I need a way to fetch the desired file path as mentioned below:
example:
file paths: (Desired input)
/root/tmp/uname/abc.txt
/root/tmp/uname/abc/abc.txt
/root/uname/abc.txt
Now, I want to print path till uname/ directory.
like: (expected output)
/root/tmp/uname
/root/tmp/uname
/root/uname
Need to extract path till any desired directory.
The variable substitution ${var%pattern} produces the value of var with any suffix matching pattern removed.
for p in /root/tmp/uname/abc.txt /root/tmp/uname/abc/abc.txt /root/uname/abc.txt
do
echo "${p%/uname*}/uname"
done
There is also ${var#pattern} to remove any matching prefix.
If the paths are in a file, use a while read instead of a for loop.
while read -r p; do
echo "${p%/uname*}/uname"
done <file
... though in that case, sed 's%\(/uname\)/.*%1%' file will be simpler and faster.
Suppose the file temp1.txt contains
root#nviewsrvr ~Competitive ] cat temp1.txt
/root/tmp/uname/abc.txt
/root/tmp/uname/abc/abc.txt
/root/uname/abc.txt
Following script takes the input from this file and gives you the desired result.
#!/bin/bash
cat temp1.txt|while read LINE
do
for p in `echo ${LINE}`
do
echo "${p%/uname*}/uname"
done
done
Results:
[xvishuk#ecamolx1820 ~/Competitive]$ ./forUname.sh
/root/tmp/uname
/root/tmp/uname
/root/uname
I have hardcoded the file as temp1.txt you can pass them as arguments to the script.
Related
Hello I am trying to get all files with Jane's name to a separate file called oldFiles.txt. In a directory called "data" I am reading from a list of file names from a file called list.txt, from which I put all the file names containing the name Jane into the files variable. Then I'm trying to test the files variable with the files in list.txt to ensure they are in the file system, then append the all the files containing jane to the oldFiles.txt file(which will be in the scripts directory), after it tests to make sure the item within the files variable passes.
#!/bin/bash
> oldFiles.txt
files= grep " jane " ../data/list.txt | cut -d' ' -f 3
if test -e ~data/$files; then
for file in $files; do
if test -e ~/scripts/$file; then
echo $file>> oldFiles.txt
else
echo "no files"
fi
done
fi
The above code gets the desired files and displays them correctly, as well as creates the oldFiles.txt file, but when I open the file after running the script I find that nothing was appended to the file. I tried changing the file assignment to a pointer instead files= grep " jane " ../data/list.txt | cut -d' ' -f 3 ---> files=$(grep " jane " ../data/list.txt) to see if that would help by just capturing raw data to write to file, but then the error comes up "too many arguments on line 5" which is the 1st if test statement. The only way I get the script to work semi-properly is when I do ./findJane.sh > oldFiles.txt on the shell command line, which is me essentially manually creating the file. How would I go about this so that I create oldFiles.txt and append to the oldFiles.txt all within the script?
The biggest problem you have is matching names like "jane" or "Jane's", etc. while not matching "Janes". grep provides the options -i (case insensitive match) and -w (whole-word match) which can tailor your search to what you appear to want without having to use the kludge (" jane ") of appending spaces before an after your search term. (to properly do that you would use [[:space:]]jane[[:space:]])
You also have the problem of what is your "script dir" if you call your script from a directory other than the one containing your script, such as calling your script from your $HOME directory with bash script/findJane.sh. In that case your script will attempt to append to $HOME/oldFiles.txt. The positional parameter $0 always contains the full pathname to the current script being run, so you can capture the script directory no matter where you call the script from with:
dirname "$0"
You are using bash, so store all the filenames resulting from your grep command in an array, not some general variable (especially since your use of " jane " suggests that your filenames contain whitespace)
You can make your script much more flexible if you take the information of your input file (e.g list.txt), the term to search for (e.g. "jane"), the location where to check for existence of the files (e.g. $HOME/data) and the output filename to append the names to (e.g. "oldFile.txt") as command line [positonal] parameters. You can give each default values so it behaves as you currently desire without providing any arguments.
