In order to simplify my work I usually do this:
for FILE in ./*.txt;
do ID=`echo ${FILE} | sed 's/^.*\///'`;
bin/Tool ${FILE} > ${ID}_output.txt;
done
Hence process loops over all *.txt files.
Now I have two file groups - my Tool uses two inputs (-a & -b). Is there any command to run Tool for every FILE_A over every FILE_B and name the output file as a combination of both them?
I imagine it should look like something like this:
for FILE_A in ./filesA/*.txt;
do for FILE_B in ./filesB/*.txt;
bin/Tool -a ${FILE_A} -b ${FILE_B} > output.txt;
done
So the process would run number of *.txt in filesA over number of *.txt in filesB.
And also the naming issue which I even don't know where to put in...
Hope it is clear what I am asking. Never had to do such task before and a command line would be really helpful.
Looking forward!
NEWNAME="${FILE_A##*/}_${FILE_B##*/}_output.txt"
Related
I have a file .txt with some informations, i need to grep the "Report:" line and save each line in a different .txt file!
it should result something like this in the end:
case1.txt
case2.txt
case3.txt
I tried to
cat cases.txt| grep Report: | while read Report; do echo $Report | > /home/kali/Desktop/allcases/case.txt done
but it didnt work and just created one file called case.txt containing the last grepped "Report:"
I dont know if i was very clear then i'll show this screenshot:
cases
I wanted to split all theses reports in a different .txt file for each report!
These case informations are from a game, so dont worry!
awk would be better suited than grep and a while loop. If acceptable, you can try;
awk '/^Report/{cnt++;close(report); report="case"cnt".txt"}/./{print > report}' file.txt
perl -ne '++$i && `printf "$_" > case$i.txt` if /Report:/' cases.txt
This is looping over cases.txt and shelling out printf "$_" > case$i.txt if the line matches /Report:/
Because it's perl there's some syntax and precedence tricks in here to make it terse and confusing.
I have a folder /jobs and I am trying to merge every files present within that folder into a new file called workflows.yaml with a new line between each file merged.
I am able to loop over the directory using
for FILE in jobs/*; do awk $$FILE > workflows.yaml
And I am also able to merge
awk '{print}' jobs/a.yaml jobs/b.yaml > workflows.yaml
What I tried but did not work:
for FILE in jobs/*; do echo $$FILE; done
You don't need awk, ed, etc. At least I don't see why based on the question. Isn't this good enough:
all:
for f in jobs/*; do cat $$f; echo; done > workflows.yml
? If not perhaps you could be clear in your question about exactly what you want to do. When you say "a new line" do you mean, a blank line?
I am trying to to provide a file for my shell as an input which in return should test if the file contains a specific word and decide what command to execute. I am not figuring out yet where the mistake might lie. Please find the shell script that i wrote:
#!/bin/(shell)
input_file="$1"
output_file="$2"
grep "val1" | awk -f ./path/to/script.awk $input_file > $output_file
grep "val2" | sh ./path/to/script.sh $input_file > $output_file
when I input the the file that uses awk everything get executed as expected, but for the second command I don't even get an output file. Any help is much appreciated
Cheers,
You haven't specified this in your question, but I'm guessing you have a file with the keyword, e.g. file cmdfile that contains x-g301. And then you run your script like:
./script "input_file" "output_file" < cmdfile
If so, the first grep command will consume the whole cmdfile on stdin while searching for the first pattern, and nothing will be left for the second grep. That's why the second grep, and then your second script, produces no output.
There are many ways to fix this, but choosing the right one depends on what exactly you are trying to do, and how does that cmdfile look like. Assuming that's a larger file with other things than just the command pattern, you could pass that file as a third argument to your script, like this:
./script "input_file" "output_file" "cmdfile"
And have your script handle it like this:
#!/bin/bash
input_file="$1"
output_file="$2"
cmdfile="$3"
if grep -q "X-G303" "$cmdfile"; then
awk -f ./mno/script.awk "$input_file" > t1.json
fi
if grep -q "x-g301" "$cmdfile"; then
sh ./mno/tm.sh "$input_file" > t2.json
fi
Here I'm also assuming that your awk and sh scripts don't really need the output from grep, since you're giving them the name of the input file.
