I have a giant list of files that are all currently numbered in sequential order with different file extensions.
3400.PDF
3401.xls
3402.doc
There are roughly 1400 of these files in a directory. What I would like to know is how to find numbers that do not exist in the sequence.
I've tried to write a bash script for this but my bash-fu is weak.
I can get a list of the files without their extensions by using
FILES=$(ls -1 | sed -e 's/\..*$//')
but a few places I've seen say to not use ls in this manner.
(15 days after asking, I couldn't relocate where I read this, if it existed at all...)
I can also get the first file via ls | head -n 1 but Im pretty sure I'm making this a whole lot more complicated that I need to.
Sounds like you want to do something like this:
shopt -s nullglob
for i in {1..1400}; do
files=($i.*)
(( ${#files[#]} > 0 )) || echo "no files beginning with $i";
done
This uses a glob to make an array of all files 1.*, 2.* etc. It then compares the length of the array to 0. If there are no files matching the pattern, the message is printed.
Enabling nullglob is important as otherwise, when there are no files matching the array will contain one element: the literal value '1.*'.
Based on deleted answer that was largely correct:
for i in $(seq 1 1400); do ls $i.* > /dev/null 2>&1 || echo $i; done
ls [0-9]* \
| awk -F. ' !seen[$1]++ { ++N }
END { for (n=1; N ; ++n) if (!seen[n]) print n; else --N }
'
Will stop when it's filled the last gap, sub in N>0 || n < 3000 to go at least that far.
Related
I have 2 files that I needed to grep in a separate file.
The two files are in this directory /var/list
TB.1234.txt
TB.135325.txt
I have to grep them in another file in another directory which is in /var/sup/. I used the command below:
for i in TB.*; do grep "$i" /var/sup/logs.txt; done
what I want to do is, if the result of the grep command contains the word "ERROR" the files which is found in /var/list will be moved to another directory /var/last.
for example I grep this file TB.1234.txt to /var/sup/logs.txt then the result is like this:
ERROR: TB.1234.txt
TB.1234.txt will be move to /var/last.
please help. I don't know how to construct the logic on how to move the files, I'm stuck in that I provided, I am also trying to use two greps in a for loop but I am encountering an error.
I am new in coding and really appreciates any help and suggestions. Thank you so much.
If you are asking how to move files which contain "ERROR", this should be extremely straightforward.
for file in TB.*; do
grep -q 'ERROR' "$file" &&
mv "$file" /var/last/
done
The notation this && that is a convenient shorthand for
if this; then
that
fi
The -q option to grep says to not print the matches, and quit as soon as you find one. Like all well-defined commands, grep sets its exit code to reflect whether it succeeded (the status is visible in $?, but usually you would not examine it directly; perhaps see also Why is testing ”$?” to see if a command succeeded or not, an anti-pattern?)
Your question is rather unclear, but if you want to find either of the matching files in a third file, perhaps something like
awk 'FNR==1 && (++n < ARGC-1) { a[n] = FILENAME; nextfile }
/ERROR/ { for(j=1; j<=n; ++j) if ($0 ~ a[j]) b[a[j]]++ }
END { for(f in b) print f }' TB*.txt /var/sup/logs.txt |
xargs -r mv -t /var/last/
This is somewhat inefficient in that it will read all the lines in the log file, and brittle in that it will only handle file names which do not contain newlines. (The latter restriction is probably unimportant here, as you are looking for file names which occur on the same line as the string "ERROR" in the first place.)
In some more detail, the Awk script collects the wildcard matches into the array a, then processes all lines in the last file, looking for ones with "ERROR" in them. On these lines, it checks if any of the file names in a are also found, and if so, also adds them to b. When all lines have been processed, print the entries in b, which are then piped to a simple shell command to move them.
xargs is a neat command to read some arguments from standard input, and run another command with those arguments added to its command line. The -r option says to not run the other command if there are no arguments.
(mv -t is a GNU extension; it's convenient, but not crucial to have here. If you need portable code, you could replace xargs with a simple while read -r loop.)