Even with the additional scripting flexibility of taking the command line arguments, the script actually has fewer lines simply filling an array using mapfile (synonymous with readarray) and then looping over the contents of the array. You also avoid the additional subshell for dirname with a simple parameter expansion and test whether the path component is empty -- to replace with '.', up to you.
If I've understood your goal correctly, you can put all the pieces together with:
#!/bin/bash
# positional parameters
src="${1:-../data/list.txt}" # 1st param - input (default: ../data/list.txt)
term="${2:-jane}" # 2nd param - search term (default: jane)
data="${3:-$HOME/data}" # 3rd param - file location (defaut: ../data)
outfn="${4:-oldFiles.txt}" # 4th param - output (default: oldFiles.txt)
# save the path to the current script in script
script="$(dirname "$0")"
# if outfn not given, prepend path to script to outfn to output
# in script directory (if script called from elsewhere)
[ -z "$4" ] && outfn="$script/$outfn"
# split names w/term into array
# using the -iw option for case-insensitive whole-word match
mapfile -t files < <(grep -iw "$term" "$src" | cut -d' ' -f 3)
# loop over files array
for ((i=0; i<${#files[#]}; i++)); do
# test existence of file in data directory, redirect name to outfn
[ -e "$data/${files[i]}" ] && printf "%s\n" "${files[i]}" >> "$outfn"
done
(note: test expression and [ expression ] are synonymous, use what you like, though you may find [ expression ] a bit more readable)
(further note: "Janes" being plural is not considered the same as the singular -- adjust the grep expression as desired)
Example Use/Output
As was pointed out in the comment, without a sample of your input file, we cannot provide an exact test to confirm your desired behavior.
Let me know if you have questions.
As far as I can tell, this is what you're going for. This is totally a community effort based on the comments, catching your bugs. Obviously credit to Mark and Jetchisel for finding most of the issues. Notable changes:
Fixed $files to use command substitution
Fixed path to data/$file, assuming you have a directory at ~/data full of files
Fixed the test to not test for a string of files, but just the single file (also using -f to make sure it's a regular file)
Using double brackets — you could also use double quotes instead, but you explicitly have a Bash shebang so there's no harm in using Bash syntax
Adding a second message about not matching files, because there are two possible cases there; you may need to adapt depending on the output you're looking for
Removed the initial empty redirection — if you need to ensure that the file is clear before the rest of the script, then it should be added back, but if not, it's not doing any useful work
Changed the shebang to make sure you're using the user's preferred Bash, and added set -e because you should always add set -e
#!/usr/bin/env bash
set -e
files=$(grep " jane " ../data/list.txt | cut -d' ' -f 3)
for file in $files; do
if [[ -f $HOME/data/$file ]]; then
if [[ -f $HOME/scripts/$file ]]; then
echo "$file" >> oldFiles.txt
else
echo "no matching file"
fi
else
echo "no files"
fi
done
I want to read a list of file names stored in a file, and the top level directory is a macro, since this is for a script that may be run in several environments.
For example, there is a file file_list.txt holding the following fully qualified file paths:
$TOP_DIR/subdir_a/subdir_b/file_1
$TOP_DIR/subdir_x/subdir_y/subdir_z/file_2
In my script, I want to tar the files, but in order to do that, tar must know the actual path.
How can I get the string containing the file path to expand the macro to get the actual path?
In the code below the string value echoed is exactly as in the file above.
I tried using actual_file_path=`eval $file_path` and while eval does evaluate the macro, it returns a status, not the evaluated path.
for file_path in `cat $input_file_list`
do
echo "$file_path"
done
With the tag ksh I think you do not have the utility envsubst.
When the number of variables in $input_file_list is very limited, you can substitute vars with awk :
awk -v top_dir="${TOP_DIR}" '{ sub(/$TOP_DIR/, top_dir); print}' "${input_file_list}"
I was using eval incorrectly. The solution is to use an assignment on the right side of eval as follows:
for file_path in `cat $input_file_list`
do
eval myfile=$file_name
echo "myfile = $myfile"
done
$myfile now has the actual expansion of the macro.
