I have a text file like this (e.g., a.txt):
1.1.t
1.2.m
If the last line consists of the character m, I want to echo Ok.
I tried this:
line=` awk '/./{line=$0} END{print line}' a.txt`
line1= `echo $line | grep "m"`
if [[ $line1= `:` ]] ; then
echo
else
echo "Ok"
fi
It does not work, and the error is:
bash: conditional binary operator expected
bash: syntax error near ``:`,
`if [[ $line1= `:` ]] ; then'
if [[ $line1=:]] is incorrect syntax in couple of ways as spaces around = are missing and backtick is used for command substitution
awk itself can handle this:
awk '/./{line=$0} END{print (line ~ /\.m/)? "ok" : "no"}' file
ok
You could also use tail and grep:
[[ -n $(tail -1 a.txt | grep "m$") ]] && echo "OK" || echo "FAILED"
You can use sed:
sed -n '${/m$/s/.*/OK/p;}' file
The option -n suppresses output by default. $ addresses the last line of input. In that case we check if the line ends with m through /m$/. If that is the case we substitute the line with the word OK and print it.
Btw, I was going trough your shell script, there are really too many errors to explain, the syntax error is because there is no space between $line1 and the = in the [[ ... ]] conditional. But hey, this is far from being the only problem with that script. ;)
http://www.shellcheck.net/ might be a good resource to enhance your scripts.
Related
I am attempting to return the line number of lines that have a break. An input example:
2938
383
3938
3
383
33333
But my script is not working and I can't see why. My script:
input="./input.txt"
declare -i count=0
while IFS= read -r line;
do
((count++))
if [ "$line" == $'\n\n' ]; then
echo "$count"
fi
done < "$input"
So I would expect, 3, 6 as output.
I just receive a blank response in the terminal when I execute. So there isn't a syntax error, something else is wrong with the approach I am taking. Bit stumped and grateful for any pointers..
Also "just use awk" doesn't help me. I need this structure for additional conditions (this is just a preliminary test) and I don't know awk syntax.
The issue is that "$line" == $'\n\n' won't match a newline as it won't be there after consuming an empty line from the input, instead you can match an empty line with regex pattern ^$:
if [[ "$line" =~ ^$ ]]; then
Now it should work.
It's also match easier with awk command:
$ awk '$0 == ""{ print NR }' test.txt
3
6
As Roman suggested, line read by read terminates with a delimiter, and that delimiter would not show up in the line the way you're testing for.
If the pattern you are searching for looks like an empty line (which I infer is how a "double newline" always manifests), then you can just test for that:
while read -r; do
((count++))
if [[ -z "$REPLY" ]]; then
echo "$count"
fi
done < "$input"
Note that IFS is for field-splitting data on lines, and since we're only interested in empty lines, IFS is moot.
Or if the file is small enough to fit in memory and you want something faster:
mapfile -t -O1 foo < i
declare -p foo
for n in "${!foo[#]}"; do
if [[ -z "${foo[$n]}" ]]; then
echo "$n"
fi
done
Reading the file all at once (mapfile) then stepping through an array may be easier on resources than stepping through a file line by line.
You can also just use GNU awk:
gawk -v RS= -F '\n' '{ print (i += NF); i += length(RT) - 1 }' input.txt
By using FS = ".+", it ensures only truly zero-length (i.e. $0 == "") line numbers get printed, while skipping rows consisting entirely of [[:space:]]'s
echo '2938
383
3938
3
383
33333' |
{m,g,n}awk -F'.+' '!NF && $!NF = NR'
3
6
This sed one-liner should do the job at once:
sed -n '/^$/=' input.txt
Simply writes the current line number (the = command) if the line read is empty (the /^$/ matches the empty line).
I am trying to collect the lines from a file which doesn't start with a # as its first caracter.
I have this code I am able to get them:
while IFS= read -r line
do
[[ -z "$line" ]] && continue
[[ "$line" =~ ^# ]] && continue
#echo "LINEREADED: $line"
done < $file
So the output I have is something like this:
modules/core_as/xxxx/xxxxxxxxxxxxxxxxxxxxxxxxxxx [100]
My question is how can I get only the string without the [100]?
I know there is some commands like sed or trim but the problem is that the string is not always that length, sometimes is different like:
cross_modules/core_as/xxxx/xxxxxxxxx [100-103]
or
cross_modules/core_as/xxxxxxxxxxxx/xxxxxxxxx [100-103]
or anything like that...
And in all this cases I only need the string without the [....] and without the last blank space at the end of last x, whichever the length of the string is, like cross_modules/core_as/xxxxxxxxxxxx/xxxxxxxxx
echo ${caseReaded:1:${#caseReaded}-7}
This also do the job but is not generic for any length.
