I have piped the output of ls command into a file. The contents are like so:
[Chihiro]_Grisaia_no_Kajitsu_-_01_[1920x816_Blu-ray_FLAC][D2B961D6].mkv
[Chihiro]_Grisaia_no_Kajitsu_-_02_[1920x816_Blu-ray_FLAC][38F88A81].mkv
[Chihiro]_Grisaia_no_Kajitsu_-_03_[1920x816_Blu-ray_FLAC][410F74F7].mkv
My attempt to rename these episodes according to episode number is as follows:
cat grisaia | while read line;
#get the episode number
do EP=$(echo $line | egrep -o "_([0-9]{2})_" | cut -d "_" -f2)
if [[ $EP ]]
#escape special characters
then line=$(echo $line | sed 's/\[/\\[/g' | sed 's/\]/\\]/g')
mv "$line" "Grisaia_no_Kajitsu_${EP}.mkv"
fi
done
The mv commands exit with code 1 with the following error:
mv: cannot stat
'\[Chihiro\]_Grisaia_no_Kajitsu_-01\[1920x816_Blu-ray_FLAC\]\[D2B961D6\].mkv':
No such file or directory
What I really don't get is that if I copy the file that could not be stat and attempt to stat the file, it works. I can even take the exact same string that is output and execute the mv command individually.
If you surround your variable ($line) with double quotes (") you don't need to escape those special characters. So you have two options there:
Remove the following assignation completely:
then # line=$(echo $line | sed 's/\[/\\[/g' | sed 's/\]/\\]/g')`
or
Remove the double quotes in the following line:
mv $line "Grisaia_no_Kajitsu_${EP}.mkv"
Further considerations
Parsing the output of ls is never a good idea. Think about filenames with spaces. See this document for more information.
The cat here is unnecessary:
cat grisaia | while read line;
...
done
Use this instead to avoid an unnecessary pipe:
while read line;
...
done < grisaia
Why is good to avoid pipes in some scenarios? (answering comment)
Pipes create subshells (which are expensive), and you can also make some mistakes as the following:
last=""
cat grisaia | while read line; do
last=$line
done
echo $last # surprise!! it outputs an empty string
The reason is that $last inside the loop belongs to another subshell.
Now, see the same approach wothout pipes:
while read line; do
last=$line
done < grisaia
echo $last # it works as expected and prints the last line
Related
it is my first question here. I have a folder called "materials", which has 40 text files in it. I am basically trying to combine the text files that contain the word "carbon"(both in capitalized and lowercase form)in it into a single file with leaving newlines between them. I used " grep -w carbon * " to identify the files that contain the word carbon. I just don't know what to do after this point. I really appreciate all your help!
grep -il carbon materials/*txt | while read line; do
echo ">> Adding $line";
cat $line >> result.out;
echo >> result.out;
done
Explanation
grep searches the strings in the files. -i ignores the case for the searched string. -l prints on the filename containing the string
while command loops over the files containing the string
cat with >> appends to the results.out
echo >> adds new line after appending each files content to result.out
Execution
$ ls -1 materials/*.txt
materials/1.txt
materials/2.txt
materials/3.txt
$ grep -i carbon materials/*.txt
materials/1.txt:carbon
materials/2.txt:CARBON
$ grep -irl carbon materials/*txt | while read line; do echo ">> Adding $line"; cat $line >> result.out; echo >> result.out; done
>> Adding materials/1.txt
>> Adding materials/2.txt
$ cat result.out
carbon
CARBON
Try this (assuming your filenames don't contain newline characters):
grep -iwl carbon ./* |
while IFS= read -r f; do cat "$f"; echo; done > /tmp/combined
If it is possible that your filenames may contain newline characters and your shell is bash, then:
grep -iwlZ carbon ./* |
while IFS= read -r -d '' f; do cat "$f"; echo; done > /tmp/combined
grep is assumed to be GNU grep (for the -w and -Z options). Note that these will leave a trailing newline character in the file /tmp/combined.
In a bash script, I am trying to remove the directory name in filenames :
documents/file.txt
direc/file5.txt
file2.txt
file3.txt
So I try to first see if there is a "/" and if yes delete everything before :
for i in **/*.scss *.scss; do
echo "$i" | sed -n '^/.*\// s/^.*\///p'
done
But it doesn't work for files in the current directory, it gives me a blank string.
I get :
file.txt
file5.txt
When you only want the filename, use basename instead of sed.
# basename /path/to/file
returns file
here is the man page
Your sed attempt is basically fine, but you should print regardless of whether you performed a substitution; take out the -n and the p at the end. (Also there was an unrelated syntax error.)
Also, don't needlessly loop over all files.
printf '%s\n' **/*.scss *.scss |
sed -n 's%^.*/%%p'
This also can be done with awk bash util.
Example:
echo "1/2/i.py" | awk 'BEGIN {FS="/"} {print $NF}'
output: i.py
Eventually, I did :
for i in **/*.scss *.scss; do
# for i in *.scss; do
# for i in _hm-globals.scss; do
name=${i##*/} # remove dir name
name=${name%.scss} # remove extension
name=`echo "$name" | sed -n "s/^_hm-//p"` # remove _hm-
if [[ $name = *"."* ]]; then
name=`echo "$name" | sed -n 's/\./-/p'` #replace . to --
fi
echo "$name" >&2
done
I am trying to remove newlines from a file. My file is like this (it contains backward slashes):
line1\|
line2\|
I am using the following script to remove newlines:
#!/bin/bash
INPUT="file1"
while read line
do
: echo -n $line
done < $INPUT
I get the following output:
line1|line2|
It removes the backslashes. How can I retain those backslashes?