I have a set of files places in a directory, and I want to read the second line from each file, extract the first substring which is placed between braces " () " and rename that file with this substing.
I'm not looking for a full bash code, I just need some hints and commands to use for each step
Example:
a file has these lines:
/* USER: 202166 (just_yousef) */
/* PROBLEM: 2954 (11854 - Egypt) */
/* SUBMISSION: 11071978 */
/* SUBMISSION TIME: 2012-12-25 15:49:25 */
/* LANGUAGE: 3 */
I need to take substring "11854 - Egypt" and rename the file with it, and proceed to the next file.
from each file
for f in /the/directory/*; do
# ...
done
read the second line [and] extract the first substring which is placed between [parentheses]
v=$(sed -n '2 { s/[^(]*(\([^)]*\)).*/\1/; p; q }' < "${f}")
rename that file
Use the mv command.
Use For each in the directory pipe the same with xargs and filter the files you want to read . and pass the file name with path to the Script which will do reading of second line.
For the script open the file and extract the String a per the logic.
THen close the file and use rename command to the file name you have as an input to this script.
rename $0 .txt
If the file structure is recurrrsive the extract the name before the last / using index of method of string.
Looks like sed may well be the tool for the job:
for i in *
do
j=`sed -e '1d;2{;s/.*(\(.*\)).*/\1/;q;}' "$i"`
test -z "$j" || test -e "$j" || mv -v "$i$ "$j"
done
I put the test for $j empty or already existing as safeguards; you might think of others. I also gave the -v flag to mv so you can see what it is doing.
You may prefer to use sed -n and just act on lines matching PROBLEM: if that's more reliable than always using the second line.
I have an arbitrary path which contains the directory mydir:
/some/path/to/mydir/further/path/file.ext
I want to get the part after mydir, in this example:
/further/path/file.ext
Please note that the levels of subdirectories are also arbitrary, so a path like
/yet/another/long/path/to/mydir/file.ext
is also possible (where the result would be "file.ext")
The first occurrence of mydir should be used, so the path
/path/mydir/some/other/path/mydir/path/file.ext
should result in
/some/other/path/mydir/path/file.ext
How can one do this with bash?
Note. It is assumed that mydir will always appear enclosed between slashes.
after=${mydir#*/mydir/}
if [ "$mydir" = "$after" ]; then
fail_with_error "Path does not contain /mydir/"
fi
after="/$after"
In line 1, the # means substring after, and the * is the usual placeholder. To be safe against directories like .../mydirectaccess/... I included the slashes at both ends of mydir. Line 5 just prepends the slash that had been taken off by line 1.
Using Shell Parameter Expansion:
$ mydir="/some/path/to/mydir/further/path/file.ext"
$ echo ${mydir#*mydir}
/further/path/file.ext
$ mydir="/path/mydir/some/other/path/mydir/path/file.ext"
$ echo ${mydir#*mydir}
/some/other/path/mydir/path/file.ext
Go through sed. Example:
echo /some/path/to/mydir/further/path/file.ext | sed 's/.*mydir/mydir/'
Using bash, you can do something like this:
V=/yet/another/long/path/to/mydir/file.ext
R=${V#*mydir/}
echo $R
file.ext
Let's say I've some files like:
samplea.txt
sampleb.txt
samplec.txt
And I want to run some command with this form:
./cmd -foo a.xml -bar samplea.txt
First I've tried to
for file in "./*.txt"
do
echo -e $file
done
But this way it will print every file in a straight line. By trying:
echo -e $file\n
It does not produce the expected (single line for each file).
Couldn't even pass through the first part of the problem, that would be running a command on each file (which it could be achieved by find (...) -exec), but what i really wanted to do was extract a substring of each name.
Doing:
echo ${file:1}
won't work since I could only do so after splitting the filenames, to get the "a","b","c" from each one.
I'm sorry if it sounds confusing, but it's my first bash script.
Do not quote the wildcard expression. You can use parameter expansion to remove parts of a string:
for file in sample*.txt ; do
part=${file#sample} # Remove "sample" at the beginning.
part=${part%.txt} # Remove ".txt" at the end.
./cmd -foo "$part".xml -bar "$file"
done