Loop to remove inverted commas (") in a dataset in bash - bash

I would like to find a way to use an iterative loop (for or while) to remove quotes (") from a dataset like the following:
",great britain,"America"
I know that there is the following way to remove them, which is actually more efficient, but it is important that it is an iterative statement in bash
sed -i -e 's/"//g' file.csv
Any idea?

You can use while read -r to read the lines. Use Parameter Expansion to remove the double quotes.
while read -r line ; do printf '%s\n' "${line//\"}" ; done < input

Related

echo inside a for loop lists files matching the output pattern

I have a problem with the following for loop:
X="*back* OLD"
for P in $X
do
echo "-$P"
done
I need it to output just:
-*back*
-OLD
However, it lists all files in the current directory matching the *back* pattern. For example it gives the following:
-backup.bkp
-backup_new.bkp
-backup_X
-OLD
How to force it to output the exact pattern?
Use an array, as unquoted parameter expansions are still subject to globbing.
X=( "*back*" OLD )
for P in "${X[#]}"; do
printf '%s\n' "$P"
done
(Use printf, as echo could try to interpret an argument as an option, for example, if you had n in the value of X.)
Use set -o noglob before your loop and set +o noglob after to disable and enable globbing.
To prevent filename expansion you could read in the string as a Here String.
To iterate over the items, you could turn them into lines using parameter expansion and read them linewise using read. In order to be able to put a - sign as the first character, use printf instead of echo.
X="*back* OLD"
while read -r x
do printf -- '-%s\n' "$x"
done <<< "${X/ /$'\n'}"
Another way could be to use tr to transform the string into lines, then use paste with the - sign as delimiter and "nothing" from /dev/null as first column.
X="*back* OLD"
tr ' ' '\n' <<< "$X" | paste -d- /dev/null -
Both should output:
-*back*
-OLD

Using bash, how to pass filename arguments to a command sorted by date and dealing with spaces and other special characters?

I am using the bash shell and want to execute a command that takes filenames as arguments; say the cat command. I need to provide the arguments sorted by modification time (oldest first) and unfortunately the filenames can contain spaces and a few other difficult characters such as "-", "[", "]". The files to be provided as arguments are all the *.txt files in my directory. I cannot find the right syntax. Here are my efforts.
Of course, cat *.txt fails; it does not give the desired order of the arguments.
cat `ls -rt *.txt`
The `ls -rt *.txt` gives the desired order, but now the blanks in the filenames cause confusion; they are seen as filename separators by the cat command.
cat `ls -brt *.txt`
I tried -b to escape non-graphic characters, but the blanks are still seen as filename separators by cat.
cat `ls -Qrt *.txt`
I tried -Q to put entry names in double quotes.
cat `ls -rt --quoting-style=escape *.txt`
I tried this and other variants of the quoting style.
Nothing that I've tried works. Either the blanks are treated as filename separators by cat, or the entire list of filenames is treated as one (invalid) argument.
Please advise!
Using --quoting-style is a good start. The trick is in parsing the quoted file names. Backticks are simply not up to the job. We're going to have to be super explicit about parsing the escape sequences.
First, we need to pick a quoting style. Let's see how the various algorithms handle a crazy file name like "foo 'bar'\tbaz\nquux". That's a file name containing actual single and double quotes, plus a space, tab, and newline to boot. If you're wondering: yes, these are all legal, albeit unusual.
$ for style in literal shell shell-always shell-escape shell-escape-always c c-maybe escape locale clocale; do printf '%-20s <%s>\n' "$style" "$(ls --quoting-style="$style" '"foo '\''bar'\'''$'\t''baz '$'\n''quux"')"; done
literal <"foo 'bar' baz
quux">
shell <'"foo '\''bar'\'' baz
quux"'>
shell-always <'"foo '\''bar'\'' baz
quux"'>
shell-escape <'"foo '\''bar'\'''$'\t''baz '$'\n''quux"'>
shell-escape-always <'"foo '\''bar'\'''$'\t''baz '$'\n''quux"'>
c <"\"foo 'bar'\tbaz \nquux\"">
c-maybe <"\"foo 'bar'\tbaz \nquux\"">
escape <"foo\ 'bar'\tbaz\ \nquux">
locale <‘"foo 'bar'\tbaz \nquux"’>
clocale <‘"foo 'bar'\tbaz \nquux"’>
The ones that actually span two lines are no good, so literal, shell, and shell-always are out. Smart quotes aren't helpful, so locale and clocale are out. Here's what's left:
shell-escape <'"foo '\''bar'\'''$'\t''baz '$'\n''quux"'>
shell-escape-always <'"foo '\''bar'\'''$'\t''baz '$'\n''quux"'>
c <"\"foo 'bar'\tbaz \nquux\"">
c-maybe <"\"foo 'bar'\tbaz \nquux\"">
escape <"foo\ 'bar'\tbaz\ \nquux">
Which of these can we work with? Well, we're in a shell script. Let's use shell-escape.
There will be one file name per line. We can use a while read loop to read a line at a time. We'll also need IFS= and -r to disable any special character handling. A standard line processing loop looks like this:
while IFS= read -r line; do ... done < file
That "file" at the end is supposed to be a file name, but we don't want to read from a file, we want to read from the ls command. Let's use <(...) process substitution to swap in a command where a file name is expected.
while IFS= read -r line; do
# process each line
done < <(ls -rt --quoting-style=shell-escape *.txt)
Now we need to convert each line with all the quoted characters into a usable file name. We can use eval to have the shell interpret all the escape sequences. (I almost always warn against using eval but this is a rare situation where it's okay.)
while IFS= read -r line; do
eval "file=$line"
done < <(ls -rt --quoting-style=shell-escape *.txt)
If you wanted to work one file at a time we'd be done. But you want to pass all the file names at once to another command. To get to the finish line, the last step is to build an array with all the file names.
files=()
while IFS= read -r line; do
eval "files+=($line)"
done < <(ls -rt --quoting-style=shell-escape *.txt)
cat "${files[#]}"
There we go. It's not pretty. It's not elegant. But it's safe.
Does this do what you want?
for i in $(ls -rt *.txt); do echo "FILE: $i"; cat "$i"; done