The -r option to read prevents backslash processing of the input.
while read -r line
do
echo -n "$line"
done < $INPUT
But if you just want to remove all newlines from the input, the tr command would be better:
tr -d '\n' < $INPUT
Try sed 's/\n//' /path/to/file
I have written the following filter as a function in my ~/.bash_profile:
hilite() {
export REGEX_SED=$(echo $1 | sed "s/[|()]/\\\&/g")
while read line
do
echo $line | egrep "$1" | sed "s/$REGEX_SED/\x1b[7m&\x1b[0m/g"
done
exit 0
}
to find lines of anything piped into it matching a regular expression, and highlight matches using ANSI escape codes on a VT100-compatible terminal.
For example, the following finds and highlights the strings bin, U or 1 which are whole words in the last 10 lines of /etc/passwd:
tail /etc/passwd | hilite "\b(bin|[U1])\b"
However, the script runs very slowly as each line forks an echo, egrep and sed.
In this case, it would be more efficient to do egrep on the entire input, and then run sed on its output.
How can I modify my function to do this? I would prefer to not create any temporary files if possible.
P.S. Is there another way to find and highlight lines in a similar way?
sed can do a bit of grepping itself: if you give it the -n flag (or #n instruction in a script) it won't echo any output unless asked. So
while read line
do
echo $line | egrep "$1" | sed "s/$REGEX_SED/\x1b[7m&\x1b[0m/g"
done
could be simplified to
sed -n "s/$REGEX_SED/\x1b[7m&\x1b[0m/gp"
EDIT:
Here's the whole function:
hilite() {
REGEX_SED=$(echo $1 | sed "s/[|()]/\\\&/g");
sed -n "s/$REGEX_SED/\x1b[7m&\x1b[0m/gp"
}
That's all there is to it - no while loop, reading, grepping, etc.
If your egrep supports --color, just put this in .bash_profile:
hilite() { command egrep --color=auto "$#"; }
(Personally, I would name the function egrep; hence the usage of command).
I think you can replace the whole while loop with simply
sed -n "s/$REGEX_SED/\x1b[7m&\x1b[0m/gp"
because sed can read from stdin line-by-line so you don't need read
I'm not sure if running egrep and piping to sed is faster than using sed alone, but you can always compare using time.
Edit: added -n and p to sed to print only highlighted lines.
Well, you could simply do this:
egrep "$1" $line | sed "s/$REGEX_SED/\x1b[7m&\x1b[0m/g"
But I'm not sure that it'll be that much faster ; )
Just for the record, this is a method using a temporary file:
hilite() {
export REGEX_SED=$(echo $1 | sed "s/[|()]/\\\&/g")
export FILE=$2
if [ -z "$FILE" ]
then
export FILE=~/tmp
echo -n > $FILE
while read line
do
echo $line >> $FILE
done
fi
egrep "$1" $FILE | sed "s/$REGEX_SED/\x1b[7m&\x1b[0m/g"
return $?
}
which also takes a file/pathname as the second argument, for case like
cat /etc/passwd | hilite "\b(bin|[U1])\b"
I'm trying to read file line by line in bash.
Every line has format as follows text|number.
I want to produce file with format as follows text,text,text etc. so new file would have just text from previous file separated by comma.
Here is what I've tried and couldn't get it to work :
FILENAME=$1
OLD_IFS=$IFSddd
IFS=$'\n'
i=0
for line in $(cat "$FILENAME"); do
array=(`echo $line | sed -e 's/|/,/g'`)
echo ${array[0]}
i=i+1;
done
IFS=$OLD_IFS
But this prints both text and number but in different format text number
here is sample input :
dsadadq-2321dsad-dasdas|4212
dsadadq-2321dsad-d22as|4322
here is sample output:
dsadadq-2321dsad-dasdas,dsadadq-2321dsad-d22as
What did I do wrong?
Not pure bash, but you could do this in awk:
awk -F'|' 'NR>1{printf(",")} {printf("%s",$1)}'
Alternately, in pure bash and without having to strip the final comma:
#/bin/bash
# You can get your input from somewhere else if you like. Even stdin to the script.
input=$'dsadadq-2321dsad-dasdas|4212\ndsadadq-2321dsad-d22as|4322\n'
# Output should be reset to empty, for safety.
output=""
# Step through our input. (I don't know your column names.)
while IFS='|' read left right; do
# Only add a field if it exists. Salt to taste.
if [[ -n "$left" ]]; then
# Append data to output string
output="${output:+$output,}$left"
fi
done <<< "$input"
echo "$output"
No need for arrays and sed:
while IFS='' read line ; do
echo -n "${line%|*}",
done < "$FILENAME"
You just have to remove the last comma :-)
Using sed:
$ sed ':a;N;$!ba;s/|[0-9]*\n*/,/g;s/,$//' file
dsadadq-2321dsad-dasdas,dsadadq-2321dsad-d22as
Alternatively, here is a bit more readable sed with tr:
$ sed 's/|.*$/,/g' file | tr -d '\n' | sed 's/,$//'
dsadadq-2321dsad-dasdas,dsadadq-2321dsad-d22as
Choroba has the best answer (imho) except that it does not handle blank lines and it adds a trailing comma. Also, mucking with IFS is unnecessary.
This is a modification of his answer that solves those problems:
while read line ; do
if [ -n "$line" ]; then
if [ -n "$afterfirst" ]; then echo -n ,; fi
afterfirst=1
echo -n "${line%|*}"
fi
done < "$FILENAME"
The first if is just to filter out blank lines. The second if and the $afterfirst stuff is just to prevent the extra comma. It echos a comma before every entry except the first one. ${line%|\*} is a bash parameter notation that deletes the end of a paramerter if it matches some expression. line is the paramter, % is the symbol that indicates a trailing pattern should be deleted, and |* is the pattern to delete.