -n "$MYNUMDOWN"p file.txt #display below keyword
The plus point of the script is you can change the keyword in VAL variable as you like and execute to get the needed output.