I have a lot of text files and I want to make a bash script in linux to print the name of file in each lines of file. For example I have file lenovo.txt and I want that every line in the file to start with lenovo.txt.
I try to make a "for" for this but didn't work.
for i in *.txt
do
awk '{print '$i' $0}' /var/SambaShare/$i > /var/SambaShare/new_$i
done
Thanks!
It doesn't work because you need to pass $i to awk with the -v option. But you can also use the FILENAME built-in variable in awk :
ls *txt
file.txt file2.txt
cat *txt
A
B
C
A2
B2
C2
for i in *txt; do
awk '{print FILENAME,$0}' $i;
done
file.txt A
file.txt B
file.txt C
file2.txt A2
file2.txt B2
file2.txt C2
An to redirect into a new file :
for i in *txt; do
awk '{print FILENAME,$0}' $i > ${i%.txt}_new.txt;
done
As for your corrected version :
for i in *.txt
do
awk -v i=$i '{print i,$0}' $i > new_$i
done
Hope this helps.
Using grep you can make use of the --with-filename (alias -H) option and use an empty pattern that always matches:
for i in *.txt
do
grep -H "" $i > new_$i
done
Awk and Bash don't share the same variables as they are different languages with separate interpreters. You should pass Bash variables to Awk with the -v option.
You should also quote your file name variables to ensure they don't get expanded as separate arguments if they contain whitespace.
for i in *.txt
do
awk -v i="$i" '{print i,$0}' "$i" > "$i"
done
Related
I am trying to change the column names to lowercase in a csv file. I found the code to do that online but I dont know how to replace the old column names(uppercase) with new column names(lowercase) in the original file. I did something like this:
$cat head -n1 xxx.csv | tr "[A-Z]" "[a-z]"
But it simply just prints out the column names in lowercase, which is not enough for me.
I tried to add sed -i but it did not do any good. Thanks!!
Using awk (readability winner) :
concise way:
awk 'NR==1{print tolower($0);next}1' file.csv
or using ternary operator:
awk '{print (NR==1) ? tolower($0): $0}' file.csv
or using if/else statements:
awk '{if (NR==1) {print tolower($0)} else {print $0}}' file.csv
To change the file for real:
awk 'NR==1{print tolower($0);next}1' file.csv | tee /tmp/temp
mv /tmp/temp file.csv
For your information, sed using the in place edit switch -i do the same: it use a temporary file under the hood.
You can check this by using :
strace -f -s 800 sed -i'' '...' file
Using perl:
perl -i -pe '$_=lc() if $.==1' file.csv
It replace the file on the fly with -i switch
You can use sed to tell it to replace the first line with all lower-case and then print the rest as-is:
sed '1s/.*/\L&/' ./xxx.csv
Redirect the output or use -i to do an in-place edit.
Proof of Concept
$ echo -e "COL1,COL2,COL3\nFoO,bAr,baZ" | sed '1s/.*/\L&/'
col1,col2,col3
FoO,bAr,baZ
I have more than 50000 files in a directory such as file1.txt, file2.txt, ....., file50000.txt. I would like to concatenate of some files whose file numbers are listed in the following text file (need.txt).
need.txt
1
4
35
45
71
.
.
.
I tried with the following. Though it is working, but I look for more simpler and short way.
n1=1
n2=$(wc -l < need.txt)
while [ $n1 -le $n2 ]
do
f1=$(awk 'NR=="$n1" {print $1}' need.txt)
cat file$f1.txt >> out.txt
(( n1++ ))
done
This might also work for you:
sed 's/.*/file&.txt/' < need.txt | xargs cat > out.txt
Something like this should work for you:
sed -e 's/.*/file&.txt/' need.txt | xargs cat > out.txt
It uses sed to translate each line into the appropriate file name and then hands the filenames to xargs to hand them to cat.
Using awk it could be done this way:
awk 'NR==FNR{ARGV[ARGC]="file"$1".txt"; ARGC++; next} {print}' need.txt > out.txt
Which adds each file to the ARGV array of files to process and then prints every line it sees.
It is possible do it without any sed or awk command. Directly using bash built-in functions and cat (of course).
for i in $(cat need.txt); do cat file${i}.txt >> out.txt; done
And as you want, it is quite simple.
Suppose I have a file input.txt with few columns and few rows, the first column is the key, and a directory dir with files which contain some of these keys. I want to find all lines in the files in dir which contain these key words. At first I tried to run the command
cat input.txt | awk '{print $1}' | xargs grep dir
This doesn't work because it thinks the keys are paths on my file system. Next I tried something like
cat input.txt | awk '{system("grep -rn dir $1")}'
But this didn't work either, eventually I have to admit that even this doesn't work
cat input.txt | awk '{system("echo $1")}'
After I tried to use \ to escape the white space and the $ sign, I came here to ask for your advice, any ideas?