The FNR==1 condition requires that the input files are non-empty.
If the text file is small, or you expect a match near its beginning most of the time, perhaps just live with grepping it multiple times:
for file in TB.*; do
grep -Eq "ERROR.*$file|$file.*ERROR" /var/sup/logs.txt &&
mv "$file" /var/last/
done
Notice how we now need double quotes, not single, around the regular expression so that the variable $file gets substituted in the string.
grep has an -l switch, showing only the filename of the file which contains a pattern. It should not be too difficult to write something like (this is pseudocode, it won't work, it's just for giving you an idea):
if $(grep -l "ERROR" <directory> | wc -l) > 0
then foreach (f in $(grep -l "ERROR")
do cp f <destination>
end if
The wc -l is to check if there are any files which contain the word "ERROR". If not, nothing needs to be done.
Edit after Tripleee's comment:
My proposal can be simplified as:
if grep -lq "ERROR" TB.*;
then foreach (f in $(grep -l "ERROR")
do cp f <destination>
end if
Edit after Tripleee's second comment:
This is even shorter:
for f in $(grep -l "ERROR" TB.*);
do cp "$f" destination;
done
How do i print out all the files of the current directory that start with the letter "k" ?Also needs to count this files.
I tried some methods but i only got errors or wrong outputs. Really stuck on this as a newbie in bash.
Try this Shellcheck-clean pure POSIX shell code:
count=0
for file in k*; do
if [ -f "$file" ]; then
printf '%s\n' "$file"
count=$((count+1))
fi
done
printf 'count=%d\n' "$count"
It works correctly (just prints count=0) when run in a directory that contains nothing starting with 'k'.
It doesn't count directories or other non-files (e.g. fifos).
It counts symlinks to files, but not broken symlinks or symlinks to non-files.
It works with 'bash' and 'dash', and should work with any POSIX-compliant shell.
Here is a pure Bash solution.
files=(k*)
printf "%s\n" "${files[#]}"
echo "${#files[#]} files total"
The shell expands the wildcard k* into the array, thus populating it with a list of matching files. We then print out the array's elements, and their count.
The use of an array avoids the various problems with metacharacters in file names (see e.g. https://mywiki.wooledge.org/BashFAQ/020), though the syntax is slightly hard on the eyes.
As remarked by pjh, this will include any matching directories in the count, and fail in odd ways if there are no matches (unless you set nullglob to true). If avoiding directories is important, you basically have to get the directories into a separate array and exclude those.
To repeat what Dominique also said, avoid parsing ls output.
Demo of this and various other candidate solutions:
https://ideone.com/XxwTxB
To start with: never parse the output of the ls command, but use find instead.
As find basically goes through all subdirectories, you might need to limit that, using the -maxdepth switch, use value 1.
In order to count a number of results, you just count the number of lines in your output (in case your output is shown as one piece of output per line, which is the case of the find command). Counting a number of lines is done using the wc -l command.
So, this comes down to the following command:
find ./ -maxdepth 1 -type f -name "k*" | wc -l
Have fun!
This should work as well:
VAR="k"
COUNT=$(ls -p ${VAR}* | grep -v ":" | wc -w)
echo -e "Total number of files: ${COUNT}\n" 1>&2
echo -e "Files,that begin with ${VAR} are:\n$(ls -p ${VAR}* | grep -v ":" )" 1>&2
Assuming that I have files with 100 lines. There are a lot of lines that repeat themselves in the file, and only one line that does not.
I want to find the line that shows only once. Is there a command for that or do I have to build some complicated loop as below?
My code so far:
#!/bin/bash
filename="repeat_lines.txt"
var="$(wc -l <$filename )"
echo "length:" $var
#cp ex4.txt ex4_copy.txt
for((index=0; index < var; index++));
do
one="$(head -n $index $filename | tail -1)"
counter=0
for((index2=0; index2 < var; index2++));
do
two="$(head -n $index2 $filename | tail -1)"
if [ "$one" == "$two" ]; then
counter=$((counter+1))
fi
done
echo $one"is "$counter" times in the text: "
done
If I understood your question correctly, then
sort repeat_lines.txt | uniq -u should do the trick.
e.g. for file containing:
a
b
a
c
b
it will output c.
