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
Related
I have read the config file which has the below variable:
export BASE_DIR="\usr\usr1"
In the same script I read a file line by line and I wanted to substitute the ${BASE_DIR} with \usr\usr1.
In the script:
while read line; do
echo $line
done <file.txt
${BASE_DIR}\path1 should be printed as \usr\usr1\path1
Tried eval echo and $(( )).
Can use sed, This command will search and replace a value. The dollar sign is the separator.
sed -ie 's$\${BASE_DIR}$\\usr\\usr1$1' hello.txt
You need to set the variable when you read the line that contains the assignment. Then you can replace it later.
#!/bin/bash
while read line; do
if [[ $line =~ ^BASE_DIR= ]]
then basedir=${line#BASE_DIR=}
fi
line=${line/'${BASE_DIR}'/$basedir}
printf "%s\n" "$line"
done < file.txt > newfile.txt
In bash script, how can I read the file line by line and assign to the variable with delimiter?
example.txt file contents:
string1
string2
string3
string4
Expected output:
string1,string2,string3,string4
Thanks in advance
Apparently my answer below leaves a comma at the end of the line. A quick workaround is to use the following builtin in Unix:
paste -sd, example.txt
Where you use the paste program to concatenate all the lines into one and then add the string delimiter ','
Using the builtin commands in unix:
tr '\n' ',' < example.txt
This can be broken down as truncating all Newline widcards and inserting a comma delimiter instead.
Other possible ways, just for fun:
mapfile -t a < example.txt
(IFS=,; echo "${a[*]}")
mapfile -t a < example.txt
foo=$(printf '%s' "${a[#]/%/,}")
echo "${foo%,}"
foo=$(<example.txt)
echo "${foo//$'\n'/,}"
{
IFS= read -r foo
while IFS= read -r line; do
foo+=,$line
done
} < example.txt
echo "$foo"
sed ':a;N;$!ba;s/\n/,/g' example.txt
It should work:
#!/bin/bash
output=''
while IFS='' read -r line || [[ -n "$line" ]]; do
output=$output:",$line"
done < "$1"
echo $output
Give the file as argument
I have piped the output of ls command into a file. The contents are like so:
[Chihiro]_Grisaia_no_Kajitsu_-_01_[1920x816_Blu-ray_FLAC][D2B961D6].mkv
[Chihiro]_Grisaia_no_Kajitsu_-_02_[1920x816_Blu-ray_FLAC][38F88A81].mkv
[Chihiro]_Grisaia_no_Kajitsu_-_03_[1920x816_Blu-ray_FLAC][410F74F7].mkv
My attempt to rename these episodes according to episode number is as follows:
cat grisaia | while read line;
#get the episode number
do EP=$(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
I need you help on this;
I am currently trying to apply a sed command to lines from a file.
2014-08-05T09:29:13+01:00 (INFO:3824.87075728): [27219] [ <email#domain.com>] A message from <user1#domain.com> source <asdfg> this is a test.
I need to apply this sed cmd to this line but keep this others that does not have 'this is a test'
pattern="this\ is\ a test"
while IFS='' read -r line; do
if [[ $line = *"${pattern}"* ]]; then
sed 's/\[ .*\(source\)/\1/g' ${line}
else
echo "${line}"
fi
done < ${INPUT} > ${OUPUT}
I have set the input and output; however ideally keeping the same file would be ideal.
Thank you for your input.
You don't need a loop for this. Use this sed:
sed -i.bak '/this is a test/s/\[ .*\(source\)/\1/g' "${INPUT}"
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.