Of course I can do something like
for x in `cat input.txt` ; do grep -rn $x dir ; done
This is not good enough, because it takes two commands, but I want only one. This also shows why xargs doesn't work, the parameter is not the last argument
You don't need grep with awk, and you don't need cat to open files:
awk 'NR==FNR{keys[$1]; next} {for (key in keys) if ($0 ~ key) {print FILENAME, $0; next} }' input.txt dir/*
Nor do you need xargs, or shell loops or anything else - just one simple awk command does it all.
If input.txt is not a file, then tweak the above to:
real_input_generating_command |
awk 'NR==FNR{keys[$1]; next} {for (key in keys) if ($0 ~ key) {print FILENAME, $0; next} }' - dir/*
All it's doing is creating an array of keys from the first file (or input stream) and then looking for each key from that array in every file in the dir directory.
Try following
awk '{print $1}' input.txt | xargs -n 1 -I pattern grep -rn pattern dir
First thing you should do is research this.
Next ... you don't need to grep inside awk. That's completely redundant. It's like ... stuffing your turkey with .. a turkey.
Awk can process input and do "grep" like things itself, without the need to launch the grep command. But you don't even need to do this. Adapting your first example:
awk '{print $1}' input.txt | xargs -n 1 -I % grep % dir
This uses xargs' -I option to put xargs' input into a different place on the command line it runs. In FreeBSD or OSX, you would use a -J option instead.
But I prefer your for loop idea, converted into a while loop:
while read key junk; do grep -rn "$key" dir ; done < input.txt
Use process substitution to create a keyword "file" that you can pass to grep via the -f option:
grep -f <(awk '{print $1}' input.txt) dir/*
This will search each file in dir for lines containing keywords printed by the awk command. It's equivalent to
awk '{print $1}' input.txt > tmp.txt
grep -f tmp.txt dir/*
grep requires parameters in order: [what to search] [where to search]. You need to merge keys received from awk and pass them to grep using the \| regexp operator.
For example:
arturcz#szczaw:/tmp/s$ cat words.txt
foo
bar
fubar
foobaz
arturcz#szczaw:/tmp/s$ grep 'foo\|baz' words.txt
foo
foobaz
Finally, you will finish with:
grep `commands|to|prepare|a|keywords|list` directory
In case you still want to use grep inside awk, make sure $1, $2 etc are outside quote.
eg. this works perfectly
cat file_having_query | awk '{system("grep " $1 " file_to_be_greped")}'
// notice the space after grep and before file name
In Linux (Cento OS) I have a file that contains a set of additional information that I want to removed. I want to generate a new file with all characters until to the first |.
The file has the following information:
ALFA12345|7890
Beta0-XPTO-2|30452|90 385|29
ZETA2334423 435; 2|2|90dd5|dddd29|dqe3
The output expected will be:
ALFA12345
Beta0 XPTO-2
ZETA2334423 435; 2
That is removed all characters after the character | (inclusive).
Any suggestion for a script that reads File1 and generates File2 with this specific requirement?
Try
cut -d'|' -f1 oldfile > newfile
And, to round out the "big 3", here's the awk version:
awk -F\| '{print $1}' in.dat
You can use a simple sed script.
sed 's/^\([^|]*\).*/\1/g' in.dat
ALFA12345
Beta0-XPTO-2
ZETA2334423 435; 2
Redirect to a file to capture the output.
sed 's/^\([^|]*\).*/\1/g' in.dat > out.dat
And with grep:
$ grep -o '^[^|]*' file1
ALFA12345
Beta0-XPTO-2
ZETA2334423 435; 2
$ grep -o '^[^|]*' file1 > file2
I have a number of log files in a directory. I am trying to write a script to search all the log files for a string and echo the name of the files and the line number that the string is found.
I figure I will probably have to use 2 grep's - piping the output of one into the other since the -l option only returns the name of the file and nothing about the line numbers. Any insight in how I can successfully achieve this would be much appreciated.
Many thanks,
Alex
$ grep -Hn root /etc/passwd
/etc/passwd:1:root:x:0:0:root:/root:/bin/bash
combining -H and -n does what you expect.
If you want to echo the required informations without the string :
$ grep -Hn root /etc/passwd | cut -d: -f1,2
/etc/passwd:1
or with awk :
$ awk -F: '/root/{print "file=" ARGV[1] "\nline=" NR}' /etc/passwd
file=/etc/passwd
line=1
if you want to create shell variables :
$ awk -F: '/root/{print "file=" ARGV[1] "\nline=" NR}' /etc/passwd | bash
$ echo $line
1
$ echo $file
/etc/passwd
Use -H. If you are using a grep that does not have -H, specify two filenames. For example:
grep -n pattern file /dev/null
My version of grep kept returning text from the matching line, which I wasn't sure if you were after... You can also pipe the output to an awk command to have it ONLY print the file name and line number
grep -Hn "text" . | awk -F: '{print $1 ":" $2}'