I want to display text file name and its size in different lines
I have tried
du *.* | cut -f 1
This give me only size of the files in given directory
du *.* | cut -f 2
This gives the filenames
But i could't figure out how to format it in way where the size comes first then the file name.
example :
4
file1.txt
5
file2.txt
I just figured it out this is working as expected.
du *.txt* | tr [:space:] '\n'
You can do some awk scripting:
for file in *
do
echo "$file $(du "$file" | awk '{print $1}')"
done
Consider:
var=`ls -l | grep TestFile.txt | awk '{print $5}'`
I am able to read file size, but how does it work?
Don't parse ls
size=$( stat -c '%s' TestFile.txt )
Yes, so basically you could divide it into 4 parts:
ls -l
List the current directory content (-l for long listing format)
| grep TestFile.txt
Pipe the result and look for the file you are interested in
| awk '{print $5}
Pipe the result to awk program which cuts (by using spaces as separator) the fifth column which happens to be the file size in this case (but this can be broken by spaces in the filename, for example)
var=`...`
The backquotes (`) enclose commands. The output of the commands gets stored in the var variable.
NOTE: You can get the file size directly by using du -b TestFile.txt or stat -c %s TestFile.txt
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
I am doing a homework which ask me to find the smallest file and biggest file under the directory, I have done that. But my output is something like
"the smallest file is xxx (xxxx -'filename' bytes)
I wish I could print something without the filename part.
I am using du -b $filename to get the size.
du -b | sort -rh | head -n 1 | awk '{print "The smallest file is " $1 " bytes"}'
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}'