Note the proper way to use grep for existence search is via its exit code (and the muted output with -q). Instead of the if we could have used shortcircuiting (grep ... && awk ...), but this way is probably more readable.
I have a folder filled with ~300 files. They are named in this form username#mail.com.pdf. I need about 40 of them, and I have a list of usernames (saved in a file called names.txt). Each username is one line in the file. I need about 40 of these files, and would like to copy over the files I need into a new folder that has only the ones I need.
Where the file names.txt has as its first line the username only:
(eg, eternalmothra), the PDF file I want to copy over is named eternalmothra#mail.com.pdf.
while read p; do
ls | grep $p > file_names.txt
done <names.txt
This seems like it should read from the list, and for each line turns username into username#mail.com.pdf. Unfortunately, it seems like only the last one is saved to file_names.txt.
The second part of this is to copy all the files over:
while read p; do
mv $p foldername
done <file_names.txt
(I haven't tried that second part yet because the first part isn't working).
I'm doing all this with Cygwin, by the way.
1) What is wrong with the first script that it won't copy everything over?
2) If I get that to work, will the second script correctly copy them over? (Actually, I think it's preferable if they just get copied, not moved over).
Edit:
I would like to add that I figured out how to read lines from a txt file from here: Looping through content of a file in bash
Solution from comment: Your problem is just, that echo a > b is overwriting file, while echo a >> b is appending to file, so replace
ls | grep $p > file_names.txt
with
ls | grep $p >> file_names.txt
There might be more efficient solutions if the task runs everyday, but for a one-shot of 300 files your script is good.
Assuming you don't have file names with newlines in them (in which case your original approach would not have a chance of working anyway), try this.
printf '%s\n' * | grep -f names.txt | xargs cp -t foldername
The printf is necessary to work around the various issues with ls; passing the list of all the file names to grep in one go produces a list of all the matches, one per line; and passing that to xargs cp performs the copying. (To move instead of copy, use mv instead of cp, obviously; both support the -t option so as to make it convenient to run them under xargs.) The function of xargs is to convert standard input into arguments to the program you run as the argument to xargs.
This should be a no-brainer, but apparently I have no brain today.
I have 50 20-gig logs that contain entries from multiple apps, one of which addes a transaction ID to its log lines. I have 42 transaction IDs I need to review, and I'd like to parse out the appropriate lines into separate files.
To do a single file, the command would be simply,
grep CDBBDEADBEEF2020X02393 server.log* > CDBBDEADBEEF2020X02393.log
that creates a log isolated to that transaction, from all 50 server.logs.
Now, I have a file with 42 txnIDs (shortening to 4 here):
CDBBDEADBEEF2020X02393
CDBBDEADBEEF6548X02302
CDBBDE15644F2020X02354
ABBDEADBEEF21014777811
And I wrote:
#/bin/sh
grep $1 server.\* > $1.log
But that is not working. Changing the shebang to #/bin/bash -xv, gives me this weird output (obviously I'm playing with what the correct escape magic must be):
$ ./xtrakt.sh B7F6E465E006B1F1A
#!/bin/bash -xv
grep - ./server\.\*
' grep - './server.*
: No such file or directory
I have also tried the command line
grep - server.* < txids.txt > $1
But OBVIOUSLY that $1 is pointless and I have no idea how to get a file named per txid using the input redirect form of the command.
Thanks in advance for any ideas. I haven't gone the route of doing a foreach in the shell script, because I want grep to put the original filename in the output lines so I can examine context later if I need to.
Also - it would be great to have the server.* files ordered numerically (server.log.1, server.log.2 NOT server.log.1, server.log.10...)
try this:
while read -r txid
do
grep "$txid" server.* > "$txid.log"
done < txids.txt
and for the file ordering - rename files with one digit to two digit, with leading zeroes, e.g. mv server.log.1 server.log.01.