Does anyone knows how I can get this?
You can strip a certain part of a string in bash
echo "${line% [*}"
cross_modules/core_as/xxxx/xxxxxxxxx
modules/core_as/xxxx/xxxxxxxxxxxxxxxxxxxxxxxxxxx
cross_modules/core_as/xxxxxxxxxxxx/xxxxxxxxx
If the spaces are only before [:
while IFS= read -r line _
do
[[ -z $line ]] && continue
[[ $line =~ ^# ]] && continue
done < "$file"
grep to match all lines not starting with # and then display the first field using cut, which works if the first field doesn't contain spaces:
grep -v ^# "$file" | cut -f1 -d' '
If the thing before [100] contains spaces, this may be the way to go:
grep -v ^# "$file" | sed -E 's/^(.*) .*$/\1/'
The last one works because the .* match in sed is greedy so only the last space will be left to match the outer condition .*$.
What I have is a file (let's call it 'xfile'), containing lines such as
file1 <- this line goes to file1
file2 <- this goes to file2
and what I want to do is run a script that does the work of actually taking the lines and writing them into the file.
The way I would do that manually could be like the following (for the first line)
(echo "this line goes to file1"; echo) >> file1
So, to automate it, this is what I tried to do
IFS=$'\n'
for l in $(grep '[a-z]* <- .*' xfile); do
$(echo $l | sed -e 's/\([a-z]*\) <- \(.*\)/(echo "\2"; echo)\>\>\1/g')
done
unset IFS
But what I get is
-bash: file1(echo "this content goes to file1"; echo)>>: command not found
-bash: file2(echo "this goes to file2"; echo)>>: command not found
(on OS X)
What's wrong?
This solves your problem on Linux
awk -F ' <- ' '{print $2 >> $1}' xfile
Take care in choosing field-separator in such a way that new files does not have leading or trailing spaces.
Give this a try on OSX
You can use the regex capabilities of bash directly. When you use the =~ operator to compare a variable to a regular expression, bash populates the BASH_REMATCH array with matches from the groups in the regex.
re='(.*) <- (.*)'
while read -r; do
if [[ $REPLY =~ $re ]]; then
file=${BASH_REMATCH[1]}
line=${BASH_REMATCH[2]}
printf '%s\n' "$line" >> "$file"
fi
done < xfile
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.
I am using sh shell script to read the files of a folder and display on the screen:
for d in `ls -1 $IMAGE_DIR | egrep "jpg$"`
do
pgm_file=$IMAGE_DIR/`echo $d | sed 's/jpg$/pgm/'`
echo "file $pgm_file";
done
the output result is reading line by line:
file file1.jpg
file file2.jpg
file file3.jpg
file file4.jpg
Because I am not familiar with this language, I would like to have the result that print first 2 results in the same row like this:
file file1.jpg; file file2.jpg;
file file3.jpg; file file4.jpg;
In other languages, I just put d++ but it does not work with this case.
Would this be doable? I will be happy if you would provide me sample code.
thanks in advance.
Let the shell do more work for you:
end_of_line=""
for d in "$IMAGE_DIR"/*.jpg
do
file=$( basename "$d" )
printf "file %s; %s" "$file" "$end_of_line"
if [[ -z "$end_of_line" ]]; then
end_of_line=$'\n'
else
end_of_line=""
fi
pgm_file=${d%.jpg}.pgm
# do something with "$pgm_file"
done
for d in "$IMAGE_DIR"/*jpg; do
pgm_file=${d%jpg}pgm
printf '%s;\n' "$d"
done |
awk 'END {
if (ORS != RS)
print RS
}
ORS = NR % n ? FS : RS
' n=2
Set n to whatever value you need.
If you're on Solaris, use nawk or /usr/xpg4/bin/awk
(do not use /usr/bin/awk).
Note also that I'm trying to use a standard shell syntax,
given your question is sh related (i.e. you didn't mention bash or ksh,
for example).
Something like this inside the loop:
echo -n "something; "
[[ -n "$oddeven" ]] && oddeven= || { echo;oddeven=x;}
should do.
Three per line would be something like
[[ "$((n++%3))" = 0 ]] && echo
(with n=1) before entering the loop.
Why use a loop at all? How about:
ls $IMAGE_DIR | egrep 'jpg$' |
sed -e 's/$/;/' -e 's/^/file /' -e 's/jpg$/pgm/' |
perl -pe '$. % 2 && chomp'
(The perl just deletes every other newline. You may want to insert a space and add a trailing newline if the last line is an odd number.)