For further reference, see sort manpage, uniq manpage.
You've got a reasonable answer that uses standard shell tools sort and uniq. That's probably the solution you want to use, if you want something that is portable and doesn't require bash.
But an alternative would be to use functionality built into your bash shell. One method might be to use an associative array, which is a feature of bash 4 and above.
$ cat file.txt
a
b
c
a
b
$ declare -A lines
$ while read -r x; do ((lines[$x]++)); done < file.txt
$ for x in "${!lines[#]}"; do [[ ${lines["$x"]} -gt 1 ]] && unset lines["$x"]; done
$ declare -p lines
declare -A lines='([c]="1" )'
What we're doing here is:
declare -A creates the associative array. This is the bash 4 feature I mentioned.
The while loop reads each line of the file, and increments a counter that uses the content of a line of the file as the key in the associative array.
The for loop steps through the array, deleting any element whose counter is greater than 1.
declare -p prints the details of an array in a predictable, re-usable format. You could alternately use another for loop to step through the remaining array elements (of which there might be only one) in order to do something with them.
Note that this solution, while fine for small files (say, up to a few thousand lines), may not scale well for very large files of, say, millions of lines. Bash isn't the fastest at reading input this way, and one must be cognizant of memory limits when using arrays.
The sort alternative has the benefit of memory optimization using files on disk for extremely large files, at the expense of speed.
If you're dealing with files of only a few hundred lines, then it's hard to predict which solution will be faster. In the end, the form of output may dictate your choice of solution. The sort | uniq pipe generates a list to standard output. The bash solution above generates the same list as keys in an array. Otherwise, they are functionally equivalent.
I want to rename all files in a directory to be sequential numbers:
1.txt
2.txt
3.txt
and so on...
Here's the code I'm currently using:
ls | cat -n | while read n f; do mv "$f" "$n.txt"; done
The code does work, but I need to start with a specific number. For example, I may want to start with the number 49 instead of the number 1.
Is there any way to do this in terminal (on a Mac)?
You could use something like nl with the -v option to set a starting line number other than 1, but instead, you can just use Bash features:
i=1
for f in *; do
[[ -f $f ]] && mv "$f" $((i++)).txt
done
where i is set to the initial value you want.
This also avoids parsing the output of ls, which is recommended to avoid. Instead, I use a glob (*) and a test (-f) to make sure that I'm actually manipulating files and not directories.
#! /bin/bash
for i in $(ls);
do
j=1
echo "$i"
not expected Output:-
autodeploy
bin
config
console-ext
edit.lok
need Output like below if give input 2 it should print "bin" based on below condition, but I want out put like Directory list
1.)autodeploy
2.)bin
3.)config
4.)console-ext
5.)edit.lok
and if i like as input:- 2 then it should print "bin"
Per BashFAQ #1, a while read loop is the correct way to read content line-by-line:
#!/usr/bin/env bash
enumerate() {
local line i
i=0
while IFS= read -r line; do
((++i))
printf '%d.) %s\n' "$i" "$line"
done
}
ls | enumerate
However, ls is not an appropriate tool for programmatic use; the above is acceptable if the results of ls are only for human consumption, but not if they're going to be parsed by a machine -- see Why you shouldn't parse the output of ls(1).
If you want to list files and let the user choose among them by number, pass the results of a glob expression to select:
select filename in *; do
echo "$filename" && break
done
I don't understand what you mean in your question by like Directory list, but following your example, you do not need to write a loop:
ls|nl -s '.)' -w 1
If you want to avoid ls, you can do the following (but be careful - this only works if the directory entries do not contain white spaces (because this would make fmt to break them into two lines):
echo *|fmt -w 1 |nl -s '.)' -w 1