Variable comparison and pattern replacement

I am trying to create a bash script that uses the sed command to replace a pattern by a variable that contains a string or put a space if there is nothing in the variable. I cannot find the good way to write it and make it work. Here is the part where I have issues:
a_flag=$(echo $a | wc -w)
if [[ $a_flag = 0 ]]; then
sed -i -e 's/#b/\\hspace{2cm}/g' source.tex
else
sed -i -e "s/#b/$a/g" source.tex
fi
When running this, the condition is always false. I tried [] or (()) for the if statement but I just can't find a way to fix it.
You only need a single parameter expansion here, to replace the expansion of $a with \hspace{2cm} if the expansion is empty.
sed -i -e "s/#b/${a:-\\\\hspace{2cm}}/g" source.tex
You need a stack of \ because there are two rounds of escaping involved. First, the shell itself reduces each \\ to a single literal backslash. Then sed also reduces each pair of \\ to a single literal backslash.
Counting the number of occurrences of someething seems like a very roundabout way to approach this anyway.
case $a in
*[A-Za-z0-9_]*) x='\\hspace{2cm}';;
*) x=$a;;
esac
sed -i "s/#b/$x/g" source.tex

bash - IFS changes behavior of echo -n in for loop

I have code that requires a response within a for loop.
Prior to the loop I set IFS="\n"
Within the loop echo -n is ignored (except for the last line).
Note: this is just an example of the behavior of echo -n
Example:
IFS='\n'
for line in `cat file`
do
echo -n $line
done
This outputs:
this is a test
this is a test
this is a test$
with the user prompt occuring only at the end of the last line.
Why is this occuring and is there a fix?
Neither IFS="\n" nor IFS='\n' set $IFS to a newline; instead they set it to literal \ followed by literal n.
You'd have to use an ANSI C-quoted string in order to assign an actual newline: IFS=$'\n'; alternatively, you could use a normal string literal that contains an actual newline (spans 2 lines).
Assigning literal \n had the effect that the output from cat file was not split into lines, because an actual newline was not present in $IFS; potentially - though not with your sample file content - the output could have been split into fields by embedded \ and n characters.
Without either, the entire file contents were passed at once, resulting in a single iteration of your for loop.
That said, your approach to looping over lines from a file is ill-advised; try something like the following instead:
while IFS= read -r line; do
echo -n "$line"
done < file
Never use for loops when parsing files in bash. Use while loops instead. Here is a really good tutorial on that.
http://mywiki.wooledge.org/BashFAQ/001

bash grep variable as pattern

I don't usually work in bash but grep could be a really fast solution in this case. I have read a lot of questions on grep and variable assignment in bash yet I do not see the error. I have tried several flavours of double quotes around $pattern, used `...`` or $(...) but nothing worked.
So here's what I try to do:
I have two files. The first contains several names. Each of them I want to use as a pattern for grep in order to search them in another file. Therefore I loop through the lines of the first file and assign the name to the variable pattern.
This step works as the variable is printed out properly.
But somehow grep does not recognize/interpret the variable. When I substitute "$pattern" with an actual name everything is fine as well. Therefore I don't think the variable assignment has a problem but the interpretation of "$pattern" as the string it should represent.
Any help is greatly appreciated!
#!/bin/bash
while IFS='' read -r line || [[ -n $line ]]; do
a=( $line )
pattern="${a[2]}"
echo "Text read from file: $pattern"
var=$(grep "$pattern" 9606.protein.aliases.v10.txt)
echo "Matched Line in Alias is: $var"
done < "$1"
> bash match_Uniprot_StringDB.sh ~/Chromatin_Computation/.../KDM.protein.tb
output:
Text read from file: "UBE2B"
Matched Line in Alias is:
Text read from file: "UTY"
Matched Line in Alias is:
EDIT
The solution drvtiny suggested works. It is necessary to get rid of the double quotes to match the string. Adding the following lines makes the script work.
pattern="${pattern#\"}"
pattern="${pattern%\"}"
Please, look at "-f FILE" option in man grep.
I advise that this option do exactly what you need without any bash loops or such other "hacks" :)
And yes, according to the output of your code, you read pattern including double quotes literally. In other words, you read from file ~/Chromatin_Computation/.../KDM.protein.tb this string:
"UBE2B"
But not
UBE2B
as you probably expect.
Maybe you need to remove double quotes on the boundaries of your $pattern?
Try to do this after reading pattern:
pattern=${pattern#\"}
pattern=${pattern%\"}

Resources